hypothesis test example questions

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Hypothesis Testing Solved Examples(Questions and Solutions)

Here is a list hypothesis testing exercises and solutions. Try to solve a question by yourself first before you look at the solution.

Question 1 In the population, the average IQ is 100 with a standard deviation of 15. A team of scientists want to test a new medication to see if it has either a positive or negative effect on intelligence, or not effect at all. A sample of 30 participants who have taken the medication  has a mean of 140. Did the medication affect intelligence? View Solution to Question 1

A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:62, 92, 75, 68, 83, 95. Can the professor have 90% confidence that the mean score for the class on the test would be above 70. Solution to Question 2

Question 3 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster? Perform  the required hypothesis test at the 5% level of significance. Solution to Question 3 

Question 4 We want to compare the heights in inches of two groups of individuals. Here are the measurements: X: 175, 168, 168, 190, 156, 181, 182, 175, 174, 179 Y:  120, 180, 125, 188, 130, 190, 110, 185, 112, 188 Solution to Question 4 

Question 5 A clinic provides a program to help their clients lose weight and asks a consumer agency to investigate the effectiveness of the program. The agency takes a sample of 15 people, weighing each person in the sample before the program begins and 3 months later. The results a tabulated below

Determine is the program is effective. Solution to Question 5

Question 6 A sample of 20 students were selected and given a diagnostic module prior to studying for a test. And then they were given the test again after completing the module. . The result of the students scores in the test before and after the test is tabulated below.

We want to see if there is significant improvement in the student’s performance due to this teaching method Solution to Question 6 

Question 7 A study was performed to test wether cars get better mileage on premium gas than on regular gas. Each of 10 cars was first filled with regular or premium gas, decided by a coin toss, and the mileage for the tank was recorded. The mileage was recorded again for the same cars using other kind of gasoline. Determine wether cars get significantly better mileage with premium gas.

Mileage with regular gas: 16,20,21,22,23,22,27,25,27,28 Mileage with premium gas: 19, 22,24,24,25,25,26,26,28,32 Solution to Question 7 

Question 8  An automatic cutter machine must cut steel strips of 1200 mm length. From a preliminary data, we checked that the lengths of the pieces produced by the machine can be considered as normal random variables  with a 3mm standard deviation. We want to make sure that the machine is set correctly. Therefore 16 pieces of the products are randomly selected and weight. The figures were in mm: 1193,1196,1198,1195,1198,1199,1204,1193,1203,1201,1196,1200,1191,1196,1198,1191 Examine wether there is any significant deviation from the required size Solution to Question 8

Question 9 Blood pressure reading of ten patients before and after medication for reducing the blood pressure are as follows

Patient: 1,2,3,4,5,6,7,8,9,10 Before treatment: 86,84,78,90,92,77,89,90,90,86 After treatment:    80,80,92,79,92,82,88,89,92,83

Test the null hypothesis of no effect agains the alternate hypothesis that medication is effective. Execute it with Wilcoxon test Solution to Question 9

Question on ANOVA Sussan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides 24 students into three groups of 8 each. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with nose that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores are tabulated below.

Group1: Constant sound: 7,4,6,8,6,6,2,9 Group 2: Random sound: 5,5,3,4,4,7,2,2 Group 3: No sound at all: 2,4,7,1,2,1,5,5 Solution to Question 10

Question 11 Using the following three groups of data, perform a one-way analysis of variance using α  = 0.05.

Solution to Question 11

Question 12 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

New Machine: 42,41,41.3,41.8,42.4,42.8,43.2,42.3,41.8,42.7 Old Machine:  42.7,43.6,43.8,43.3,42.5,43.5,43.1,41.7,44,44.1

Perform an F-test to determine if the null hypothesis should be accepted. Solution to Question 12

Question 13 A random sample 500 U.S adults are questioned about their political affiliation and opinion on a tax reform bill. We need to test if the political affiliation and their opinon on a tax reform bill are dependent, at 5% level of significance. The observed contingency table is given below.

Solution to Question 13

Question 14 Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?

Solution to Question 14

Solution to Question 15

Question 16 A newly developed muesli contains five types of seeds (A, B, C, D and E). The percentage of which is 35%, 25%, 20%, 10% and 10% according to the product information. In a randomly selected muesli, the following volume distribution was found.

Lets us decide about the null hypothesis whether the composition of the sample corresponds to the distribution indicated on the packaging at alpha = 0.1 significance level. Solution to Question 16

Question 17 A research team investigated whether there was any significant correlation between the severity of a certain disease runoff and the age of the patients. During the study, data for n = 200 patients were collected and grouped according to the severity of the disease and the age of the patient. The table below shows the result

Let us decided about the correlation between the age of the patients and the severity of disease progression. Solution to Question 17

Question 18 A publisher is interested in determine which of three book cover is most attractive. He interviews 400 people in each of the three states (California, Illinois and New York), and asks each person which of the  cover he or she prefers. The number of preference for each cover is as follows:

Do these data indicate that there are regional differences in people’s preferences concerning these covers? Use the 0.05 level of significance. Solution to Question 18

Question 19 Trees planted along the road were checked for which ones are healthy(H) or diseased (D) and the following arrangement of the trees were obtained:

H H H H D D D H H H H H H H D D H H D D D

Test at the    = 0.05 significance wether this arrangement may be regarded as random

Solution to Question 19 

Question 20 Suppose we flip a coin n = 15 times and come up with the following arrangements

H T T T H H T T T T H H T H H

(H = head, T = tail)

Test at the alpha = 0.05 significance level whether this arrangement may be regarded as random.

Solution to Question 20

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You might also like, chi-square goodness of fit test – question 16 ( a newly developed muesli…), hypothesis testing problem – question 2 ( a professor wants to know if…), kolmogorov-smirnov goodness of fit test.

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S.3.3 hypothesis testing examples.

• Example: Right-Tailed Test
• Example: Left-Tailed Test
• Example: Two-Tailed Test

Brinell Hardness Scores

An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:

The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:

H 0 : μ = 170 H A : μ > 170

The engineer entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

Descriptive Statistics

$\mu$: mean of Brinelli

Null hypothesis    H₀: $\mu$ = 170 Alternative hypothesis    H₁: $\mu$ > 170

The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t * is 1.22, and the P -value is 0.117.

If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were greater than 1.7109 (determined using statistical software or a t -table):

Since the engineer's test statistic, t * = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

If the engineer used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 24 curve and to the right of the test statistic t * = 1.22:

In the output above, Minitab reports that the P -value is 0.117. Since the P -value, 0.117, is greater than \(\alpha\) = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Height of Sunflowers

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:

The biologist's hypotheses are:

H 0 : μ = 15.7 H A : μ < 15.7

The biologist entered her data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. She obtained the following output:

$\mu$: mean of Height

Null hypothesis    H₀: $\mu$ = 15.7 Alternative hypothesis    H₁: $\mu$ < 15.7

The output tells us that the average height of the n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t * is -4.60, and the P -value, 0.000, is to three decimal places.

Minitab Note. Minitab will always report P -values to only 3 decimal places. If Minitab reports the P -value as 0.000, it really means that the P -value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P -value as 0.000, you should report the P -value as being "< 0.001."

If the biologist set her significance level \(\alpha\) at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t * were less than -1.6939 (determined using statistical software or a t -table):s-3-3

Since the biologist's test statistic, t * = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

If the biologist used the P -value approach to conduct her hypothesis test, she would determine the area under a t n - 1 = t 32 curve and to the left of the test statistic t * = -4.60:

In the output above, Minitab reports that the P -value is 0.000, which we take to mean < 0.001. Since the P -value is less than 0.001, it is clearly less than \(\alpha\) = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

Gum Thickness

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of n = 10 pieces of gum and measured their thickness. He obtained:

The quality control specialist's hypotheses are:

H 0 : μ = 7.5 H A : μ ≠ 7.5

The quality control specialist entered his data into Minitab and requested that the "one-sample t -test" be conducted for the above hypotheses. He obtained the following output:

$\mu$: mean of Thickness

Null hypothesis    H₀: $\mu$ = 7.5 Alternative hypothesis    H₁: $\mu \ne$ 7.5

The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t * is 1.54, and the P -value is 0.158.

If the quality control specialist sets his significance level \(\alpha\) at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t * were less than -2.2616 or greater than 2.2616 (determined using statistical software or a t -table):

Since the quality control specialist's test statistic, t * = 1.54, is not less than -2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.

If the quality control specialist used the P -value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 9 curve, to the right of 1.54 and to the left of -1.54:

In the output above, Minitab reports that the P -value is 0.158. Since the P -value, 0.158, is greater than \(\alpha\) = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.

Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean \(\mu\). The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the P -value approach — generally extend to all of the hypothesis tests you will encounter.

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Practice Problems on Hypothesis Testing

In this post I have put together the practice problems (from my academics study notes) to explain how in practical Hypothesis Testing works. This post is written mostly for the learners who want to deep dive into the statistics for data science. Focus will be on problem solving. For concepts please refer my previous posts on testing of hypothesis.

Prerequisite to understand Hypothesis testing examples:

• Understanding of hypothesis testing concepts
• How to use z-table, t-table and chi square table.

Formula list:

Critical Regions

In hypothesis testing, critical region is represented by set of values, where null hypothesis is rejected. So it is also know as region of rejection. It takes different boundary values for different level of significance. Below info graphics shows the region of rejection that is critical region and region of acceptance with respect to the level of significance 1%.

A Telecom service provider claims that individual customers pay on an average 400 rs. per month with standard deviation of 25 rs. A random sample of 50 customers bills during a given month is taken with a mean of 250 and standard deviation of 15. What to say with respect to the claim made by the service provider?

From the data available, it is observed that 400 out of 850 customers purchased the groceries online. Can we say that most of the customers are moving towards online shopping even for groceries?

It is found that 250 errors in the randomly selected 1000 lines of code from Team A and 300 errors in 800 lines of code from Team B. Can we assume that team B’s performance is superior to that of A.

Following is the record of number of accidents took place during the various days of the week.

Can we conclude that accident s are independent of the day of week?

Analyze the below data and tell whether you can conclude that smoking causes cancer or not?

It is claimed that the mean of the population is 67 at 5% level of significance. Mean obtained from a random sample of size 100 is 64 with SD 3. Validate the claim.

There is an assumption that there is no significant difference between boys and girls with respect to intelligence. Tests are conducted on two groups and the following are the observations

Validate the claim with 5% LoS (Level of Significance)

An automobile tyre manufacturer claims that the average life of a particular grade of tyre is more than 20,000 km. A random sample of 16 tyres is having mean 22,000 km with a standard deviation of 5000 km.

Validate the claim of the manufacturer at 5% LoS.

That is all for now. Please share your thoughts using the comment section below.

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I think smoking problem is wrongly concluded. There the H0 is assumed of dependency(smoking and cancer are dependent) while for chi square test, the null hypothesis is always for independence. So the H0 should be “Smoking and Cancer are independent”.

Please how can I download this page?

In the last sum the alternate hypothesis is less than 22,000 and the null hypothesis is more than 20,000. If the value turns out to be 21,000 then which hypothesis will you accept? I guess there’s an error, the alternate hypothesis should be less than equal to 20,000 and not 22,000. Correct me if I’m wrong.

Thanks for presenting above test cases, it really really helps to understand the tail concept. I am reading z-test and refer your hypothesis page as an example. I go through z-test example and in example no.8 last one it is “One tail – Left tailed test” but in the diag below it shows the right tailed. Not sure am I interpret wrong or diag error ? pls . correct me. Thanks.

The alternate hypothesis is the opposite of null hypothesis so it’s less less than or left tailed. Since the null hypothesis was accepted the graph is Right Tailed had the null hypothesis been rejected or the alternate hypothesis been accepted the graph would have been left tailed. I hope I cleared your doubt

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Statistics By Jim

Making statistics intuitive

Hypothesis Testing: Uses, Steps & Example

By Jim Frost 4 Comments

What is Hypothesis Testing?

Hypothesis testing in statistics uses sample data to infer the properties of a whole population . These tests determine whether a random sample provides sufficient evidence to conclude an effect or relationship exists in the population. Researchers use them to help separate genuine population-level effects from false effects that random chance can create in samples. These methods are also known as significance testing.

For example, researchers are testing a new medication to see if it lowers blood pressure. They compare a group taking the drug to a control group taking a placebo. If their hypothesis test results are statistically significant, the medication’s effect of lowering blood pressure likely exists in the broader population, not just the sample studied.

Using Hypothesis Tests

A hypothesis test evaluates two mutually exclusive statements about a population to determine which statement the sample data best supports. These two statements are called the null hypothesis and the alternative hypothesis . The following are typical examples:

• Null Hypothesis : The effect does not exist in the population.
• Alternative Hypothesis : The effect does exist in the population.

Hypothesis testing accounts for the inherent uncertainty of using a sample to draw conclusions about a population, which reduces the chances of false discoveries. These procedures determine whether the sample data are sufficiently inconsistent with the null hypothesis that you can reject it. If you can reject the null, your data favor the alternative statement that an effect exists in the population.

Statistical significance in hypothesis testing indicates that an effect you see in sample data also likely exists in the population after accounting for random sampling error , variability, and sample size. Your results are statistically significant when the p-value is less than your significance level or, equivalently, when your confidence interval excludes the null hypothesis value.

Conversely, non-significant results indicate that despite an apparent sample effect, you can’t be sure it exists in the population. It could be chance variation in the sample and not a genuine effect.

Learn more about Failing to Reject the Null .

5 Steps of Significance Testing

Hypothesis testing involves five key steps, each critical to validating a research hypothesis using statistical methods:

• Formulate the Hypotheses : Write your research hypotheses as a null hypothesis (H 0 ) and an alternative hypothesis (H A ).
• Data Collection : Gather data specifically aimed at testing the hypothesis.
• Conduct A Test : Use a suitable statistical test to analyze your data.
• Make a Decision : Based on the statistical test results, decide whether to reject the null hypothesis or fail to reject it.
• Report the Results : Summarize and present the outcomes in your report’s results and discussion sections.

While the specifics of these steps can vary depending on the research context and the data type, the fundamental process of hypothesis testing remains consistent across different studies.

Let’s work through these steps in an example!

Hypothesis Testing Example

Researchers want to determine if a new educational program improves student performance on standardized tests. They randomly assign 30 students to a control group , which follows the standard curriculum, and another 30 students to a treatment group, which participates in the new educational program. After a semester, they compare the test scores of both groups.

Download the CSV data file to perform the hypothesis testing yourself: Hypothesis_Testing .

The researchers write their hypotheses. These statements apply to the population, so they use the mu (μ) symbol for the population mean parameter .

• Null Hypothesis (H 0 ) : The population means of the test scores for the two groups are equal (μ 1 = μ 2 ).
• Alternative Hypothesis (H A ) : The population means of the test scores for the two groups are unequal (μ 1 ≠ μ 2 ).

Choosing the correct hypothesis test depends on attributes such as data type and number of groups. Because they’re using continuous data and comparing two means, the researchers use a 2-sample t-test .

Here are the results.

The treatment group’s mean is 58.70, compared to the control group’s mean of 48.12. The mean difference is 10.67 points. Use the test’s p-value and significance level to determine whether this difference is likely a product of random fluctuation in the sample or a genuine population effect.

Because the p-value (0.000) is less than the standard significance level of 0.05, the results are statistically significant, and we can reject the null hypothesis. The sample data provides sufficient evidence to conclude that the new program’s effect exists in the population.

Limitations

Hypothesis testing improves your effectiveness in making data-driven decisions. However, it is not 100% accurate because random samples occasionally produce fluky results. Hypothesis tests have two types of errors, both relating to drawing incorrect conclusions.

• Type I error: The test rejects a true null hypothesis—a false positive.
• Type II error: The test fails to reject a false null hypothesis—a false negative.

Learn more about Type I and Type II Errors .

Our exploration of hypothesis testing using a practical example of an educational program reveals its powerful ability to guide decisions based on statistical evidence. Whether you’re a student, researcher, or professional, understanding and applying these procedures can open new doors to discovering insights and making informed decisions. Let this tool empower your analytical endeavors as you navigate through the vast seas of data.

Learn more about the Hypothesis Tests for Various Data Types .

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Reader Interactions

June 10, 2024 at 10:51 am

Thank you, Jim, for another helpful article; timely too since I have started reading your new book on hypothesis testing and, now that we are at the end of the school year, my district is asking me to perform a number of evaluations on instructional programs. This is where my question/concern comes in. You mention that hypothesis testing is all about testing samples. However, I use all the students in my district when I make these comparisons. Since I am using the entire “population” in my evaluations (I don’t select a sample of third grade students, for example, but I use all 700 third graders), am I somehow misusing the tests? Or can I rest assured that my district’s student population is only a sample of the universal population of students?

June 10, 2024 at 1:50 pm

I hope you are finding the book helpful!

Yes, the purpose of hypothesis testing is to infer the properties of a population while accounting for random sampling error.

In your case, it comes down to how you want to use the results. Who do you want the results to apply to?

If you’re summarizing the sample, looking for trends and patterns, or evaluating those students and don’t plan to apply those results to other students, you don’t need hypothesis testing because there is no sampling error. They are the population and you can just use descriptive statistics. In this case, you’d only need to focus on the practical significance of the effect sizes.

On the other hand, if you want to apply the results from this group to other students, you’ll need hypothesis testing. However, there is the complicating issue of what population your sample of students represent. I’m sure your district has its own unique characteristics, demographics, etc. Your district’s students probably don’t adequately represent a universal population. At the very least, you’d need to recognize any special attributes of your district and how they could bias the results when trying to apply them outside the district. Or they might apply to similar districts in your region.

However, I’d imagine your 3rd graders probably adequately represent future classes of 3rd graders in your district. You need to be alert to changing demographics. At least in the short run I’d imagine they’d be representative of future classes.

Think about how these results will be used. Do they just apply to the students you measured? Then you don’t need hypothesis tests. However, if the results are being used to infer things about other students outside of the sample, you’ll need hypothesis testing along with considering how well your students represent the other students and how they differ.

I hope that helps!

June 10, 2024 at 3:21 pm

Thank you so much, Jim, for the suggestions in terms of what I need to think about and consider! You are always so clear in your explanations!!!!

June 10, 2024 at 3:22 pm

You’re very welcome! Best of luck with your evaluations!

Comments and Questions Cancel reply

9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43   H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

• H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
• Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
• Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50   H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

• H 0 : μ ≤ 1
• H a : μ > 1
• Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations and draw the graph .
• State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

• H 0 : p ≤ 0.00034
• H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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Hypothesis Testing – A Complete Guide with Examples

Published by Alvin Nicolas at August 14th, 2021 , Revised On October 26, 2023

In statistics, hypothesis testing is a critical tool. It allows us to make informed decisions about populations based on sample data. Whether you are a researcher trying to prove a scientific point, a marketer analysing A/B test results, or a manufacturer ensuring quality control, hypothesis testing plays a pivotal role. This guide aims to introduce you to the concept and walk you through real-world examples.

What is a Hypothesis and a Hypothesis Testing?

A hypothesis is considered a belief or assumption that has to be accepted, rejected, proved or disproved. In contrast, a research hypothesis is a research question for a researcher that has to be proven correct or incorrect through investigation.

What is Hypothesis Testing?

Hypothesis testing  is a scientific method used for making a decision and drawing conclusions by using a statistical approach. It is used to suggest new ideas by testing theories to know whether or not the sample data supports research. A research hypothesis is a predictive statement that has to be tested using scientific methods that join an independent variable to a dependent variable.  

Example: The academic performance of student A is better than student B

Characteristics of the Hypothesis to be Tested

A hypothesis should be:

• Clear and precise
• Capable of being tested
• Able to relate to a variable
• Stated in simple terms
• Consistent with known facts
• Limited in scope and specific
• Tested in a limited timeframe
• Explain the facts in detail

What is a Null Hypothesis and Alternative Hypothesis?

A  null hypothesis  is a hypothesis when there is no significant relationship between the dependent and the participants’ independent  variables . 

In simple words, it’s a hypothesis that has been put forth but hasn’t been proved as yet. A researcher aims to disprove the theory. The abbreviation “Ho” is used to denote a null hypothesis.

If you want to compare two methods and assume that both methods are equally good, this assumption is considered the null hypothesis.

Example: In an automobile trial, you feel that the new vehicle’s mileage is similar to the previous model of the car, on average. You can write it as: Ho: there is no difference between the mileage of both vehicles. If your findings don’t support your hypothesis and you get opposite results, this outcome will be considered an alternative hypothesis.

If you assume that one method is better than another method, then it’s considered an alternative hypothesis. The alternative hypothesis is the theory that a researcher seeks to prove and is typically denoted by H1 or HA.

If you support a null hypothesis, it means you’re not supporting the alternative hypothesis. Similarly, if you reject a null hypothesis, it means you are recommending the alternative hypothesis.

Example: In an automobile trial, you feel that the new vehicle’s mileage is better than the previous model of the vehicle. You can write it as; Ha: the two vehicles have different mileage. On average/ the fuel consumption of the new vehicle model is better than the previous model.

If a null hypothesis is rejected during the hypothesis test, even if it’s true, then it is considered as a type-I error. On the other hand, if you don’t dismiss a hypothesis, even if it’s false because you could not identify its falseness, it’s considered a type-II error.

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How to Conduct Hypothesis Testing?

Here is a step-by-step guide on how to conduct hypothesis testing.

Step 1: State the Null and Alternative Hypothesis

Once you develop a research hypothesis, it’s important to state it is as a Null hypothesis (Ho) and an Alternative hypothesis (Ha) to test it statistically.

A null hypothesis is a preferred choice as it provides the opportunity to test the theory. In contrast, you can accept the alternative hypothesis when the null hypothesis has been rejected.

Example: You want to identify a relationship between obesity of men and women and the modern living style. You develop a hypothesis that women, on average, gain weight quickly compared to men. Then you write it as: Ho: Women, on average, don’t gain weight quickly compared to men. Ha: Women, on average, gain weight quickly compared to men.

Step 2: Data Collection

Hypothesis testing follows the statistical method, and statistics are all about data. It’s challenging to gather complete information about a specific population you want to study. You need to  gather the data  obtained through a large number of samples from a specific population. 

Example: Suppose you want to test the difference in the rate of obesity between men and women. You should include an equal number of men and women in your sample. Then investigate various aspects such as their lifestyle, eating patterns and profession, and any other variables that may influence average weight. You should also determine your study’s scope, whether it applies to a specific group of population or worldwide population. You can use available information from various places, countries, and regions.

Step 3: Select Appropriate Statistical Test

There are many  types of statistical tests , but we discuss the most two common types below, such as One-sided and two-sided tests.

Note: Your choice of the type of test depends on the purpose of your study 

One-sided Test

In the one-sided test, the values of rejecting a null hypothesis are located in one tail of the probability distribution. The set of values is less or higher than the critical value of the test. It is also called a one-tailed test of significance.

Example: If you want to test that all mangoes in a basket are ripe. You can write it as: Ho: All mangoes in the basket, on average, are ripe. If you find all ripe mangoes in the basket, the null hypothesis you developed will be true.

Two-sided Test

In the two-sided test, the values of rejecting a null hypothesis are located on both tails of the probability distribution. The set of values is less or higher than the first critical value of the test and higher than the second critical value test. It is also called a two-tailed test of significance. 

Example: Nothing can be explicitly said whether all mangoes are ripe in the basket. If you reject the null hypothesis (Ho: All mangoes in the basket, on average, are ripe), then it means all mangoes in the basket are not likely to be ripe. A few mangoes could be raw as well.

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Step 4: Select the Level of Significance

When you reject a null hypothesis, even if it’s true during a statistical hypothesis, it is considered the  significance level . It is the probability of a type one error. The significance should be as minimum as possible to avoid the type-I error, which is considered severe and should be avoided. 

If the significance level is minimum, then it prevents the researchers from false claims. 

The significance level is denoted by  P,  and it has given the value of 0.05 (P=0.05)

If the P-Value is less than 0.05, then the difference will be significant. If the P-value is higher than 0.05, then the difference is non-significant.

Example: Suppose you apply a one-sided test to test whether women gain weight quickly compared to men. You get to know about the average weight between men and women and the factors promoting weight gain.

Step 5: Find out Whether the Null Hypothesis is Rejected or Supported

After conducting a statistical test, you should identify whether your null hypothesis is rejected or accepted based on the test results. It would help if you observed the P-value for this.

Example: If you find the P-value of your test is less than 0.5/5%, then you need to reject your null hypothesis (Ho: Women, on average, don’t gain weight quickly compared to men). On the other hand, if a null hypothesis is rejected, then it means the alternative hypothesis might be true (Ha: Women, on average, gain weight quickly compared to men. If you find your test’s P-value is above 0.5/5%, then it means your null hypothesis is true.

Step 6: Present the Outcomes of your Study

The final step is to present the  outcomes of your study . You need to ensure whether you have met the objectives of your research or not. 

In the discussion section and  conclusion , you can present your findings by using supporting evidence and conclude whether your null hypothesis was rejected or supported.

In the result section, you can summarise your study’s outcomes, including the average difference and P-value of the two groups.

If we talk about the findings, our study your results will be as follows:

Example: In the study of identifying whether women gain weight quickly compared to men, we found the P-value is less than 0.5. Hence, we can reject the null hypothesis (Ho: Women, on average, don’t gain weight quickly than men) and conclude that women may likely gain weight quickly than men.

Did you know in your academic paper you should not mention whether you have accepted or rejected the null hypothesis? 

Always remember that you either conclude to reject Ho in favor of Haor   do not reject Ho . It would help if you never rejected  Ha  or even  accept Ha .

Suppose your null hypothesis is rejected in the hypothesis testing. If you conclude  reject Ho in favor of Haor   do not reject Ho,  then it doesn’t mean that the null hypothesis is true. It only means that there is a lack of evidence against Ho in favour of Ha. If your null hypothesis is not true, then the alternative hypothesis is likely to be true.

Example: We found that the P-value is less than 0.5. Hence, we can conclude reject Ho in favour of Ha (Ho: Women, on average, don’t gain weight quickly than men) reject Ho in favour of Ha. However, rejected in favour of Ha means (Ha: women may likely to gain weight quickly than men)

Frequently Asked Questions

What are the 3 types of hypothesis test.

The 3 types of hypothesis tests are:

• One-Sample Test : Compare sample data to a known population value.
• Two-Sample Test : Compare means between two sample groups.
• ANOVA : Analyze variance among multiple groups to determine significant differences.

What is a hypothesis?

A hypothesis is a proposed explanation or prediction about a phenomenon, often based on observations. It serves as a starting point for research or experimentation, providing a testable statement that can either be supported or refuted through data and analysis. In essence, it’s an educated guess that drives scientific inquiry.

What are null hypothesis?

A null hypothesis (often denoted as H0) suggests that there is no effect or difference in a study or experiment. It represents a default position or status quo. Statistical tests evaluate data to determine if there’s enough evidence to reject this null hypothesis.

What is the probability value?

The probability value, or p-value, is a measure used in statistics to determine the significance of an observed effect. It indicates the probability of obtaining the observed results, or more extreme, if the null hypothesis were true. A small p-value (typically <0.05) suggests evidence against the null hypothesis, warranting its rejection.

What is p value?

The p-value is a fundamental concept in statistical hypothesis testing. It represents the probability of observing a test statistic as extreme, or more so, than the one calculated from sample data, assuming the null hypothesis is true. A low p-value suggests evidence against the null, possibly justifying its rejection.

What is a t test?

A t-test is a statistical test used to compare the means of two groups. It determines if observed differences between the groups are statistically significant or if they likely occurred by chance. Commonly applied in research, there are different t-tests, including independent, paired, and one-sample, tailored to various data scenarios.

When to reject null hypothesis?

Reject the null hypothesis when the test statistic falls into a predefined rejection region or when the p-value is less than the chosen significance level (commonly 0.05). This suggests that the observed data is unlikely under the null hypothesis, indicating evidence for the alternative hypothesis. Always consider the study’s context.

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A case study is a detailed analysis of a situation concerning organizations, industries, and markets. The case study generally aims at identifying the weak areas.

A variable is a characteristic that can change and have more than one value, such as age, height, and weight. But what are the different types of variables?

Disadvantages of primary research – It can be expensive, time-consuming and take a long time to complete if it involves face-to-face contact with customers.

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Statistics and probability

Course: statistics and probability   >   unit 12, hypothesis testing and p-values.

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Methodology

• How to Write a Strong Hypothesis | Steps & Examples

How to Write a Strong Hypothesis | Steps & Examples

Published on May 6, 2022 by Shona McCombes . Revised on November 20, 2023.

A hypothesis is a statement that can be tested by scientific research. If you want to test a relationship between two or more variables, you need to write hypotheses before you start your experiment or data collection .

Example: Hypothesis

Daily apple consumption leads to fewer doctor’s visits.

Table of contents

What is a hypothesis, developing a hypothesis (with example), hypothesis examples, other interesting articles, frequently asked questions about writing hypotheses.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess – it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Variables in hypotheses

Hypotheses propose a relationship between two or more types of variables .

• An independent variable is something the researcher changes or controls.
• A dependent variable is something the researcher observes and measures.

If there are any control variables , extraneous variables , or confounding variables , be sure to jot those down as you go to minimize the chances that research bias  will affect your results.

In this example, the independent variable is exposure to the sun – the assumed cause . The dependent variable is the level of happiness – the assumed effect .

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Step 1. ask a question.

Writing a hypothesis begins with a research question that you want to answer. The question should be focused, specific, and researchable within the constraints of your project.

Step 2. Do some preliminary research

Your initial answer to the question should be based on what is already known about the topic. Look for theories and previous studies to help you form educated assumptions about what your research will find.

At this stage, you might construct a conceptual framework to ensure that you’re embarking on a relevant topic . This can also help you identify which variables you will study and what you think the relationships are between them. Sometimes, you’ll have to operationalize more complex constructs.

Step 3. Formulate your hypothesis

Now you should have some idea of what you expect to find. Write your initial answer to the question in a clear, concise sentence.

4. Refine your hypothesis

You need to make sure your hypothesis is specific and testable. There are various ways of phrasing a hypothesis, but all the terms you use should have clear definitions, and the hypothesis should contain:

• The relevant variables
• The specific group being studied
• The predicted outcome of the experiment or analysis

5. Phrase your hypothesis in three ways

To identify the variables, you can write a simple prediction in  if…then form. The first part of the sentence states the independent variable and the second part states the dependent variable.

In academic research, hypotheses are more commonly phrased in terms of correlations or effects, where you directly state the predicted relationship between variables.

If you are comparing two groups, the hypothesis can state what difference you expect to find between them.

6. Write a null hypothesis

If your research involves statistical hypothesis testing , you will also have to write a null hypothesis . The null hypothesis is the default position that there is no association between the variables. The null hypothesis is written as H 0 , while the alternative hypothesis is H 1 or H a .

• H 0 : The number of lectures attended by first-year students has no effect on their final exam scores.
• H 1 : The number of lectures attended by first-year students has a positive effect on their final exam scores.

If you want to know more about the research process , methodology , research bias , or statistics , make sure to check out some of our other articles with explanations and examples.

• Sampling methods
• Simple random sampling
• Stratified sampling
• Cluster sampling
• Likert scales
• Reproducibility

 Statistics

• Null hypothesis
• Statistical power
• Probability distribution
• Effect size
• Poisson distribution

Research bias

• Optimism bias
• Cognitive bias
• Implicit bias
• Hawthorne effect
• Anchoring bias
• Explicit bias

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A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

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McCombes, S. (2023, November 20). How to Write a Strong Hypothesis | Steps & Examples. Scribbr. Retrieved August 5, 2024, from https://www.scribbr.com/methodology/hypothesis/

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