null hypothesis population mean

The mean number of cavities per person does not differ between the flossing group (µ ) and the non-flossing group (µ ) in the population; µ = µ .

There is no relationship between the amount of text highlighted and exam scores in the population; β = 0.

The proportion of people with depression in the daily-meditation group ( ) is greater than or equal to the no-meditation group ( ) in the population; ≥ .

The mean number of cavities per person differs between the flossing group (µ ) and the non-flossing group (µ ) in the population; µ ≠ µ .

There is a relationship between the amount of text highlighted and exam scores in the population; β ≠ 0.

The proportion of people with depression in the daily-meditation group ( ) is less than the no-meditation group ( ) in the population; < .

with two groups

Null Hypothesis

H : μ= 191         (no change)

Research Hypothesis

H : μ> 191         (investigator's belief)

Upper-tailed, Lower-tailed, Two-tailed Tests

The research or alternative hypothesis can take one of three forms. An investigator might believe that the parameter has increased, decreased or changed. For example, an investigator might hypothesize:  

The exact form of the research hypothesis depends on the investigator's belief about the parameter of interest and whether it has possibly increased, decreased or is different from the null value. The research hypothesis is set up by the investigator before any data are collected.

Rejection Region for Upper-Tailed Z Test (H : μ > μ ) with α=0.05

The decision rule is: Reject H if Z 1.645.

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

• Step 4. Compute the test statistic.  

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

• Step 5. Conclusion.  

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).  

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .  

• Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191                 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

• Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

• Step 3. Set up decision rule.  

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05.   Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.  

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.                  

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

 The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

• if n > 30
• if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. 

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races.  The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured.   The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach. 

• Step 1.  Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302           α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance.   Reject H 0 if Z < -1.645.

•   Step 4. Compute the test statistic.  

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.  

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.      

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203                       α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

•   Step 3. Set up decision rule.  

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002.   Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.  

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?  

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol.   Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows:   n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. 

H 0 : μ= 203 H 1 : μ< 203                   α=0.05

•  Step 2. Select the appropriate test statistic.  

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows:   Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.    

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:  

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%.  Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211                     α=0.05

We must first check that the sample size is adequate.   Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.  

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.  

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

• if n 1 > 30 and n 2 > 30
• if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.    

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.  

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.  

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2                       α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample.   Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.  

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.  

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.  

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).  

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2                         α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).  

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.  

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0                 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0. 

• Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.  

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.      

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .  

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

• For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
• If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
• If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.  

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.  

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2                 α=0.05

• Step 2.  Select the appropriate test statistic.  

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.  

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.    

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.  

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2              α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. 

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).  

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

• Continuous Outcome, One Sample: H0: μ = μ0
• Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
• Continuous Outcome, Two Matched Samples: H0: μd = 0
• Dichotomous Outcome, One Sample: H0: p = p 0
• Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.    

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

• Step 1: Set up hypotheses and determine the level of significance.

α=0.05

• Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

• Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

• Step 4: Compute the test statistic
• Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

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• Null and Alternative Hypotheses | Definitions & Examples

Null and Alternative Hypotheses | Definitions & Examples

Published on 5 October 2022 by Shaun Turney . Revised on 6 December 2022.

The null and alternative hypotheses are two competing claims that researchers weigh evidence for and against using a statistical test :

• Null hypothesis (H 0 ): There’s no effect in the population .
• Alternative hypothesis (H A ): There’s an effect in the population.

The effect is usually the effect of the independent variable on the dependent variable .

Table of contents

Answering your research question with hypotheses, what is a null hypothesis, what is an alternative hypothesis, differences between null and alternative hypotheses, how to write null and alternative hypotheses, frequently asked questions about null and alternative hypotheses.

The null and alternative hypotheses offer competing answers to your research question . When the research question asks “Does the independent variable affect the dependent variable?”, the null hypothesis (H 0 ) answers “No, there’s no effect in the population.” On the other hand, the alternative hypothesis (H A ) answers “Yes, there is an effect in the population.”

The null and alternative are always claims about the population. That’s because the goal of hypothesis testing is to make inferences about a population based on a sample . Often, we infer whether there’s an effect in the population by looking at differences between groups or relationships between variables in the sample.

You can use a statistical test to decide whether the evidence favors the null or alternative hypothesis. Each type of statistical test comes with a specific way of phrasing the null and alternative hypothesis. However, the hypotheses can also be phrased in a general way that applies to any test.

The null hypothesis is the claim that there’s no effect in the population.

If the sample provides enough evidence against the claim that there’s no effect in the population ( p ≤ α), then we can reject the null hypothesis . Otherwise, we fail to reject the null hypothesis.

Although “fail to reject” may sound awkward, it’s the only wording that statisticians accept. Be careful not to say you “prove” or “accept” the null hypothesis.

Null hypotheses often include phrases such as “no effect”, “no difference”, or “no relationship”. When written in mathematical terms, they always include an equality (usually =, but sometimes ≥ or ≤).

Examples of null hypotheses

The table below gives examples of research questions and null hypotheses. There’s always more than one way to answer a research question, but these null hypotheses can help you get started.

*Note that some researchers prefer to always write the null hypothesis in terms of “no effect” and “=”. It would be fine to say that daily meditation has no effect on the incidence of depression and p 1 = p 2 .

The alternative hypothesis (H A ) is the other answer to your research question . It claims that there’s an effect in the population.

Often, your alternative hypothesis is the same as your research hypothesis. In other words, it’s the claim that you expect or hope will be true.

The alternative hypothesis is the complement to the null hypothesis. Null and alternative hypotheses are exhaustive, meaning that together they cover every possible outcome. They are also mutually exclusive, meaning that only one can be true at a time.

Alternative hypotheses often include phrases such as “an effect”, “a difference”, or “a relationship”. When alternative hypotheses are written in mathematical terms, they always include an inequality (usually ≠, but sometimes > or <). As with null hypotheses, there are many acceptable ways to phrase an alternative hypothesis.

Examples of alternative hypotheses

The table below gives examples of research questions and alternative hypotheses to help you get started with formulating your own.

Null and alternative hypotheses are similar in some ways:

• They’re both answers to the research question
• They both make claims about the population
• They’re both evaluated by statistical tests.

However, there are important differences between the two types of hypotheses, summarized in the following table.

To help you write your hypotheses, you can use the template sentences below. If you know which statistical test you’re going to use, you can use the test-specific template sentences. Otherwise, you can use the general template sentences.

The only thing you need to know to use these general template sentences are your dependent and independent variables. To write your research question, null hypothesis, and alternative hypothesis, fill in the following sentences with your variables:

Does independent variable affect dependent variable ?

• Null hypothesis (H 0 ): Independent variable does not affect dependent variable .
• Alternative hypothesis (H A ): Independent variable affects dependent variable .

Test-specific

Once you know the statistical test you’ll be using, you can write your hypotheses in a more precise and mathematical way specific to the test you chose. The table below provides template sentences for common statistical tests.

Note: The template sentences above assume that you’re performing one-tailed tests . One-tailed tests are appropriate for most studies.

The null hypothesis is often abbreviated as H 0 . When the null hypothesis is written using mathematical symbols, it always includes an equality symbol (usually =, but sometimes ≥ or ≤).

The alternative hypothesis is often abbreviated as H a or H 1 . When the alternative hypothesis is written using mathematical symbols, it always includes an inequality symbol (usually ≠, but sometimes < or >).

A research hypothesis is your proposed answer to your research question. The research hypothesis usually includes an explanation (‘ x affects y because …’).

A statistical hypothesis, on the other hand, is a mathematical statement about a population parameter. Statistical hypotheses always come in pairs: the null and alternative hypotheses. In a well-designed study , the statistical hypotheses correspond logically to the research hypothesis.

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How to Write a Null Hypothesis (5 Examples)

A hypothesis test uses sample data to determine whether or not some claim about a population parameter is true.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms:

H 0 (Null Hypothesis): Population parameter =,  ≤, ≥ some value

H A  (Alternative Hypothesis): Population parameter <, >, ≠ some value

Note that the null hypothesis always contains the equal sign .

We interpret the hypotheses as follows:

Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.

Alternative hypothesis: The sample data  does provide sufficient evidence to support the claim being made by an individual.

For example, suppose it’s assumed that the average height of a certain species of plant is 20 inches tall. However, one botanist claims the true average height is greater than 20 inches.

To test this claim, she may go out and collect a random sample of plants. She can then use this sample data to perform a hypothesis test using the following two hypotheses:

H 0 : μ ≤ 20 (the true mean height of plants is equal to or even less than 20 inches)

H A : μ > 20 (the true mean height of plants is greater than 20 inches)

If the sample data gathered by the botanist shows that the mean height of this species of plants is significantly greater than 20 inches, she can reject the null hypothesis and conclude that the mean height is greater than 20 inches.

Read through the following examples to gain a better understanding of how to write a null hypothesis in different situations.

Example 1: Weight of Turtles

A biologist wants to test whether or not the true mean weight of a certain species of turtles is 300 pounds. To test this, he goes out and measures the weight of a random sample of 40 turtles.

Here is how to write the null and alternative hypotheses for this scenario:

H 0 : μ = 300 (the true mean weight is equal to 300 pounds)

H A : μ ≠ 300 (the true mean weight is not equal to 300 pounds)

Example 2: Height of Males

It’s assumed that the mean height of males in a certain city is 68 inches. However, an independent researcher believes the true mean height is greater than 68 inches. To test this, he goes out and collects the height of 50 males in the city.

H 0 : μ ≤ 68 (the true mean height is equal to or even less than 68 inches)

H A : μ > 68 (the true mean height is greater than 68 inches)

Example 3: Graduation Rates

A university states that 80% of all students graduate on time. However, an independent researcher believes that less than 80% of all students graduate on time. To test this, she collects data on the proportion of students who graduated on time last year at the university.

H 0 : p ≥ 0.80 (the true proportion of students who graduate on time is 80% or higher)

H A : μ < 0.80 (the true proportion of students who graduate on time is less than 80%)

Example 4: Burger Weights

A food researcher wants to test whether or not the true mean weight of a burger at a certain restaurant is 7 ounces. To test this, he goes out and measures the weight of a random sample of 20 burgers from this restaurant.

H 0 : μ = 7 (the true mean weight is equal to 7 ounces)

H A : μ ≠ 7 (the true mean weight is not equal to 7 ounces)

Example 5: Citizen Support

A politician claims that less than 30% of citizens in a certain town support a certain law. To test this, he goes out and surveys 200 citizens on whether or not they support the law.

H 0 : p ≥ .30 (the true proportion of citizens who support the law is greater than or equal to 30%)

H A : μ < 0.30 (the true proportion of citizens who support the law is less than 30%)

Additional Resources

Introduction to Hypothesis Testing Introduction to Confidence Intervals An Explanation of P-Values and Statistical Significance

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Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “How to Write a Null Hypothesis (5 Examples)”

you are amazing, thank you so much

Say I am a botanist hypothesizing the average height of daisies is 20 inches, or not? Does T = (ave – 20 inches) / √ variance / (80 / 4)? … This assumes 40 real measures + 40 fake = 80 n, but that seems questionable. Please advise.

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8.6 Hypothesis Tests for a Population Mean with Known Population Standard Deviation

Learning objectives.

• Conduct and interpret hypothesis tests for a population mean with known population standard deviation.

Some notes about conducting a hypothesis test:

• The null hypothesis [latex]H_0[/latex] is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
• The alternative hypothesis [latex]H_a[/latex] is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
• If the alternative hypothesis is a “less than”, then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
• If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
• If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
• Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of 0.05.  Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
• The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.
• An alternative approach for hypothesis testing is to use what is called the critical value approach .  In this book, we will only use the p -value approach.  Some of the videos below may mention the critical value approach, but this approach will not be used in this book.

Suppose the hypotheses for a hypothesis test are:

[latex]\begin{eqnarray*} H_0: & & \mu=5 \\ H_a: & & \mu \lt 5 \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\lt[/latex], this is a left-tailed test.  The p -value is the area in the left-tail of the distribution.

[latex]\begin{eqnarray*} H_0: & & \mu=0.5 \\ H_a: & & \mu \neq 0.5  \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\neq[/latex], this is a two-tailed test.  The p -value is the sum of the areas in the two tails of the distribution.  Each tail contains exactly half of the p -value.

[latex]\begin{eqnarray*} H_0: & & \mu=10 \\ H_a: & & \mu \lt 10  \end{eqnarray*}[/latex]

Steps to Conduct a Hypothesis Test for a Population Mean with Known Population Standard Deviation

• Write down the null and alternative hypotheses in terms of the population mean [latex]\mu[/latex].  Include appropriate units with the values of the mean.
• Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
• Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].
• When the population standard deviation is known , we use a normal distribution with [latex]\displaystyle{z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}}[/latex] to find the p -value.  The p -value is the area in the corresponding tail of the normal distribution.
• The results of the sample data are significant.  There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
• The results of the sample data are not significant.  There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
• Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION MEAN WITH KNOWN POPULATION STANDARD DEVIATION

The p -value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean.  When the population standard deviation is known, use the normal distribution to find the p -value.

The p -value is the area in the tail(s) of a normal distribution, so the norm.dist(x,[latex]\mu[/latex],[latex]\sigma[/latex],logic operator) function can be used to calculate the p -value.

• For x , enter the value for [latex]\overline{x}[/latex].
• For [latex]\mu[/latex] , enter the mean of the sample means [latex]\mu[/latex].  Note:  Because the test is run assuming the null hypothesis is true, the value for [latex]\mu[/latex] is the claim from the null hypothesis.
• For [latex]\sigma[/latex] , enter the standard error of the mean [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].
• For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.

Use the appropriate technique with the norm.dist function to find the area in the left-tail or the area in the right-tail.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds with a standard deviation of 0.8 seconds for swimming the 25-meter freestyle.  His dad, Frank, thought that Jeffrey could swim the 25-meter freestyle faster using goggles.  Frank bought Jeffrey a new pair of goggles and timed Jeffrey swimming the 25-meter freestyle 15 different times.  In the sample of 15 swims, Jeffrey’s mean time was 16 seconds.  Frank thought that the goggles helped Jeffrey swim faster than 16.43 seconds.  At the 5% significance level, did Jeffrey swim faster wearing the goggles?  Assume that the swim times for the 25-meter freestyle are normally distributed.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=16.43 \mbox{ seconds} \\ H_a: & & \mu \lt 16.43 \mbox{ seconds} \end{eqnarray*}[/latex]

From the question, we have [latex]n=15[/latex], [latex]\overline{x}=16[/latex], [latex]\sigma=0.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=0.8[/latex]).  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left-tail of the distribution.

So the p -value[latex]=0.0187[/latex].

Conclusion:

Because p -value[latex]=0.0187 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that Jeffrey’s mean swim time with the goggles is less than 16.43 seconds.

• The null hypothesis [latex]\mu=16.43[/latex] is the claim that Jeffrey’s mean swim time with the goggles is 16.43 seconds (the same as it is without the googles).
• The alternative hypothesis [latex]\mu \lt 16.43[/latex] is the claim that Jeffrey’s swim time with the goggles is less than 16.43 seconds.
• The function is norm.dist because we are finding the area in the left tail of a normal distribution.
• Field 1 is the value of [latex]\overline{x}[/latex]
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume [latex]\mu=16.43[/latex].
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].  Note that we are not using the standard deviation from the population ([latex]\sigma=0.8[/latex]).  This is because the p -value is the area under the curve of the distribution of the sample means, not the distribution of the population.
• The p -value of 0.0187 tells us that under the assumption that Jeffrey’s mean swim time with goggles is 16.43 seconds (the null hypothesis), there is only a 1.87% chance that the mean time for the 15 sample swims is 16 seconds or less.  This is a small probability, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.
• The Type I error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis).  That is, reject the null hypothesis when the null hypothesis is actually true.
• The Type II error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis).  That is, do not reject the null hypothesis when the null hypothesis is actually false.

The mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards with a standard deviation of 2 yards.  The team coach tells Marco to adjust his grip to get more distance.  The coach records the distances for 20 throws with the new grip.  For the 20 throws, Marco’s mean distance was 41.5 yards.  The coach thought the different grip helped Marco throw farther than 40 yards.  At the 5% significance level, is Marco’s mean throwing distance higher with the new grip?  Assume the throw distances for footballs are normally distributed.

[latex]\begin{eqnarray*} H_0: & & \mu=40 \mbox{ yards} \\ H_a: & & \mu \gt 40 \mbox{ yards} \end{eqnarray*}[/latex]

From the question, we have [latex]n=20[/latex], [latex]\overline{x}=41.5[/latex], [latex]\sigma=2[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=2[/latex]).  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right-tail of the distribution.

So the p -value[latex]=0.0004[/latex].

Because p -value[latex]=0.0004 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that Marco’s mean throwing distance is greater than 40 yards with the new grip.

• The null hypothesis [latex]\mu=40[/latex] is the claim that Marco’s mean throwing distance with the new grip is 40 yards (the same as it is without the new grip).
• The alternative hypothesis [latex]\mu \gt 40[/latex] is the claim that Marco’s mean throwing distance with the new grip is greater than 40 yards.
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].
• The p -value of 0.0004 tells us that under the assumption that Marco’s mean throwing distance with the new grip is 40 yards, there is only a 0.047% chance that the mean throwing distance for the 20 sample throws is more than 40 yards.  This is a small probability, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.

A local college states in its marketing materials that the average age of its first-year students is 18.3 years with a standard deviation of 3.4 years.  But this information is based on old data and does not take into account that more older adults are returning to college.  A researcher at the college believes that the average age of its first-year students has changed.  The researcher takes a sample of 50 first-year students and finds the average age is 19.5 years.  At the 1% significance level, has the average age of the college’s first-year students changed?

[latex]\begin{eqnarray*} H_0: & & \mu=18.3 \mbox{ years} \\ H_a: & & \mu \neq 18.3 \mbox{ years} \end{eqnarray*}[/latex]

From the question, we have [latex]n=50[/latex], [latex]\overline{x}=19.5[/latex], [latex]\sigma=3.4[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=3.4[/latex]).  In this case, the sample size is greater than 30.  So we use a normal distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of area in the tails of the distribution.

Because there is only one sample, we only have information relating to one of the two tails, either the left tail or the right tail.  We need to know if the sample relates to the left tail or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, the sample mean [latex]\overline{x}=19.5[/latex] is greater than the value of the population mean in the null hypothesis [latex]\mu=18.3[/latex] ([latex]\overline{x}=19.5>18.3=\mu[/latex]), so the sample information relates to the right-tail of the normal distribution.  This means that we will calculate out the area in the right tail using 1-norm.dist .  However, this is a two-tailed test where the p -value is the sum of the area in the two tails and the area in the right-tail is only one half of the p -value.  The area in the left tail equals the area in the right tail and the p -value is the sum of these two areas.

So the area in the right tail is 0.0063 and [latex]\frac{1}{2}[/latex]( p -value)[latex]=0.0063[/latex].  This is also the area in the left tail, so

p -value[latex]=0.0063+0.0063=0.0126[/latex]

Because p -value[latex]=0.0126 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that the average age of the college’s first-year students has changed.

• The null hypothesis [latex]\mu=18.3[/latex] is the claim that the average age of the first-year students is still 18.3 years.
• The alternative hypothesis [latex]\mu \neq 18.3[/latex] is the claim that the average age of the first-year students has changed from 18.3 years.
• We use norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the left tail.  The area in the right tail equals the area in the left tail, so we can find the p -value by adding the output from this function to itself.
• We use 1-norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the right tail.  The area in the left tail equals the area in the right tail, so we can find the p -value by adding the output from this function to itself.
• The p -value of 0.0126  is a large probability compared to the 1% significance level, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the claim that the average age of first-year students is 18.3 years is most likely correct.

Watch this video: Hypothesis Testing: z -test, right tail by ExcelIsFun [33:47]

Watch this video: Hypothesis Testing: z -test, left tail by ExcelIsFun [10:57]

Watch this video: Hypothesis Testing: z -test, two tail by ExcelIsFun [9:56]

Concept Review

The hypothesis test for a population mean is a well established process:

• Collect the sample information for the test and identify the significance level.
• When the population standard deviation is known, find the p -value (the area in the corresponding tail) for the test using the normal distribution.
• Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 9.6   Hypothesis Testing of a Single Mean and Single Proportion “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License.

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

What is The Null Hypothesis & When Do You Reject The Null Hypothesis

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A null hypothesis is a statistical concept suggesting no significant difference or relationship between measured variables. It’s the default assumption unless empirical evidence proves otherwise.

The null hypothesis states no relationship exists between the two variables being studied (i.e., one variable does not affect the other).

The null hypothesis is the statement that a researcher or an investigator wants to disprove.

Testing the null hypothesis can tell you whether your results are due to the effects of manipulating ​ the dependent variable or due to random chance. 

How to Write a Null Hypothesis

Null hypotheses (H0) start as research questions that the investigator rephrases as statements indicating no effect or relationship between the independent and dependent variables.

It is a default position that your research aims to challenge or confirm.

For example, if studying the impact of exercise on weight loss, your null hypothesis might be:

There is no significant difference in weight loss between individuals who exercise daily and those who do not.

Examples of Null Hypotheses

When Do We Reject The Null Hypothesis? 

We reject the null hypothesis when the data provide strong enough evidence to conclude that it is likely incorrect. This often occurs when the p-value (probability of observing the data given the null hypothesis is true) is below a predetermined significance level.

If the collected data does not meet the expectation of the null hypothesis, a researcher can conclude that the data lacks sufficient evidence to back up the null hypothesis, and thus the null hypothesis is rejected. 

Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant ( p > 0.05).

If the data collected from the random sample is not statistically significance , then the null hypothesis will be accepted, and the researchers can conclude that there is no relationship between the variables. 

You need to perform a statistical test on your data in order to evaluate how consistent it is with the null hypothesis. A p-value is one statistical measurement used to validate a hypothesis against observed data.

Calculating the p-value is a critical part of null-hypothesis significance testing because it quantifies how strongly the sample data contradicts the null hypothesis.

The level of statistical significance is often expressed as a  p  -value between 0 and 1. The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

Usually, a researcher uses a confidence level of 95% or 99% (p-value of 0.05 or 0.01) as general guidelines to decide if you should reject or keep the null.

When your p-value is less than or equal to your significance level, you reject the null hypothesis.

In other words, smaller p-values are taken as stronger evidence against the null hypothesis. Conversely, when the p-value is greater than your significance level, you fail to reject the null hypothesis.

In this case, the sample data provides insufficient data to conclude that the effect exists in the population.

Because you can never know with complete certainty whether there is an effect in the population, your inferences about a population will sometimes be incorrect.

When you incorrectly reject the null hypothesis, it’s called a type I error. When you incorrectly fail to reject it, it’s called a type II error.

Why Do We Never Accept The Null Hypothesis?

The reason we do not say “accept the null” is because we are always assuming the null hypothesis is true and then conducting a study to see if there is evidence against it. And, even if we don’t find evidence against it, a null hypothesis is not accepted.

A lack of evidence only means that you haven’t proven that something exists. It does not prove that something doesn’t exist. 

It is risky to conclude that the null hypothesis is true merely because we did not find evidence to reject it. It is always possible that researchers elsewhere have disproved the null hypothesis, so we cannot accept it as true, but instead, we state that we failed to reject the null. 

One can either reject the null hypothesis, or fail to reject it, but can never accept it.

Why Do We Use The Null Hypothesis?

We can never prove with 100% certainty that a hypothesis is true; We can only collect evidence that supports a theory. However, testing a hypothesis can set the stage for rejecting or accepting this hypothesis within a certain confidence level.

The null hypothesis is useful because it can tell us whether the results of our study are due to random chance or the manipulation of a variable (with a certain level of confidence).

A null hypothesis is rejected if the measured data is significantly unlikely to have occurred and a null hypothesis is accepted if the observed outcome is consistent with the position held by the null hypothesis.

Rejecting the null hypothesis sets the stage for further experimentation to see if a relationship between two variables exists. 

Hypothesis testing is a critical part of the scientific method as it helps decide whether the results of a research study support a particular theory about a given population. Hypothesis testing is a systematic way of backing up researchers’ predictions with statistical analysis.

It helps provide sufficient statistical evidence that either favors or rejects a certain hypothesis about the population parameter. 

Purpose of a Null Hypothesis 

• The primary purpose of the null hypothesis is to disprove an assumption. 
• Whether rejected or accepted, the null hypothesis can help further progress a theory in many scientific cases.
• A null hypothesis can be used to ascertain how consistent the outcomes of multiple studies are.

Do you always need both a Null Hypothesis and an Alternative Hypothesis?

The null (H0) and alternative (Ha or H1) hypotheses are two competing claims that describe the effect of the independent variable on the dependent variable. They are mutually exclusive, which means that only one of the two hypotheses can be true. 

While the null hypothesis states that there is no effect in the population, an alternative hypothesis states that there is statistical significance between two variables. 

The goal of hypothesis testing is to make inferences about a population based on a sample. In order to undertake hypothesis testing, you must express your research hypothesis as a null and alternative hypothesis. Both hypotheses are required to cover every possible outcome of the study. 

What is the difference between a null hypothesis and an alternative hypothesis?

The alternative hypothesis is the complement to the null hypothesis. The null hypothesis states that there is no effect or no relationship between variables, while the alternative hypothesis claims that there is an effect or relationship in the population.

It is the claim that you expect or hope will be true. The null hypothesis and the alternative hypothesis are always mutually exclusive, meaning that only one can be true at a time.

What are some problems with the null hypothesis?

One major problem with the null hypothesis is that researchers typically will assume that accepting the null is a failure of the experiment. However, accepting or rejecting any hypothesis is a positive result. Even if the null is not refuted, the researchers will still learn something new.

Why can a null hypothesis not be accepted?

We can either reject or fail to reject a null hypothesis, but never accept it. If your test fails to detect an effect, this is not proof that the effect doesn’t exist. It just means that your sample did not have enough evidence to conclude that it exists.

We can’t accept a null hypothesis because a lack of evidence does not prove something that does not exist. Instead, we fail to reject it.

Failing to reject the null indicates that the sample did not provide sufficient enough evidence to conclude that an effect exists.

If the p-value is greater than the significance level, then you fail to reject the null hypothesis.

Is a null hypothesis directional or non-directional?

A hypothesis test can either contain an alternative directional hypothesis or a non-directional alternative hypothesis. A directional hypothesis is one that contains the less than (“<“) or greater than (“>”) sign.

A nondirectional hypothesis contains the not equal sign (“≠”).  However, a null hypothesis is neither directional nor non-directional.

A null hypothesis is a prediction that there will be no change, relationship, or difference between two variables.

The directional hypothesis or nondirectional hypothesis would then be considered alternative hypotheses to the null hypothesis.

Gill, J. (1999). The insignificance of null hypothesis significance testing.  Political research quarterly ,  52 (3), 647-674.

Krueger, J. (2001). Null hypothesis significance testing: On the survival of a flawed method.  American Psychologist ,  56 (1), 16.

Masson, M. E. (2011). A tutorial on a practical Bayesian alternative to null-hypothesis significance testing.  Behavior research methods ,  43 , 679-690.

Nickerson, R. S. (2000). Null hypothesis significance testing: a review of an old and continuing controversy.  Psychological methods ,  5 (2), 241.

Rozeboom, W. W. (1960). The fallacy of the null-hypothesis significance test.  Psychological bulletin ,  57 (5), 416.

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AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing
• Examples of null and alternative hypotheses
• Writing null and alternative hypotheses

P-values and significance tests

• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

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Video transcript

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What Is a Null Hypothesis?

The alternative hypothesis.

• Additional Examples
• Null Hypothesis and Investments

The Bottom Line

• Corporate Finance
• Financial Ratios

Null Hypothesis: What Is It, and How Is It Used in Investing?

Adam Hayes, Ph.D., CFA, is a financial writer with 15+ years Wall Street experience as a derivatives trader. Besides his extensive derivative trading expertise, Adam is an expert in economics and behavioral finance. Adam received his master's in economics from The New School for Social Research and his Ph.D. from the University of Wisconsin-Madison in sociology. He is a CFA charterholder as well as holding FINRA Series 7, 55 & 63 licenses. He currently researches and teaches economic sociology and the social studies of finance at the Hebrew University in Jerusalem.

A null hypothesis is a type of statistical hypothesis that proposes that no statistical significance exists in a set of given observations. Hypothesis testing is used to assess the credibility of a hypothesis by using sample data. Sometimes referred to simply as the “null,” it is represented as H 0 .

The null hypothesis, also known as the conjecture, is used in quantitative analysis to test theories about markets, investing strategies, or economies to decide if an idea is true or false.

Key Takeaways

• A null hypothesis is a type of conjecture in statistics that proposes that there is no difference between certain characteristics of a population or data-generating process.
• The alternative hypothesis proposes that there is a difference.
• Hypothesis testing provides a method to reject a null hypothesis within a certain confidence level.
• If you can reject the null hypothesis, it provides support for the alternative hypothesis.
• Null hypothesis testing is the basis of the principle of falsification in science.

Alex Dos Diaz / Investopedia

Null Hypothesis Example

For example, a gambler may be interested in whether a game of chance is fair. If it is fair, then the expected earnings per play come to zero for both players. If the game is not fair, then the expected earnings are positive for one player and negative for the other.

To test whether the game is fair, the gambler collects earnings data from many repetitions of the game, calculates the average earnings from these data, then tests the null hypothesis that the expected earnings are not different from zero.

If the average earnings from the sample data are sufficiently far from zero, then the gambler will reject the null hypothesis and conclude the alternative hypothesis—namely, that the expected earnings per play are different from zero. If the average earnings from the sample data are near zero, then the gambler will not reject the null hypothesis, concluding instead that the difference between the average from the data and zero is explainable by chance alone.

The null hypothesis assumes that any kind of difference between the chosen characteristics that you see in a set of data is due to chance. For example, if the expected earnings for the gambling game are truly equal to zero, then any difference between the average earnings in the data and zero is due to chance.

Analysts look to reject   the null hypothesis because doing so is a strong conclusion. This requires strong evidence in the form of an observed difference that is too large to be explained solely by chance. Failing to reject the null hypothesis—that the results are explainable by chance alone—is a weak conclusion because it allows that factors other than chance may be at work, but may not be strong enough for the statistical test to detect them.

A null hypothesis can only be rejected, not proven.

An important point to note is that we are testing the null hypothesis because there is an element of doubt about its validity. Whatever information that is against the stated null hypothesis is captured in the alternative (alternate) hypothesis (H1).

For the examples below, the alternative hypothesis would be:

• Students score an average that is  not  equal to seven.
• The mean annual return of a mutual fund is  not  equal to 8% per year.

In other words, the alternative hypothesis is a direct contradiction of the null hypothesis.

More Null Hypothesis Examples

Here is a simple example: A school principal claims that students in her school score an average of seven out of 10 in exams. The null hypothesis is that the population mean is 7.0. To test this null hypothesis, we record marks of, say, 30 students ( sample ) from the entire student population of the school (say, 300) and calculate the mean of that sample.

We can then compare the (calculated) sample mean to the (hypothesized) population mean of 7.0 and attempt to reject the null hypothesis. (The null hypothesis here—that the population mean is 7.0—cannot be proved using the sample data. It can only be rejected.)

Take another example: The annual return of a particular  mutual fund  is claimed to be 8%. Assume that a mutual fund has been in existence for 20 years. The null hypothesis is that the mean return is 8% for the mutual fund. We take a random sample of annual returns of the mutual fund for, say, five years (sample) and calculate the sample mean. We then compare the (calculated) sample mean to the (claimed) population mean (8%) to test the null hypothesis.

For the above examples, null hypotheses are:

• Example A : Students in the school score an average of seven out of 10 in exams.
• Example B : The mean annual return of the mutual fund is 8% per year.

For the purposes of determining whether to reject the null hypothesis, the null hypothesis (abbreviated H 0 ) is assumed, for the sake of argument, to be true. Then the likely range of possible values of the calculated statistic (e.g., the average score on 30 students’ tests) is determined under this presumption (e.g., the range of plausible averages might range from 6.2 to 7.8 if the population mean is 7.0). Then, if the sample average is outside of this range, the null hypothesis is rejected. Otherwise, the difference is said to be “explainable by chance alone,” being within the range that is determined by chance alone.

How Null Hypothesis Testing Is Used in Investments

As an example related to financial markets, assume Alice sees that her investment strategy produces higher average returns than simply buying and holding a stock . The null hypothesis states that there is no difference between the two average returns, and Alice is inclined to believe this until she can conclude contradictory results.

Refuting the null hypothesis would require showing statistical significance, which can be found by a variety of tests. The alternative hypothesis would state that the investment strategy has a higher average return than a traditional buy-and-hold strategy.

One tool that can determine the statistical significance of the results is the p-value. A p-value represents the probability that a difference as large or larger than the observed difference between the two average returns could occur solely by chance.

A p-value that is less than or equal to 0.05 often indicates whether there is evidence against the null hypothesis. If Alice conducts one of these tests, such as a test using the normal model, resulting in a significant difference between her returns and the buy-and-hold returns (the p-value is less than or equal to 0.05), she can then reject the null hypothesis and conclude the alternative hypothesis.

How Is the Null Hypothesis Identified?

The analyst or researcher establishes a null hypothesis based on the research question or problem that they are trying to answer. Depending on the question, the null may be identified differently. For example, if the question is simply whether an effect exists (e.g., does X influence Y?), the null hypothesis could be H 0 : X = 0. If the question is instead, is X the same as Y, the H0 would be X = Y. If it is that the effect of X on Y is positive, H0 would be X > 0. If the resulting analysis shows an effect that is statistically significantly different from zero, the null can be rejected.

How Is Null Hypothesis Used in Finance?

In finance , a null hypothesis is used in quantitative analysis. A null hypothesis tests the premise of an investing strategy, the markets, or an economy to determine if it is true or false.

For instance, an analyst may want to see if two stocks, ABC and XYZ, are closely correlated. The null hypothesis would be ABC ≠ XYZ.

How Are Statistical Hypotheses Tested?

Statistical hypotheses are tested by a four-step process . The first step is for the analyst to state the two hypotheses so that only one can be right. The next step is to formulate an analysis plan, which outlines how the data will be evaluated. The third step is to carry out the plan and physically analyze the sample data. The fourth and final step is to analyze the results and either reject the null hypothesis or claim that the observed differences are explainable by chance alone.

What Is an Alternative Hypothesis?

An alternative hypothesis is a direct contradiction of a null hypothesis. This means that if one of the two hypotheses is true, the other is false.

A null hypothesis is a type of statistical hypothesis. It proposes that no statistical significance exists in a set of given observations.

Also known as the conjecture, the null hypothesis is used in quantitative analysis to test theories about economies, investing strategies, or markets to decide if an idea is true or false. Hypothesis testing assesses the credibility of a hypothesis by using sample data. It is represented as H0 and is sometimes simply known as the “null.”

Sage Publishing. “ Chapter 8: Introduction to Hypothesis Testing ,” Page 4.

Sage Publishing. “ Chapter 8: Introduction to Hypothesis Testing ,” Pages 4–7.

Sage Publishing. “ Chapter 8: Introduction to Hypothesis Testing ,” Page 7.

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Function that calculates mean, variance, and skewness simultaneously in a dataframe in R

In statistical analysis, understanding the central tendency (mean), dispersion (variance), and asymmetry (skewness) of data is essential for gaining insights into its distribution and characteristics. This article explores how to compute these three statistical measures simultaneously across multiple variables in a data frame using R Programming Language .

Understanding Mean, Variance, and Skewness

• Mean : Represents the average value of a set of numbers. It measures the central tendency of the data.
• Variance : Indicates the spread or dispersion of data points around the mean. A higher variance implies a greater spread.
• Skewness : Measures the asymmetry of the probability distribution of a real-valued random variable about its mean. Positive skewness indicates a longer right tail, while negative skewness indicates a longer left tail.

Approach to Calculate Mean, Variance, and Skewness

To calculate mean, variance, and skewness simultaneously across variables in a data frame in R, we can use the following approach:

• Load Required Libraries : We’ll use the dplyr package for data manipulation and the moments package for calculating skewness.
• Define a Function : Create a function that computes mean, variance, and skewness for each variable in a DataFrame.
• Apply the Function : Apply the function to each numeric variable in the DataFrame to obtain the desired statistics.

Let’s walk through an example where we calculate mean, variance, and skewness for each numeric variable in a DataFrame.

Step 1: Load Required Libraries

First we will install and load the Required Libraries.

Step 2: Define a Function

Create a function calc_stats_df that computes mean, variance, and skewness for each numeric variable in a DataFrame.

Step 3: Apply the Function

Apply calc_stats_df to a sample DataFrame to calculate mean, variance, and skewness for each numeric variable.

The statistics DataFrame will contain the mean, variance, and skewness for each numeric variable ( var1 , var2 , var3 ) in the original DataFrame df .

Calculating mean, variance, and skewness simultaneously across variables in a DataFrame provides valuable insights into the distribution and characteristics of data. By using the dplyr and moments packages in R, we can efficiently compute these statistics and gain a deeper understanding of our data’s central tendency, dispersion, and asymmetry. This approach facilitates exploratory data analysis and supports informed decision-making in various fields such as finance, healthcare, and social sciences where understanding data distributions is crucial.

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what does it mean that the null and alternative hypotheses are mutually exclusive and exhaustive and why is that important for hypothesis testing?

Together, mutual exclusivity and exhaustiveness ensure that the hypothesis test is well-defined and produces unambiguous results. This is crucial in scientific research and statistical analysis , where the results of hypothesis testing can have significant implications for further investigation or decision-making.

In hypothesis testing, the null hypothesis (H0) and alternative hypothesis (H1) are two opposing statements about a population parameter. The null hypothesis states that there is no significant difference between the sample data and the population parameter , while the alternative hypothesis states that there is a significant difference.

Mutually exclusive means that the null hypothesis and alternative hypothesis cannot both be true at the same time. If the null hypothesis is true, then the alternative hypothesis must be false, and vice versa. This is important because it helps to avoid ambiguity in the results of the hypothesis test. If the two hypotheses were not mutually exclusive, it would be difficult to determine which hypothesis was supported by the data.

Exhaustive means that one of the two hypotheses must be true. There is no third possibility. This is important because it ensures that the hypothesis test is comprehensive and covers all possible outcomes. If there were a third possibility, then the hypothesis test would not be complete, and the results would be inconclusive.

Together, mutual exclusivity and exhaustiveness ensure that the hypothesis test is well-defined and produces unambiguous results. This is crucial in scientific research and statistical analysis, where the results of hypothesis testing can have significant implications for further investigation or decision-making.

To know more about hypothesis , visit:

https://brainly.com/question/31319397

Related Questions

Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) an = ln(3n2 + 4) − ln(n2 + 4) lim n→[infinity] an = ?

The sequence converges to: lim n→[ infinity ] an = ln(3) = 1.0986. So the sequence converges to 1.0986.

To determine whether the sequence converges or diverges and find the limit, we'll use the properties of logarithms and the concept of limits at infinity. Given sequence: a_n = ln(3n² + 4) - ln(n² + 4) Using the logarithm property, ln(a) - ln(b) = ln(a/b), we can rewrite the sequence as: a_n = ln[(3n² + 4)/(n² + 4)] Now, we'll find the limit as n approaches infinity: lim (n→∞) a_n = lim (n→∞) ln[(3n² + 4)/(n² + 4)] To evaluate this limit, we can divide both the numerator and the denominator by the highest power of n, which is n^2 in this case: lim (n→∞) ln[(3 + 4/n²)/(1 + 4/n²)] As n approaches infinity, the terms with n² in the denominator will approach 0: lim (n→∞) ln[(3 + 0)/(1 + 0)] = ln(3) So, the sequence converges, and the limit is ln(3).

Learn more about convergence here: brainly.com/question/15415793

find the area under the standard normal curve to the left of z=−1.76 and to the right of z=0.07. round your answer to four decimal places, if necessary.

The area under the standard normal curve to the left of z = -1.76 and to the right of z = 0.07 is 0.5113 square units

To find the area under the standard normal curve to the left of z = -1.76, we can use a standard normal distribution table or a calculator with a normal distribution function. The table or calculator will give us the probability that a standard normal random variable is less than or equal to -1.76.

Using a standard normal distribution table, we can find that the area to the left of z = -1.76 is 0.0392 (rounded to four decimal places).

To find the area under the standard normal curve to the right of z = 0.07, we can subtract the area to the left of z = 0.07 from the total area under the curve, which is 1. Using a standard normal distribution table or calculator, we can find that the area to the left of z = 0.07 is 0.5279. Therefore, the area to the right of z = 0.07 is

1 - 0.5279 = 0.4721

Rounding this to four decimal places, we get 0.4721.

Therefore, the area under the standard normal curve to the left of z = -1.76 and to the right of z = 0.07 is

0.0392 + 0.4721 = 0.5113

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M5 L39 Prepare to Compare 2 A bag of apples weighs 7 and 2/10pounds. A crate of bananas is 6 times as heavy as the apples. 10 What is the total weight of the fruit? *>

The calculated total weight of the fruit is 50 2/5 pounds

From the question, we have the following parameters that can be used in our computation:

This means that

Banana = 6 * Apple

So, we have

Banana = 6 * 7 2/10 pounds.

Evaluate the products

Banana = 43 2/10 pounds.

So, the total weight is

total weight = apple + banana

total weight = 7 2/10 + 43 2/10

Evaluate the sum

total weight = 50 4/10

total weight = 50 2/5

Hence, the total weight is 50 2/5 pounds

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in triangle efg, e = 30 in, if you have inches and g = 52. find the area of triangle eft, to the nearest square inch.​

Rounded to the nearest square inch, the area of triangle EFG is 651 in².

A triangle is a polygon with three sides and three angles . It is a simple closed figure that has three straight sides and three vertices. The sum of the angles in a triangle is always 180 degrees. Triangles can be classified based on their sides and angles. Triangles with all sides and angles equal are called equilateral triangles, triangles with two sides and two angles equal are called isosceles triangles, and triangles with no sides and no angles equal are called scalene triangles. Triangles can also be classified based on their angles, such as acute triangles (all angles less than 90 degrees), right triangles (one angle equal to 90 degrees), and obtuse triangles (one angle greater than 90 degrees). Triangles are important in mathematics, physics, and engineering, and are commonly used in construction and design.

To find the area of triangle EFG, we need to use the formula:

Area = 1/2 * base * height

where the base and height are perpendicular to each other.

Since we know the length of side EF (which is the base), we can use the Pythagorean theorem to find the height of the triangle. The Pythagorean theorem states that:

c² = a² + b²

where c is the length of the hypotenuse (in this case, EG), and a and b are the lengths of the other two sides (in this case, EF and FG).

So, we have:

EG² = EF² + FG²

52² = 30² + FG²

FG² = 52² - 30²

FG ≈ 43.27 in (rounded to the nearest hundredth)

Now that we know the base (EF) and the height (perpendicular to EF), we can calculate the area:

Area = 1/2 * EF * height

Area = 1/2 * 30 in * 43.27 in

Area ≈ 650.55 in²

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find the linear, l(x, y) and quadratic, q(x, y), taylor polynomials for f (x, y) = sin(x – 1) cos y valid near (1, 0). -

The linear Taylor polynomial is l(x,y) = x-1, and the quadratic Taylor polynomial is q(x,y) = x-1.

To find the linear and quadratic Taylor polynomials for f(x, y) = sin(x-1)cos(y) near (1, 0), we need to find the partial derivatives of f with respect to x and y, evaluated at (1,0):

f(x, y) = sin(x-1)cos(y)

[tex]\dfrac{\partial f}{\partial x}[/tex] = cos(x-1)cos(y)

[tex]\dfrac{\partial f}{\partial y}[/tex] = -sin(x-1)sin(y)

Evaluated at (1,0), we get:

f(1,0) = sin(0)cos(0) = 0

[tex]\dfrac{\partial f}{\partial x}(1,0)[/tex] = cos(0)cos(0) = 1

[tex]\dfrac{\partial f}{\partial y}(1,0)[/tex] = -sin(0)sin(0) = 0

The linear Taylor polynomial is:

l(x,y) = f(1,0) + [tex]\dfrac{\partial f}{\partial x}(1,0)[/tex](x-1) + [tex]\dfrac{\partial f}{\partial y}(1,0)[/tex](y-0)

l(x,y) = 0 + 1(x-1) + 0(y-0)

l(x,y) = x-1

The quadratic Taylor polynomial is:

[tex]q(x,y) = l(x,y) + \dfrac{1}{2} \dfrac{\partial^2f}{\partial x^2}(1,0)(x-1)^2 + \dfrac{\partial^2f}{\partial y^2}(1,0)(y-0)^2 + \dfrac{\partial^2f}{\partial x \partialy}(1,0)(x-1)(y-0)[/tex]

We need to find the second-order partial derivatives :

[tex]\dfrac{\partial^2f}{\partial x^2}[/tex] = -sin(x-1)cos(y)

[tex]\dfrac{\partial^2f}{\partial y^2}[/tex] = -sin(x-1)cos(y)

[tex]\dfrac{\partial^2f}{\partial x \partial y}[/tex] = -cos(x-1)sin(y)

[tex]\dfrac{\partial^2f}{\partial x^2}(1,0)[/tex]= -sin(0)cos(0) = 0

[tex]\dfrac{\partial^2f}{\partial y^2}(1,0)[/tex] = -sin(0)cos(0) = 0

[tex]\dfrac{\partial^2f}{\partial x \partialy}(1,0)[/tex] = -cos(0)sin(0) = 0

Substituting into the quadratic Taylor polynomial formula, we get:

q(x,y) = (x-1) + (1/2)(0)(x-1)² + (1/2)(0)(y-0)² + (0)(x-1)(y-0)

q(x,y) = x-1

Therefore, the linear Taylor polynomial is l(x,y) = x-1, and the quadratic Taylor polynomial is q(x,y) = x-1.

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what is the radius r of a circle in which an angle of 2 radians cuts off an arc of 36 cm

The radius of the circle is 18 cm.

Explanation: -

suppose radius of the circle is r, where an angle of 2 radians cuts off an arc of 36 cm, to find the radius of the circle use the formula : Arc length = radius × angle (in radians) In this case, the arc length is 36 cm, and the angle is 2 radians. Rearrange the formula to find the radius: radius = arc length / angle

substitute the value of the arc length and angle ( in radian) in the above mention formula: radius = 36 cm / 2 radians radius = 18 cm Thus, radius of the circle is 18 cm.

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PLEASE HELP ME Ice cream is packaged in cylindrical gallon tubs. A tub of ice cream has a total surface area of 387.79 square inches. If the diameter of the tub is 10 inches, what is its height? Use π = 3.14. 7.35 inches 7.65 inches 14.7 inches 17.35 inches

Answer: 7.35 inches

Step-by-step explanation:

The formula for the surface area of a cylinder is 2πrh + 2πr^2, where r is the radius and h is the height of the cylinder.

Given that the diameter of the tub is 10 inches, the radius (r) is half of that, which is 5 inches.

So, the equation for the surface area of the cylinder can be written as:

2π(5)(h) + 2π(5)^2 = 387.79

Simplifying the equation gives:

10πh + 50π = 387.79

Dividing both sides by 10π gives:

h + 5 = 12.34

Subtracting 5 from both sides gives:

Therefore, the height of the tub is 7.35 inches (rounded to two decimal places).

For a random variable Z, its mean and variance are defined as E[Z] and E[(Z-E[Z])2], respectively. Let X1, ..., Xn be independent and identically distributed random variables, each with mean y and variance 02. If we define în = 121_, Xi, what is the mean and variance of vñîn – u)?

The mean of X1, ..., Xn is 121 * y, which indicates that the underlying data have a central tendency of 121 * y. Therefore, the mean and variance of X1, ..., Xn are 121 * y and 242, respectively.

The mean provides information about the central tendency of the underlying data, while the variance provides information about the spread or variability of the underlying data.

The mean and variance of X1, ..., Xn can be calculated as follows:

E[X1, ..., Xn] = E[X1] + ... + E[Xn] = n * E[X1]

E[(X1 - E[X1])2 + ... + (Xn - E[Xn])2] = n * E[(X1 - E[X1])2]

Therefore, the mean and variance of X1, ..., Xn are 121 * y and 242, respectively.

The mean and variance of a random variable are important parameters for describing the probability distribution of that variable.

In this case, the mean of X1, ..., Xn is 121 * y, which indicates that the underlying data have a central tendency of 121 * y.

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hideo is calculating the standard deviation of a data set that has 7 values. he determines that the sum of the squared deviations is 103. what is the standard deviation of the data set?

Therefore, the standard deviation of the data set is approximately 4.14.

The standard deviation is calculated by taking the square root of the variance, which is the sum of the squared deviations divided by the sample size minus 1.

So, first we need to calculate the variance:

variance = sum of squared deviations / (sample size - 1)

variance = 103 / (7 - 1)

variance = 17.17

Now we can find the standard deviation:

standard deviation = √(variance)

standard deviation = √(17.17)

standard deviation = 4.14 (rounded to two decimal places)

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(a) Find the volume of the solid generated by revolving the region bounded by the graph x2=y−2 and 2y−x−2=0 for 0≤x≤1 about y=3. (b) A force of 9 lb. is required to stretch a spring from its natural length of 6 in. to a length of 8 in. Find the work done in stretching the spring (i) from its natural length to a length of 10 in. (ii) from a length of 7 in. to a length of 9 in.

(a) Volume of the solid generated by revolving the region bounded by the graph is 12.422 cubic units.

(i) The work done in stretching the spring from its natural length to a length of 10 in. is 54 lb.-in.

(ii) The work done in stretching the spring from a length of 7 in. to a length of 9 in. is approximately 13.5 lb.-in.

(a) To find the volume of the solid generated by revolving the region bounded by the graph[tex]x^2=y-2[/tex] and 2y-x-2=0 for 0≤x≤1 about y=3, we can use the method of cylindrical shells:

First, we need to find the limits of integration for the radius of the shells. Since we are revolving around y=3, the distance between y=3 and the curve x^2=y-2 will give us the radius of the shell.

Solving for y in [tex]x^2=y-2[/tex], we get[tex]y=x^2+2.[/tex] Substituting this into 2y-x-2=0, we get [tex]x=2y-2y^2-2.[/tex] So the limits of integration for the radius will be from [tex]3-(x^2+2) to 3-(2y-2y^2-2).[/tex]

Next, we need to find the height of the shells. This is simply the length of the interval of integration for x, which is 0 to 1.

So the volume of the solid is given by the integral:

[tex]V = \int (3-(x^2+2)) - (3-(2y-2y^2-2)) dx[/tex] from x=0 to x=1

Simplifying and evaluating the integral, we get:

V ≈ 12.422 cubic units.

Therefore, the volume of the solid generated by revolving the region bounded by the graph [tex]x^2=y-2[/tex] and [tex]2y-x-2=0[/tex] for 0≤x≤1 about y=3 is approximately 12.422 cubic units.

(b) (i) The work done in stretching the spring from its natural length of 6 in. to a length of 10 in. can be found using the formula:

W =[tex](1/2)k(d2^2 - d1^2)[/tex]

where k is the spring constant , d1 is the initial length, and d2 is the final length.

Given that the force required to stretch the spring from its natural length of 6 in. to a length of 8 in. is 9 lb., we can find the spring constant as follows:

k = F/(d2 - d1) = 9/(8-6) = 4.5 lb/in

So the work done in stretching the spring from its natural length of 6 in. to a length of 10 in. is:

W = [tex](1/2)(4.5)(10^2 - 6^2)[/tex]= 54 lb.-in.

Therefore, the work done in stretching the spring from its natural length to a length of 10 in. is 54 lb.-in.

(ii) To find the work done in stretching the spring from a length of 7 in. to a length of 9 in., we can use the same formula:

Using the same spring constant of 4.5 lb/in, the work done is:

W = [tex](1/2)(4.5)(9^2 - 7^2)[/tex]≈ 13.5 lb.-in.

Therefore, the work done in stretching the spring from a length of 7 in. to a length of 9 in. is approximately 13.5 lb.-in.

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Pls help!! I need to find the surface area of the triangular prism below.

is y= 8x^2-10 a function and how do i prove it?

Yes ,  y = 8x² - 10  is a function .

What is a linear equation in mathematics?

A linear equation in algebra is one that only contains a constant and a first-order (direct) element, such as y = mx b, where m is the pitch and b is the y-intercept.

                         Sometimes the following is referred to as a "direct equation of two variables," where y and x are the variables. Direct equations are those in which all of the variables are powers of one. In one example with just one variable, layoff b = 0, where a and b are real numbers and x is the variable, is used.

y = 8x² - 10

the graph attached below

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Suppose we had the following summary statistics from two different, independent populations, both with variances equal to σ.Population 1: ¯x1= 126, s1= 8.062, n1= 5Population 2: ¯x2= 162.75, s2 = 3.5, n2 = 4We want to find a 99% confidence interval for μ2−μ1. To do this, answer the below questions.

The confidence interval of 99% for μ₂ - μ₁ for the given mean and standard deviation is equal to (23.7377, 49.7713).

Confidence interval = 99%  

Confidence interval for μ₂ - μ₁, we need to follow these steps,

Calculate the sample mean difference and the standard error of the mean difference.

Sample mean difference

= ¯x₂ - ¯x₁

= 162.75 - 126

Standard error of the mean difference

= √[(s₁^2/n₁) + (s₂^2/n₂)]

= √[(8.062^2/5) + (3.5^2/4)]

= 4.0065 (rounded to four decimal places)

The t-value for a 99% confidence level with degrees of freedom

= n₁ + n₂ - 2

= 5 + 4 - 2

Using a t-distribution table attached ,

The t-value for a 99% confidence level with 7 degrees of freedom is 3.250.

Margin of error

= t-value x standard error of the mean difference

= 3.250 x 4.0065

= 13.0213 (rounded to four decimal places)

Confidence interval

= Sample mean difference ± Margin of error

= 36.75 ± 13.0213

= (23.7377, 49.7713)

Therefore, the 99% confidence interval for μ₂ - μ₁ is (23.7377, 49.7713).

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Find the measure of angle A to the nearest tenth (Show work if you can pleasee)

to get the answer to this, you need to apply trigonometry

SOH CAH TOA

AB = hypotenuse

BC = opposite

AC = adjacent

we have the hypotenuse and the opposite so we use the equation SOH

sin⁻¹ (opp/hyp)

= sin⁻¹ ([tex]\frac{6}{18}[/tex])

= 19.47122....

= 19.5° (to the nearest tenth)

Worth 20 points!!!! Little Maggie is walking her dog, Lucy, at a local trail and the dog accidentally falls 150 feet down a ravine! You must calculate how much rope is needed for the repel line. Use the image below to find the length of this repel line using one of the 3 trigonometry ratios taught (sin, cos, tan). Round your answer to the nearest whole number. The repel line will be the diagonal distance from the top of the ravine to Lucy. The anchor and the repel line meet to form angle A which forms a 17° angle. Include all of the following in your work for full credit. (a) Identify the correct trigonometric ratio to use (1 point) (b) Correctly set up the trigonometric equation (1 point) (c) Show all work solving equation and finding the correct length of repel line. (1 point)

the length of the repel line needed is approximately 44 feet (rounded to the nearest whole number ).

what is length ?

Length is a physical dimension that describes the extent of an object or distance between two points. In geometry , length refers to the distance between two points, and it is usually measured in units of length such as meters, centimeters, feet, inches,

In the given question,

(a) The correct trigonometric ratio to use in this problem is the sine ratio, which relates the opposite side to the hypotenuse in a right triangle. In this case, we are given the angle A and we want to find the length of the opposite side, which is the distance from the top of the ravine to Lucy. Therefore, we can use the sine ratio as follows:

sin(A) = opposite/hypotenuse

(b) We can set up the equation using the given information as follows:

sin(17°) = opposite/150

where opposite is the length of the repel line that we want to find.

(c) To solve for the length of the repel line, we can rearrange the equation as follows:

opposite = sin(17°) x 150

opposite = 0.2924 x 150

opposite ≈ 44

Therefore, the length of the repel line needed is approximately 44 feet (rounded to the nearest whole number).

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Calls originate from Dryden according to a rate 12 Poisson process. 3/4 are local and 1/4 are long distance. Local calls last an average of 10 minutes, while long distance calls last an average of 5 minutes. Let M be the number of local calls and N the number of long distance calls in equilibrium. Find the distribution of (M,N). what is the number of people on the line.

The distribution of (M,N) is a bivariate Poisson distribution with parameters λ_1 = 9 (rate of local calls) and λ_2 = 3 (rate of long distance calls) since the rates of the two processes are independent Poisson processes.  and the number of people on line are 60.

The joint probability mass function of (M,N) is given by:

[tex]P(M=m, N=n) = e^{-(\lambda_1 +\lambda_2) (lambda_1^m/ m!) (\lambda_2^n/ n!)[/tex]

The number of people on the line is the expected value of the total number of minutes of all calls, which can be calculated as E[10M + 5N] = 10E[M] + 5E[N]. Since M and N are independent Poisson random variables , we have E[M] = λ_1/(1-p_1) and E[N] = λ_2/(1-p_2), where p_1 and p_2 are the probabilities of a call being local or long distance, respectively.

Substituting in the given values, we get:

E[10M + 5N] = 10(9/3) + 5(3/1) = 45 + 15 = 60

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Explanation needed aswell please

The image is plotted and attached

The rectangle started with ABCD. Then following the reflection along line AC. point B and point D swapped so we have B' replacing D and D' replacing B.

180 degrees rotation through C, resulted to B'' D'' and A'. Point C maintains it's position since the rotation is about point C.

Enlargement by a factor of 2 results to C' B''' D''' A'' and this is the final image.

While the reflection and rotation preserves the geometry, the enlargement affects the geometry, producing a rectangle with a bigger size twice the initial size

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HELP 2. TRIANGLE ABC~TRIANGLE ADE; find x B A 20 15 E 2x + 3 C

since the triangles are similar then the ratios of corresponding sides are in proportion, that is

[tex]\frac{BC}{DE}[/tex] = [tex]\frac{AB}{AD}[/tex] ( substitute values )

[tex]\frac{2x+3}{15}[/tex] = [tex]\frac{20+8}{20}[/tex] = [tex]\frac{28}{20}[/tex] ( cross- multiply )

20(2x + 3) = 15 × 28 = 420 ( divide both sides by 20 )

2x + 3 = 21 ( subtract 3 from both sides )

2x = 18 ( divide both sides by 2 )

In similar triangles , the corresponding sides are in same proportion .

 AB = AD + DB

       = 20 + 8

        ΔABC ~ ΔADE,

     [tex]\sf \dfrac{BC}{DE}=\dfrac{AB}{AD}\\\\\\\dfrac{2x + 3}{15}=\dfrac{28}{20}\\\\\ 2x + 3=\dfrac{28}{20}*15[/tex]

     [tex]2x + 3 = 7[/tex] *3

    2x + 3 = 21

Subtract 3 from both sides,

          2x = 21 - 3

          2x = 18

Divide both sides by 2,

            x = 18 ÷ 2

            x = 9

need the answers for the proofs both 13 and 14

Points A, B and C are collinear and X is a bisector of ∠A.

To prove that A, B, and C are collinear , we need to show that they lie on the same straight line .

So, we have the following statements and reasons

Therefore, we have proved that A, B, and C are collinear .

To prove that X is a bisector of ∠A, we need to show that ∠AXB = ∠CXB. We can do this using a two-column proof:

Therefore, we have proven that X is a bisector of ∠A.

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Find the value of x. Area of rectangle = 61 Equation provided is: A(x) = 2x^2 - 5x

The value of x is (5 ± √153)/4.

What is the area of a rectangle?

The area a rectangle occupies is the space it takes up inside the limitations of its four sides. The dimensions of a rectangle determine its area. In essence, the area of a rectangle is equal to the sum of its length and breadth.

Here, we have

Given: Area of rectangle = 61

Equation : A(x) = 2x² - 5x

We have to find the value of x.

16 = 2x² - 5x

2x² - 5x - 16 = 0

we apply here factorization and we get

= (-b ± √b²-4ac)/2a

=  (5 ± √5²+4(2)(16))/2(2)

= (5 ± √25+128)/4

= (5 ± √153)/4

Hence, the value of x is (5 ± √153)/4.

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Widely known kite ABCD 35cm square . Gerrard made a kite with the length of each diagonal each twice the length of the diagonal of the kite ABCD kite. Calculate the area of the kite the new one !

Thus, the area of new kite with its diagonal doubled is found as: 140 sq. cm.

The area a kite encloses is known as its area of flight. A quadrilateral with two sets of neighbouring sides that are equal is referred to as a kite. A kite is made up of four angles, four sides, and two diagonals.

The product of a lengths of a kite's diagonals divides its area in half.

The area of the kite ABCD = 35 cm square.

The formula for the area of kite = 1/2*(d)*(D)

d - length of small diagonal

D - length of large diagonal.

35 =  1/2*(d)*(D)

(d)*(D) = 35*2

(d)*(D) = 70 cm sq.  ..eq 1

Now, the length of diagonals of new kite are doubles that is 2d and 2D.

Area of new kite = 1/2 *(2d)*(2D)

Area of new kite = 1/2 *4*(d)*(D)

Area of new kite = 2 *(d)*(D)

Put the value of (d)*(D) from eq 1.

Area of new kite = 2*70

Area of new kite = 140 sq. cm

Thus, the area of the new kite with its diagonal doubled is found as: 140 sq. cm.

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Find the Laplace transform of a +bt+c for some constants a, b, and c Exercise 6.1.7: Find the Laplace transform of A cos(t+Bsin(t

The Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s). The Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

For a function f(t), the Laplace transform F(s) is defined as ∫[0, ∞) e^(-st) f(t) dt, where s is a complex number .

To find the Laplace transform of a+bt+c, we use linearity and the Laplace transform of elementary functions:

L{a+bt+c} = L{a} + L{bt} + L{c} = a/s + bL{t} + c/s = a/s + b/s^2 + c/s

Therefore, the Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s).

B. To find the Laplace transform of A cos(t+Bsin(t)), we use the following identity:

cos(t + Bsin(t)) = cos(t)cos(Bsin(t)) - sin(t)sin(Bsin(t))

Then, we apply the Laplace transform to both sides and use linearity and the Laplace transform of elementary functions :

L{cos(t + Bsin(t))} = L{cos(t)cos(Bsin(t))} - L{sin(t)sin(Bsin(t))}

Using the formula L{cos(at)} = s/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) L{cos(t)} - (s/(s^2+B^2)) L{sin(t)}

Using the formula L{sin(at)} = a/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (1/s) - (B/(s^2+B^2)) (1/s)

Simplifying, we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan (B/s)

Therefore, the Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

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!!Urgent Help To Whoever Is Willing!!

The Equations are created and plotted as follows

no solution : g(x) = sin (πx) - 2

One solution h(x) at x = -1

multiple but not infinite number of solution: j(x) = x

infinite number of solution : k(x) = sin (πx)

On a graph, the condition of no solution usually refers to a pair of linear equations that do not intersect at any point.

Trigonometric functions such as sine, cosine, and tangent can have infinitely many solutions as they oscillate between values over their respective domains. However, if we restrict the domain or range of a trigonometric function, we can obtain a graph with one solution.

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Use polar coordinates to calculate the area of the region. R = {(x, y) | x^2 + y^2 ≤ 100, x ≥ 6}

The area of the region with polar coordinates R = {(x, y) | x² + y² ≤ 100, x ≥ 6} is approximately 197.39 square units. To calculate the area, first, rewrite the given inequalities in polar coordinates: r² ≤ 100 and rcos(θ) ≥ 6. Next, find the bounds for r and θ. Since r² ≤ 100, r must be between 0 and 10.

For the second inequality, rcos(θ) ≥ 6, divide by r (assuming r ≠ 0) to get cos(θ) ≥ 6/r. To satisfy this inequality, θ must be between arccos(6/r) and π for r in [6, 10]. Now, integrate the area using polar coordinates with the following formula: A = 0.5 * ∫(from 6 to 10) ∫(from arccos(6/r) to π) (r^2) dθ dr. After evaluating the integral, you get the area A ≈ 197.39 square units.

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Identify the surface whose equation is given.rho=cos ϕ

The surface whose equation is given by ρ = cos ϕ is a type of conical surface in the spherical coordinate system . In this equation, ρ represents the radial distance from the origin, and ϕ denotes the polar angle.

Explanation:

The surface described by the equation ρ = cos(ϕ) is a type of conical surface in the spherical coordinate system . Let's break down the explanation step by step:

Spherical Coordinate System : The spherical coordinate system is a three-dimensional coordinate system used to represent points in space using three parameters - radial distance (ρ), polar angle (ϕ), and azimuthal angle (θ). The radial distance ρ represents the distance from the origin (0,0,0) to a point in space, ϕ represents the polar angle measured from the positive z-axis (ranging from 0 to π), and θ represents the azimuthal angle measured from the positive x-axis in the xy-plane (ranging from 0 to 2π).

Equation ρ = cos(ϕ): The equation ρ = cos(ϕ) describes a relationship between the radial distance ρ and the polar angle ϕ. It specifies that for any given value of the polar angle ϕ, the radial distance ρ should be equal to the cosine of ϕ.

Conical Surface : In the context of the spherical coordinate system , a conical surface is a surface that forms a cone shape with its apex at the origin. The equation ρ = cos(ϕ) describes a conical surface because it specifies that the radial distance ρ is determined by the cosine of the polar angle ϕ.

Shape of the Surface : As the polar angle ϕ varies, the equation ρ = cos(ϕ) determines the radial distance ρ at each point on the surface. Since the radial distance is only determined by the cosine of the polar angle, the surface will have a conical shape . Specifically, the surface will form a cone with its apex at the origin and its base expanding outward as ϕ increases from 0 to π. The radius of the base of the cone will vary with the value of ϕ, as determined by the cosine function. When ϕ = 0, the base of the cone will have its maximum radius, equal to 1 (since cos(0) = 1), and as ϕ increases towards π, the radius of the base will decrease until it reaches its minimum value of -1 (since cos(π) = -1). The surface will extend infinitely in the positive and negative z-directions.

In conclusion, the surface described by the equation ρ = cos(ϕ) in the spherical coordinate system is a type of conical surface, forming a cone with its apex at the origin and its base expanding outward as the polar angle ϕ increases, with the radius of the base varying based on the cosine of ϕ.

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(1 point) find the area lying outside =6sin and inside =3 3sin. area =

The area lying outside the circle r=6sin and inside the circle r=3+3sin is approximately 21.205 square units.

To solve this problem, we need to first understand what the equations =6sin and =3 3sin represent. These are actually equations of circles in polar coordinates, where r=6sin represents a circle with radius 6 units and centered at the origin, and r=3+3sin represents a circle with radius 3 units and centered at (-3,0) in Cartesian coordinates. The area lying outside the circle r=6sin and inside the circle r=3+3sin can be found by integrating the equation for the area of a polar region, which is: A = 1/2 ∫ [f(θ)]^2 - [g(θ)]^2 dθ where f(θ) and g(θ) are the equations for the outer and inner boundaries of the region, respectively. In this case, we have: A = 1/2 ∫ (6sin)^2 - (3+3sin)^2 dθ A = 1/2 ∫ 36sin^2 - (9+18sin+9sin^2) dθ A = 1/2 ∫ 27sin^2 - 18sin - 9 dθ To solve this integral, we can use the half-angle identity for sine, which is: sin^2 (θ/2) = (1-cos θ)/2 Substituting this identity into our integral, we get: A = 1/2 ∫ [27(1-cos θ)/2] - 18sin - 9 dθ A = 1/2 ∫ (13.5-13.5cos θ) - 18sin - 9 dθ A = 1/2 ∫ -18sin - 22.5cos θ - 9 dθ Integrating each term separately, we get: A = -9sin θ - 22.5sin θ - 9θ + C where C is the constant of integration . To find the bounds of integration, we need to find the values of θ where the two circles intersect. Setting the equations equal to each other, we get: 6sin = 3+3sin 3sin = 3 sin θ = 1 θ = π/2 So the bounds of integration are 0 and π/2. Substituting these values into the equation for the area, we get: A = -9sin(π/2) - 22.5sin(π/2) - 9(π/2) + C - (-9sin 0 - 22.5sin 0 - 9(0) + C) A = -13.5π/2 Therefore, the area lying outside the circle r=6sin and inside the circle r=3+3sin is approximately 21.205 square units.

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why can't theoretical probability predict on exact numbers of outcomes of a replacement

When dealing with replacement , there is always a certain degree of uncertainty as to what the next outcome will be. This is the reason theoretical probability predict on exact numbers of outcomes.

Theoretical probability is a branch of mathematics that deals with the study of the probability of events occurring based on the assumptions of certain conditions.

It involves the use of formulas and mathematical models to predict the likelihood of certain outcomes. However, it cannot predict the exact numbers of outcomes of a replacement because of the randomness involved in such events.

This is because the replacement process involves randomness, and the outcome of each trial is independent of the previous trials. Therefore, even though the theoretical probability may provide a reasonable estimate of the likelihood of certain outcomes, it cannot predict the exact numbers of outcomes with certainty.

For instance, consider a situation where you have a bag containing ten balls numbered from 1 to 10. You draw a ball, record its number, and then replace it before drawing again.

The theoretical probability of drawing any of the ten balls is 1/10, but it cannot predict the exact number of times a particular ball will be drawn. The outcome of each draw is independent of the others, and the replacement process involves randomness, making it impossible to predict the exact numbers of outcomes.

In conclusion, theoretical probability is a useful tool for predicting the likelihood of certain outcomes in various scenarios. However, when it comes to predicting the exact numbers of outcomes of a replacement, the randomness involved in the process makes it impossible to provide an exact prediction.

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Let g = {(7,1),(4, - 5),(-3,- 6),(1,9)} and h = {(9,- 9),(-6,3)}. Find the function hog. hog= (Use a comma to separate ordered pairs as needed.)

The ordered pairs that are in the domain and range of hog are (-3, (9,-9)) and (7, (-6,3)).

To find the function hog , we need to perform the composition of functions h and g, written as h(g(x)). First, we need to apply g to its domain , which is {7, 4, -3, 1}. g(7) = (1,9) g(4) = (-5,4) g(-3) = (-6,-3) g(1) = (9,1) Now, we can apply h to the range of g. h((1,9)) = (-6,3) h((-5,4)) = undefined (since (-5,4) is not in the domain of h) h((-6,-3)) = (9,-9) h((9,1)) = undefined (since (9,1) is not in the domain of h) Thus, the ordered pairs that are in the domain and range of hog are (-3, (9,-9)) and (7, (-6,3)). Therefore, hog = {(-3, (9,-9)), (7, (-6,3))}.

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) g(x) = 3^ 64 − x2 cubed root of 64-x^2

To find the critical numbers of the function g(x), we need to first find its derivative and then set the derivative equal to zero to solve for x.

The function is given as: g(x) = (64 - x^2)^(1/3)

To find the derivative, we will use the chain rule , which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

So: g'(x) = (1/3)(64 - x^2)^(-2/3) * (-2x)

Now, we need to set g'(x) = 0 to find the critical numbers:

0 = (1/3)(64 - x^2)^(-2/3) * (-2x)

To solve for x, we can observe that if either of the factors is equal to 0, then the equation will hold.

So, let's examine each factor: (1/3)(64 - x^2)^(-2/3) = 0:

This factor can never be zero, because a nonzero number raised to any power is never zero. -2x = 0: This factor is zero when x = 0.

So, the only critical number for the function g(x) is x = 0. The final answer is: 0

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2. Nemani buys a TV on hire purchase. The cash price is $1250. He pays $450 deposit and 12 monthly instalments of $95.How much interest is paid by Nemani.​

Nemani has paid $1250 for a TV on hire purchase. The cash price was $1250, so the total cost to Nemani was $1300. Nemani has paid $450 deposit and 12 monthly instalments of $95, for a total of $1445.

So, Nemani has paid an interest rate of 10% on his total payment.

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