hypothesis testing for proportions in excel

How to Perform a One-Proportion Z-Test in Excel

• by Deion Menor
• November 21, 2022

This guide will explain how to perform a one-proportion Z-test in Excel.

We can use the one proportion Z-test to test if the proportions of categories in a single variable differ significantly from a known population proportion.

Table of Contents

A real example of using the one-proportion z-test, how to perform a one-proportion z test in excel, frequently asked questions (faq).

The one proportion Z-test requires your dataset to have a variable that could be one of two values.

The Z-test also requires a few assumptions about your data.

First, the sample being tested must have come from a random selection. Next, each of your data points must be independent of the others.

Let’s take a look at a quick example where we can perform a one-proportion Z-test.

Suppose you want to prove the claim that at least 95% of cars produced by a factory contain zero initial defects. In any given year, the factory produces 100,000 vehicles of different types. 

To investigate the truth of this claim, you conducted a random check of 5,000 vehicles manufactured by the factory. You later determine that the sample population includes 4,850 cars without any initial defects. How can we determine the likelihood that our claim is true in Excel?

We can perform a one-proportion Z-test in Excel by finding the test statistic given the hypothesized proportion and the actual sample proportion.

The t-statistic tells us how far a given value is from the hypothesized value. We’ll later compute the corresponding p-value to determine whether we should accept or reject our earlier claim.

Now that we know when to use a one-proportion Z-test in Excel, let’s learn how to set it up on an actual sample spreadsheet.

The following section provides two examples of a one-proportion Z-test. We will also explain the formulas and tools used in these examples.

In the first example, we want to prove that a given sample has a positivity rate of 85%. When we took a random sample with a sample size of 50, we found 40 positive observations.

First, we divided the frequency of positive observations by the sample size to get a sample proportion of 0.8 or 80%.

Next, we computed the test statistic and p-values to see if there is a statistically significant difference between the observed sample proportion and the hypothesis.

We can find the test statistic using the following formula:

We can find the p-value with this formula:

The first argument of the NORM.S.DIST function must be the result of our test statistic formula. Since the p-value is not less than 0.05, we can say that we do not have sufficient evidence yet to reject the claim that our population has a positivity rate of 85%.

Do you want to take a closer look at our examples? You can make your own copy of the spreadsheet above using the link attached below. 

Use our sample spreadsheet to test out how the p-value and test statistic changes given different results.

If you’re ready to try performing your own one-proportion Z-test, head over to the next section to read our step-by-step breakdown on how to do it!

This section will guide you through each you need to perform a one-proportion Z-test in Excel.

We’ll first learn how to find the sample proportion that we’ll compare with the overall population proportion. Next, we’ll use a custom formula to find the test statistic given the sample results. We will then find the p-value of our measurements using the NORM.S.DIST function.

Follow these steps to perform the one-proportion Z-test in Excel:

Here are some frequently asked questions about the one-proportion Z-test:

• When is it appropriate to use the one-proportion Z-test? We can use this test if we want to know the difference between two categorical variables. The categorical variable must only have two options. For example, we can have a variable labeled ‘clicked_on_ad’ that can either be ‘Yes’ or ‘No’

This step-by-step guide is a quick introduction to performing a one-proportion Z-test in Microsoft Excel.

You may use the one-proportion Z-test to accept or reject a hypothesis about a given population proportion.

The Z-test is just one example of the many statistical methods you can use in your spreadsheets. Our website offers hundreds of other functions and methods to help you get more out of Microsoft Excel.

With so many other Excel functions available, you can find one appropriate for your use case.

Don’t miss out on our team’s new spreadsheet tips, tricks, and best practices. Subscribe to our newsletter to stay updated on the latest guides from us!

Get emails from us about Excel.

Our goal this year is to create lots of rich, bite-sized tutorials for Excel users like you. If you liked this one, you'd love what we are working on! Readers receive ✨ early access ✨ to new content.

• Microsoft Excel

Deion Menor

Deion is a Python Automation Developer with a passion for data analytics. He uses both Google Sheets and Excel daily to empower his technical teams to make data-driven decisions. GitHub LinkedIn

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

How to Perform a Sign Test in Excel

How to perform cubic regression in excel, you may also like.

How to Group by Month Using PivotTable in Excel

• July 5, 2022

How to Display References Dialog Box in Excel

• October 25, 2022

How to Make Employees Schedule in Excel

• August 17, 2023

How to Delete Calculated Field in Pivot Table in Excel

• by Kaith Etcuban
• December 4, 2022

How to Trace Precedents and Dependents in Excel

• September 9, 2022

How to Use Wildcard in FILTER Function in Excel

• November 20, 2022

The Complete Guide: Hypothesis Testing in Excel

In statistics, a hypothesis test is used to test some assumption about a population parameter .

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

This tutorial explains how to perform the following types of hypothesis tests in Excel:

• One sample t-test
• Two sample t-test
• Paired samples t-test
• One proportion z-test
• Two proportion z-test

Let’s jump in!

Example 1: One Sample t-test in Excel

A one sample t-test is used to test whether or not the mean of a population is equal to some value.

For example, suppose a botanist wants to know if the mean height of a certain species of plant is equal to 15 inches.

To test this, she collects a random sample of 12 plants and records each of their heights in inches.

She would write the hypotheses for this particular one sample t-test as follows:

• H 0 :  µ = 15
• H A :  µ ≠15

Refer to this tutorial for a step-by-step explanation of how to perform this hypothesis test in Excel.

Example 2: Two Sample t-test in Excel

A two sample t-test is used to test whether or not the means of two populations are equal.

For example, suppose researchers want to know whether or not two different species of plants have the same mean height.

To test this, they collect a random sample of 20 plants from each species and measure their heights.

The researchers would write the hypotheses for this particular two sample t-test as follows:

• H 0 :  µ 1 = µ 2
• H A :  µ 1 ≠ µ 2

Example 3: Paired Samples t-test in Excel

A paired samples t-test is used to compare the means of two samples when each observation in one sample can be paired with an observation in the other sample.

For example, suppose we want to know whether a certain study program significantly impacts student performance on a particular exam.

To test this, we have 20 students in a class take a pre-test. Then, we have each of the students participate in the study program for two weeks. Then, the students retake a post-test of similar difficulty.

We would write the hypotheses for this particular two sample t-test as follows:

• H 0 :  µ pre = µ post
• H A :  µ pre ≠ µ post

Example 4: One Proportion z-test in Excel

A  one proportion z-test  is used to compare an observed proportion to a theoretical one.

For example, suppose a phone company claims that 90% of its customers are satisfied with their service.

To test this claim, an independent researcher gathered a simple random sample of 200 customers and asked them if they are satisfied with their service.

• H 0 : p = 0.90
• H A : p ≠ 0.90

Example 5: Two Proportion z-test in Excel

A two proportion z-test is used to test for a difference between two population proportions.

For example, suppose a s uperintendent of a school district claims that the percentage of students who prefer chocolate milk over regular milk in school cafeterias is the same for school 1 and school 2.

To test this claim, an independent researcher obtains a simple random sample of 100 students from each school and surveys them about their preferences.

• H 0 : p 1 = p 2
• H A : p 1  ≠ p 2

How to Change Axis Scales in Google Sheets Plots

Statistics vs. analytics: what’s the difference, related posts, how to create a stem-and-leaf plot in spss, how to create a correlation matrix in spss, how to convert date of birth to age..., excel: how to highlight entire row based on..., how to add target line to graph in..., excel: how to use if function with negative..., excel: how to use if function with text..., excel: how to use greater than or equal..., excel: how to use if function with multiple..., how to extract number from string in pandas.

One proportion test in Excel tutorial

This tutorial shows how to test the difference between an observed proportion and a theoretical one, using the one proportion test , in Excel with XLSTAT.

Dataset for comparing one proportion to a value

The data are the results of the launch of a coin 30 times. We would like to assess if the coin is biased or not so we want to compare the results to the proportion 0.5 so half of the time a tail, half of the time a head.

Setting up a test for comparing one proportion to a value

Frequency: 19

Sample size: 30

Test proportion: 0.5

Data format: Frequency

When ready, click on OK .

Results of a test for comparing one proportion to a value

Was this article useful?

Similar articles

• Compare two proportions in Excel tutorial
• Student's t test on two independent samples tutorial
• Compare ≥ 2 samples described by several variables
• Compare k proportions in Excel tutorial
• One sample t-test or z-test in Excel tutorial
• One sample variance test in Excel tutorial

Expert Software for Better Insights, Research, and Outcomes

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

8.4: Hypothesis Test Examples for Proportions

• Last updated
• Save as PDF
• Page ID 11533

• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he 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 the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

Full Hypothesis Test Examples

Example \(\PageIndex{7}\)

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.

Set up the hypothesis test:

The 1% level of significance means that α = 0.01. This is a test of a single population proportion .

\(H_{0}: p = 0.50\)  \(H_{a}: p \neq 0.50\)

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: \(P′ =\) the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion.

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber \]

\[P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber \]

where \(p = 0.50, q = 1−p = 0.50\), and \(n = 100\)

Calculate the p -value using the normal distribution for proportions:

\[p\text{-value} = P(p′ < 0.47 or p′ > 0.53) = 0.5485\nonumber \]

where \[x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber \].

Interpretation of the \(p\text{-value})\: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \(p'\) is 0.53 or more OR 0.47 or less (see the graph in Figure).

\(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis.

\(p′ = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p′\) for the left tail is equal to \(0.50 – 0.03 = 0.47\) where \(\mu = p = 0.50\). (0.03 is the difference between 0.53 and 0.50.)

Compare \(\alpha\) and the \(p\text{-value}\):

Since \(\alpha = 0.01\) and \(p\text{-value} = 0.5485\). \(\alpha < p\text{-value}\).

Make a decision: Since \(\alpha < p\text{-value}\), you cannot reject \(H_{0}\).

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

The \(p\text{-value}\) can easily be calculated.

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for \(p_{0}\), 53 for \(x\) and 100 for \(n\). Arrow down to Prop and arrow to not equals \(p_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the \(p\text{-value}\) (\(p = 0.5485\)) and the test statistic (\(z\)-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(\(z\) = 0.6\) (test statistic) and \(p = 0.5485\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms 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 brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{7}\)

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She 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.

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

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

• \(H_{0} : p = 0.85\)
• \(H_{a}: p \neq 0.85\)
• \(p = 0.7554\)

Because \(p > \alpha\), we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Example \(\PageIndex{8}\)

Suppose a consumer group suspects that the proportion of households that have three 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 cell phones.

Set up the Hypothesis Test:

\(H_{0}: p = 0.30, H_{a}: p \neq 0.30\)

Determine the distribution needed:

The random variable is \(P′ =\) proportion of households that have three cell phones.

The distribution for the hypothesis test is \(P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)\)

Exercise 9.6.8.2

a. The value that helps determine the \(p\text{-value}\) is \(p′\). Calculate \(p′\).

a. \(p' = \frac{x}{n}\) where \(x\) is the number of successes and \(n\) is the total number in the sample.

\(x = 43, n = 150\)

\(p′ = 43150\)

Exercise 9.6.8.3

b. What is a success for this problem?

b. A success is having three cell phones in a household.

Exercise 9.6.8.4

c. What is the level of significance?

c. The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume that \(\alpha = 0.05\).

Exercise 9.6.8.5

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.

Calculate the \(p\text{-value}\).

d. \(p\text{-value} = 0.7216\)

Exercise 9.6.8.6

e. Make a decision. _____________(Reject/Do not reject) \(H_{0}\) because____________.

e. Assuming that \(\alpha = 0.05, \alpha < p\text{-value}\). The decision is do not reject \(H_{0}\) because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

Exercise \(\PageIndex{8}\)

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.

• \(H_{0}: p = 0.92\)
• \(H_{a}: p < 0.92\)
• \(p\text{-value} = 0.0046\)

Because \(p < 0.05\), we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter \(p\). The distribution for the test is normal. The estimated proportion \(p′\) is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived \(\alpha = 0.01\), for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

Example \(\PageIndex{9}\)

My dog has so many fleas,

They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased.

I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe.

A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas.

I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42.

At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased.

Now it is time for you to have some fun With the level of significance being .01, You must help me figure out

Use the new shampoo or go without?

\(H_{0}: p \leq 0.25\)   \(H_{a}: p > 0.25\)

In words, CLEARLY state what your random variable \(\bar{X}\) or \(P′\) represents.

\(P′ =\) The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

\[N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber \]

Test Statistic: \(z = 2.3163\)

Calculate the \(p\text{-value}\) using the normal distribution for proportions:

\[p\text{-value} = 0.0103\nonumber \]

In one to two complete sentences, explain what the p -value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \(\left(\frac{17}{42}\right)\) or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the \(p\text{-value}\).

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.

This test result is not very definitive since the \(p\text{-value}\) is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.

Example \(\PageIndex{11}\)

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.

We will follow the four-step process.

• \(H_{0}: p \leq 0.00034\)
• \(H_{a}: p > 0.00034\)

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.

Figure 9.6.11.

Figure 9.6.12.

• Since the \(p\text{-value} = 0.0073\) is greater than our alpha value \(= 0.005\), we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Example \(\PageIndex{12}\)

According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.

We will follow the four-step plan.

• We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
• \(H_{0}: p = 0.00078\)
• \(H_{a}: p \neq 0.00078\)

Figure 9.6.13.

Figure 9.6.14.

• Since the \(p\text{-value}\), \(p = 0.00063\), is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

8.8 Hypothesis Tests for a Population Proportion

Learning objectives.

• Conduct and interpret hypothesis tests for a population proportion.

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%.

Suppose the hypotheses for a hypothesis test are:

[latex]\begin{eqnarray*} H_0: & & p=20 \% \\ H_a: & & p \gt 20\% \end{eqnarray*}[/latex]

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

[latex]\begin{eqnarray*} H_0: & & p=50 \% \\ H_a: & & p \neq  50\% \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\neq[/latex], this is a two-tail 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: & & p=10\% \\ H_a: & & p \lt  10\% \end{eqnarray*}[/latex]

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

Steps to Conduct a Hypothesis Test for a Population Proportion

• Write down the null and alternative hypotheses in terms of the population proportion [latex]p[/latex].  Include appropriate units with the values of the proportion.
• 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.
• If [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p) \geq 5[/latex], use the normal distribution with [latex]\displaystyle{z=\frac{\hat{p}-p}{\sqrt{\frac{p \times (1-p)}{n}}}}[/latex].
• If one of [latex]n \times p \lt 5[/latex] or [latex]n \times (1-p) \lt 5[/latex], use a binomial 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 PROPORTION

The p -value for a hypothesis test on a population proportion is the area in the tail(s) of distribution of the sample proportion.  If both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p) \geq 5[/latex], use the normal distribution to find the p -value.  If at least one of [latex]n \times p \lt 5[/latex] or [latex]n \times (1-p) \lt 5[/latex], use the binomial distribution to find the p -value.

If both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p) \geq 5[/latex]:

• For x , enter the value for [latex]\hat{p}[/latex].
• For [latex]\mu[/latex] , enter the mean of the sample proportions [latex]p[/latex].  Note:  Because the test is run assuming the null hypothesis is true, the value for [latex]p[/latex] is the claim from the null hypothesis.
• For [latex]\sigma[/latex] , enter the standard error of the proportions [latex]\displaystyle{\sqrt{\frac{p \times (1-p)}{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.

If at least one of [latex]n \times p \lt 5[/latex] or [latex]n \times (1-p) \lt 5[/latex]:

• The p -value is found using the binomial distribution.
• For x , enter the number of successes.
• For n , enter the sample size.
• For p , enter the the value of the population proportion [latex]p[/latex] from the null hypothesis.
• For the logic operator , enter true .  Note:  Because we are calculating an at most probability, the logic operator is always true.
• For p , enter the the value of the population proportion [latex]p[/latex] in the null hypothesis.
• For the logic operator , enter true .  Note:  Because we are calculating an at least probability, the logic operator is always true.

Marketers believe that 92% of adults own a cell phone.  A cell phone manufacturer believes that number is actually lower.  In a sample of 200 adults, 87% own a cell phone.  At the 1% significance level, determine if the proportion of adults that own a cell phone is lower than the marketers’ claim.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & p=92\% \mbox{ of adults own a cell phone} \\ H_a: & & p \lt 92\% \mbox{ of adults own a cell phone} \end{eqnarray*}[/latex]

From the question, we have [latex]n=200[/latex], [latex]\hat{p}=0.87[/latex], and [latex]\alpha=0.01[/latex].

To determine the distribution, we check [latex]n \times p[/latex] and [latex]n \times (1-p)[/latex].  For the value of [latex]p[/latex], we use the claim from the null hypothesis ([latex]p=0.92[/latex]).

[latex]\begin{eqnarray*} n \times p & = & 200 \times 0.92=184 \geq 5 \\ n \times (1-p) & = & 200 \times (1-0.92)=16 \geq 5\end{eqnarray*}[/latex]

Because both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p)  \geq 5[/latex] 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]=0.0046[/latex].

Conclusion:

Because p -value[latex]=0.0046 \lt 0.01=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 1% significance level there is enough evidence to suggest that the proportion of adults who own a cell phone is lower than 92%.

• The null hypothesis [latex]p=92\%[/latex] is the claim that 92% of adults own a cell phone.
• The alternative hypothesis [latex]p \lt 92\%[/latex] is the claim that less than 92% of adults own a cell phone.
• 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]\hat{p}[/latex].
• Field 2 is the value of [latex]p[/latex] from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume [latex]p=0.92[/latex].
• Field 3 is the standard deviation for the sample proportions [latex]\displaystyle{\sqrt{\frac{p \times (1-p)}{n}}}[/latex].
• The p -value of 0.0046 tells us that under the assumption that 92% of adults own a cell phone (the null hypothesis), there is only a 0.46% chance that the proportion of adults who own a cell phone in a sample of 200 is 87% 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.  In other words, the proportion of adults who own a cell phone is most likely less than 92%.

A consumer group claims that the proportion of households that have at least three cell phones is 30%.  A cell phone company has reason to believe that the proportion of households with at least three cell phones is much higher.  Before they start a big advertising campaign based on the proportion of households that have at least three cell phones, they want to test their claim.  Their marketing people survey 150 households with the result that 54 of the households have at least three cell phones.  At the 1% significance level, determine if the proportion of households that have at least three cell phones is less than 30%.

[latex]\begin{eqnarray*} H_0: & & p=30\% \mbox{ of household have at least 3 cell phones} \\ H_a: & & p \gt 30\% \mbox{ of household have at least 3 cell phones} \end{eqnarray*}[/latex]

From the question, we have [latex]n=150[/latex], [latex]\displaystyle{\hat{p}=\frac{54}{150}=0.36}[/latex], and [latex]\alpha=0.01[/latex].

To determine the distribution, we check [latex]n \times p[/latex] and [latex]n \times (1-p)[/latex].  For the value of [latex]p[/latex], we use the claim from the null hypothesis ([latex]p=0.3[/latex]).

[latex]\begin{eqnarray*} n \times p & = & 150 \times 0.3=45 \geq 5 \\ n \times (1-p) & = & 150 \times (1-0.3)=105 \geq 5\end{eqnarray*}[/latex]

Because both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p)  \geq  5[/latex] 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]=0.0544[/latex].

Because p -value[latex]=0.0544 \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 proportion of households with at least three cell phones is more than 30%.

• The null hypothesis [latex]p=30\%[/latex] is the claim that 30% of households have at least three cell phones.
• The alternative hypothesis [latex]p \gt 30\%[/latex] is the claim that more than 30% of households have at least three cell phones.
• The function is 1-norm.dist because we are finding the area in the right tail of a normal distribution.
• Field 2 is the value of [latex]p[/latex] from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume [latex]p=0.3[/latex].
• The p -value of 0.0544 tells us that under the assumption that 30% of households have at least three cell phones (the null hypothesis), there is a 5.44% chance that the proportion of households with at least three cell phones in a sample of 150 is 36% or more.  Compared to the 1% significance level, this is a large probability, 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 30% of households have at least three cell phones is most likely correct.

A teacher believes that 70% of students in the class will want to go on a field trip to the local zoo.  The students in the class believe the proportion is much higher and ask the teacher to verify her claim.  The teacher samples 50 students and 39 reply that they would want to go to the zoo.  At the 5% significance level, determine if the proportion of students who want to go on the field trip is higher than 70%.

[latex]\begin{eqnarray*} H_0: & & p = 70\% \mbox{ of students want to go on the field trip}  \\ H_a: & & p \gt 70\% \mbox{ of students want to go on the field trip}   \end{eqnarray*}[/latex]

From the question, we have [latex]n=50[/latex], [latex]\displaystyle{\hat{p}=\frac{39}{50}=0.78}[/latex], and [latex]\alpha=0.05[/latex].

[latex]\begin{eqnarray*} n \times p & = & 50 \times 0.7=35 \geq 5 \\ n \times (1-p) & = & 50 \times (1-0.7)=15 \geq 5\end{eqnarray*}[/latex]

Because both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p)  \geq 5[/latex] 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]=0.1085[/latex].

Because p -value[latex]=0.1085 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that the proportion of students who want to go on the field trip is higher than 70%.

• The null hypothesis [latex]p=70\%[/latex] is the claim that 70% of the students want to go on the field trip.
• The alternative hypothesis [latex]p \gt 70\%[/latex] is the claim that more than 70% of students want to go on the field trip.
• The p -value of 0.1085 tells us that under the assumption that 70% of students want to go on the field trip (the null hypothesis), there is a 10.85% chance that the proportion of students who want to go on the field trip in a sample of 50 students is 78% or more.  Compared to the 5% significance level, this is a large probability, 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 teacher’s claim that 70% of students want to go on the field trip is most likely correct.

Joan believes that 50% of first-time brides in the United States are younger than their grooms.  She performs a hypothesis test to determine if the percentage is the same or different from 50%.  Joan samples 100 first-time brides and 56 reply that they are younger than their grooms.  Use a 5% significance level.

[latex]\begin{eqnarray*} H_0: & & p=50\% \mbox{ of first-time brides are younger than the groom} \\ H_a: & & p \neq 50\% \mbox{ of first-time brides are younger than the groom} \end{eqnarray*}[/latex]

From the question, we have [latex]n=100[/latex], [latex]\displaystyle{\hat{p}=\frac{56}{100}=0.56}[/latex], and [latex]\alpha=0.05[/latex].

To determine the distribution, we check [latex]n \times p[/latex] and [latex]n \times (1-p)[/latex].  For the value of [latex]p[/latex], we use the claim from the null hypothesis ([latex]p=0.5[/latex]).

[latex]\begin{eqnarray*} n \times p & = & 100 \times 0.5=50 \geq 5 \\ n \times (1-p) & = & 100 \times (1-0.5)=50 \geq 5\end{eqnarray*}[/latex]

Because both [latex]n \times p \geq 5[/latex] and [latex]n \times (1-p)  \geq 5[/latex] 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 or the right.  We need to know if the sample relates to the left 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 proportion [latex]\hat{p}=0.56[/latex] is greater than the value of the population proportion in the null hypothesis [latex]p=0.5[/latex] ([latex]\hat{p}=0.56>0.5=p[/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 0.1151 and  [latex]\frac{1}{2}[/latex]( p -value)[latex]=0.1151[/latex].  This is also the area in the left tail, so

p -value[latex]=0.1151+0.1151=0.2302[/latex]

Because p -value[latex]=0.2302 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that the proportion of first-time brides that are younger than the groom is different from 50%.

• The null hypothesis [latex]p=50\%[/latex] is the claim that the proportion of first-time brides that are younger than the groom is 50%.
• The alternative hypothesis [latex]p \neq 50\%[/latex] is the claim that the proportion of first-time brides that are younger than the groom is different from 50%.
• We use norm.dist([latex]\hat{p}[/latex],[latex]p[/latex],[latex]\mbox{sqrt}(p*(1-p)/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]\hat{p}[/latex],[latex]p[/latex],[latex]\mbox{sqrt}(p*(1-p)/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 0.2302  is a large probability compared to the 5% 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 proportion of first-time brides who are younger than the groom is most likely correct.

Watch this video: Hypothesis Testing for Proportions: z -test by ExcelIsFun [7:27] 

An online retailer believes that 93% of the visitors to its website will make a purchase.   A researcher in the marketing department thinks the actual percent is lower than claimed.  The researcher examines a sample of 50 visits to the website and finds that 45 of the visits resulted in a purchase.  At the 1% significance level, determine if the proportion of visits to the website that result in a purchase is lower than claimed.

[latex]\begin{eqnarray*} H_0: & & p=93\% \mbox{ of visitors make a purchase} \\ H_a: & & p \lt 93\% \mbox{ of visitors make a purchase} \end{eqnarray*}[/latex]

From the question, we have [latex]n=50[/latex], [latex]x=45[/latex], and [latex]\alpha=0.01[/latex].

To determine the distribution, we check [latex]n \times p[/latex] and [latex]n \times (1-p)[/latex].  For the value of [latex]p[/latex], we use the claim from the null hypothesis ([latex]p=0.93[/latex]).

[latex]\begin{eqnarray*} n \times p & = & 50 \times 0.93=46.5 \geq 5 \\ n \times (1-p) & = & 50 \times (1-0.93)=3.5 \lt 5\end{eqnarray*}[/latex]

Because [latex]n \times (1-p)  \lt 5[/latex] we use a binomial distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the probability of getting at most 45 successes in 50 trials.

So the p -value[latex]=0.2710[/latex].

Because p -value[latex]=0.2710 \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 proportion of visitors who make a purchase is lower than 93%.

• The null hypothesis [latex]p=93\%[/latex] is the claim that 93% of visitors to the website make a purchase.
• The alternative hypothesis [latex]p \lt 93\%[/latex] is the claim that less than 93% of visitors to the website make a purchase.
• The function is binom.dist because we are finding the probability of at most 45 successes.
• Field 1 is the number of successes [latex]x[/latex].
• Field 2 is the sample size [latex]n[/latex].
• Field 3 is the probability of success [latex]p[/latex].  This is the claim about the population proportion made in the null hypothesis, so that means we assume [latex]p=0.93[/latex].
• The p -value of 0.2710 tells us that under the assumption that 93% of visitors make a purchase (the null hypothesis), there is a 27.10% chance that the number of visitors in a sample of 50 who make a purchase is 45 or less.  This is a large probability compared to the 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 proportion of visitors to the website who make a purchase adults is most likely 93%.

A drug company claims that only 4% of people who take their new drug experience any side effects from the drug.  A researcher believes that the percent is higher than drug company’s claim.  The researcher takes a sample of 80 people who take the drug and finds that 10% of the people in the sample experience side effects from the drug.  At the 5% significance level, determine if the proportion of people who experience side effects from taking the drug is higher than claimed.

[latex]\begin{eqnarray*} H_0: & & p=4\% \mbox{ of people experience side effects} \\ H_a: & & p \gt 4\% \mbox{ of people experience side effects} \end{eqnarray*}[/latex]

From the question, we have [latex]n=80[/latex], [latex]\hat{p}=0.1[/latex], and [latex]\alpha=0.05[/latex].

To determine the distribution, we check [latex]n \times p[/latex] and [latex]n \times (1-p)[/latex].  For the value of [latex]p[/latex], we use the claim from the null hypothesis ([latex]p=0.04[/latex]).

[latex]\begin{eqnarray*} n \times p & = & 80 \times 0.04=3.2 \lt 5\end{eqnarray*}[/latex]

Because [latex]n \times p  \lt 5[/latex] we use a binomial distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the probability of getting at least 8 successes in 80 trials.  (Note:  In the sample of size 80, 10% have the characteristic of interest, so this means that [latex]80 \times 0.1=8[/latex] people in the sample have the characteristic of interest.)

So the p -value[latex]=0.0147[/latex].

Because p -value[latex]=0.0147 \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 the proportion of people who experience side effects from taking the drug is higher than 4%.

• The null hypothesis [latex]p=4\%[/latex] is the claim that 4% of the people experience side effects from taking the drug.
• The alternative hypothesis [latex]p \gt 4\%[/latex] is the claim that more than 4% of the people experience side effects from taking the drug.
• The function is 1-binom.dist because we are finding the probability of at least 8 successes.
• Field 1 is [latex]x-1[/latex] where [latex]x[/latex] is the number of successes.  In this case, we are using the compliment rule to change the probability of at least 8 successes into 1 minus the probability of at most 7 successes.
• Field 3 is the probability of success [latex]p[/latex].  This is the claim about the population proportion made in the null hypothesis, so that means we assume [latex]p=0.04[/latex].
• The p -value of 0.0147 tells us that under the assumption that 4% of people experience side effects (the null hypothesis), there is a 1.47% chance that the number of people in a sample of 80 who experience side effects is 8 or more.  This is a small probability compared to the significance level, 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.  In other words, the proportion of people who experience side effects is most likely greater than 4%.

Concept Review

The hypothesis test for a population proportion is a well-established process:

• Find the p -value (the area in the corresponding tail) for the test using the appropriate distribution (normal or binomial).
• 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.

Statistics Made Easy

How to Perform One Sample & Two Sample Z-Tests in Excel

A one sample z-test is used to test whether a population mean is significantly different than some hypothesized value.

A two sample z-test is used to test whether two population means are significantly different from each other.

The following examples show how to perform each type of test in Excel.

Example 1: One Sample Z-Test in Excel

Suppose the IQ in a population is normally distributed with a mean of μ = 100 and standard deviation of σ = 15.

A scientist wants to know if a new medication affects IQ levels, so she recruits 20 patients to use it for one month and records their IQ levels at the end of the month.

We can use the following formula in Excel to perform a one sample z-test to determine if the new medication causes a significant difference in IQ levels:

The following screenshot shows how to use this formula in practice:

The one-tailed p-value is 0.181587 . Since we’re performing a two-tailed test, we can multiply this value by 2 to get p = 0.363174 .

Since this p-value is not less than .05, we do not have sufficient evidence to reject the null hypothesis.

Thus, we conclude that the new medication does not significantly affect IQ level.

Example 2: Two Sample Z-Test in Excel

Suppose the IQ levels among individuals in two different cities are known to be normally distributed each with population standard deviations of 15.

A scientist wants to know if the mean IQ level between individuals in city A and city B are different, so she selects a simple random sample of  20 individuals from each city and records their IQ levels.

The following screenshot shows the IQ levels for the individuals in each sample:

To perform a two sample z-test to determine if the mean IQ level is different between the two cities, click the Data tab along the top ribbon, then click the Data Analysis button within the Analysis  group.

If you don’t see Data Analysis as an option, you need to first load the Analysis ToolPak in Excel.

Once you click this button, select z-Test: Two Sample for Means in the new window that appears:

Once you click OK , you can fill in the following information:

Once you click OK , the results will appear in cell E1:

The test statistic for the two sample z-test is -1.71817 and the corresponding p-value is .085765.

Thus, we conclude that the mean IQ level is not significantly different between the two cities.

Additional Resources

The following tutorials explain how to perform other common statistical tests in Excel:

How to Conduct a One Sample t-Test in Excel How to Conduct a Two Sample t-Test in Excel How to Conduct a Paired Samples t-Test in Excel How to Perform Welch’s t-Test in Excel

Hey there. My name is Zach Bobbitt. I have a Master 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.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game New
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics
• Probability and Statistics

How to Perform Hypothesis Testing for a Proportion

Last Updated: July 31, 2023

This article was co-authored by Joseph Quinones . Joseph Quinones is a High School Physics Teacher working at South Bronx Community Charter High School. Joseph specializes in astronomy and astrophysics and is interested in science education and science outreach, currently practicing ways to make physics accessible to more students with the goal of bringing more students of color into the STEM fields. He has experience working on Astrophysics research projects at the Museum of Natural History (AMNH). Joseph recieved his Bachelor's degree in Physics from Lehman College and his Masters in Physics Education from City College of New York (CCNY). He is also a member of a network called New York City Men Teach. This article has been viewed 34,253 times.

Hypothesis testing for a proportion is used to determine if a sampled proportion is significantly different from a specified population proportion. For example, if you expect the proportion of male births to be 50 percent, but the actual proportion of male births is 53 percent in a sample of 1000 births. Is this significantly different from the hypothesized population parameter? To find out, follow these steps.

• Are there more than 50 percent of Americans who self-identify as liberal?
• Is the percentage of defects in a given manufacturing plant more than 5%?
• Is the proportion of babies born male different from 50 percent?
• Are there more Americans who self-identify as liberal than as conservative? (Use hypothesis testing for 2 proportions instead.)
• Is the mean number of defects in a given manufacturing plant more than 50 per month? (Use hypothesis testing for one sample t-test instead.)
• Are male births related to paternal age? (Use chi-square test for independence instead.)

• Simple random sampling is used.
• Each sample point can result in only one of two possible outcomes. These outcomes are called successes and failures.
• The sample includes at least 10 successes and 10 failures.
• The population size is at least 20 times as big as the sample size.

• Right-tailed: Research question: Is the sample proportion greater than the hypothesized population proportion? Your hypotheses would be stated as follows: H0: p<=p0; Ha: p>p0.
• Left-tailed: Research question: Is the sample proportion less than the hypothesized population proportion? Your hypotheses would be stated as follows: H0: p>=p0; Ha: p<p0.
• Two-tailed: Research question: Is the sample proportion different from the hypothesized population proportion? Your hypotheses would be stated as follows: H0: p=p0; Ha: p<>p0.
• In your example, you can use a two-tailed test to see if the sample proportion of male births, 0.53, is different from the hypothesized population proportion of 0.50. So H0: p=0.50; Ha: p<>0.50. Typically, if there is no a priori reason to believe that any differences must be unidirectional, the two-tailed test is preferred as it is a more stringent test.

• In our example, p=0.53, p0=0.50, and n=1000. s = sqrt(0.50*(1-0.50)/1000) = 0.0158. the test statistic is z = (0.53-0.50)/0.0158 = 1.8974.

• Normal distribution probability z table. It is important to read the table description to note what probability is listed by the table. Some tables list cumulative (left side) area, others list right tail area, still others list only area from mean up to a positive z value.
• Excel. The excel function =norm.s.dist(z,cumulative). Substitute the numeric value for z and "true" for cumulative. This excel formula gives cumulative area to the left of a given z value. For your example, you would use the formula =norm.s.dist(1.8974,true) to find the cumulative left side area, which includes the left tail and the body. (Body is the area from -z to z.) You can subtract this from 1 to find the right tail area. Since your example is 2-tailed, you would then multiply by 2. A formula for p can be =2*(1-norm.s.dist(1.8974,true)). The output is 0.0578.
• Texas Instrument calculator, such as TI-83 or TI-84.
• Online normal distribution calculators.

Expert Q&A

You Might Also Like

Expert Interview

Thanks for reading our article! If you’d like to learn more about teaching, check out our in-depth interview with Joseph Quinones .

• ↑ http://stattrek.com/hypothesis-test/proportion.aspx?tutorial=ap
• ↑ http://blog.minitab.com/blog/michelle-paret/alphas-p-values-confidence-intervals-oh-my

About This Article

• Send fan mail to authors

Reader Success Stories

Shanaya Malik

Jun 29, 2019

Did this article help you?

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

Six Sigma & SPC Excel Add-in

• Questions? Contact Us
• 888-468-1537

Statistical Analysis in QI Macros

Statistics wizard, data normality, hypothesis tests, test of means, equivalence tests, test of variances, test for outliers, test of proportion, test relationship, non-parametric tests.

Hypothesis Testing Cheat Sheet

Knowledge Base | Online User Guide

• Free 30-Day Trial
• Powerful SPC Software for Excel
• SPC - Smart Performance Charts
• Who Uses QI Macros?
• What Do Our Customers Say?
• QI Macros SPC Software Reviews
• SPC Software Comparison
• Control Chart
• Histogram with Cp Cpk
• Pareto Chart
• Automated Fishbone Diagram
• Gage R&R MSA
• Data Mining Tools
• Statistical Analysis - Hypothesis Testing
• Chart and Stat Wizards
• Lean Six Sigma Excel Templates
• Technical Support - PC & Mac
• QI Macros FAQs
• Upgrade History
• Submit Enhancement Request
• Data Analysis Services
• Free QI Macros Webinar
• Free QI Macros Video Tutorials
• How to Setup Excel for QI Macros
• Free Healthcare Data Analytics Course
• Free Lean Six Sigma Webinars
• Animated Lean Six Sigma Video Tutorials
• Free Agile Lean Six Sigma Trainer Training
• Free White Belt Training
• Free Yellow Belt Training
• Free Green Belt Training
• QI Macros Resources
• QI Macros Knowledge Base | User Guide
• Excel Tips and Tricks
• Lean Six Sigma Resources
• QI Macros Monthly Newsletter
• Improvement Insights Blog
• Buy QI Macros
• Quantity Discounts and W9
• Hassle Free Guarantee

QI Macros Reviews CNET Five Star Review Industry Leaders Our Customers

Home » Statistical Analysis Excel » Test of a Proportion

Looking for a Test of Proportion Template in Excel?

Qi macros has a ready made test of proportions template for you..

Use a test of proportions template to compute a confidence interval and perform a hypothesis test of the proportion.

Example of Test of Proportion Test Using QI Macros Template

Let's say a manufacturer claims that his products are less than 3% defective. You can take a sample of the products and determine whether or not the actual percent defective is consistent with his claim.

• If the sample data isn't summarized, use PivotTables to summarize the trials and successes

• Enter desired proportion in A3. The number should be between 0 and 1. (e.g., 3% is 0.03)
• Enter number of trials in B3 (100 products)
• Enter number of successes in C3 (e.g., defects in 100 products)
• Enter confidence Level in E1 (.95 = 95%)
• If Cell H3:H5 (or J3:J5) is green: Cannot reject the null hypothesis (accept the null hypothesis)
• If Cell H3:H5 is red: Reject the null hypothesis

In this example with five defects in 100 samples, you cannot reject the null hypothesis (accept the null hypothesis).

QI Macros also contains a Test of Two Proportions template.

Test of a proportion are one of many statistical tests included in QI Macros add-in for Excel.

QI Macros adds a new tab to Excel's menu, making it easy to find any tool you need. You will find the statistical tools and templates on the far left side of QI Macros menu.

QI Macros draws these SPC charts as well

• SPC Software for Excel
• Free 30 Day Trial
• On-line Tech Support
• QI Macros Reviews
• Free QI Macros Training
• Privacy Policy

KnowWare International, Inc. 2696 S. Colorado Blvd., Ste. 555 Denver, CO 80222 USA Toll-Free: 1-888-468-1537 Local: (303) 756-9144

How to Do Hypothesis Tests With the Z.TEST Function in Excel

• Statistics Tutorials
• Probability & Games
• Descriptive Statistics
• Inferential Statistics
• Applications Of Statistics
• Math Tutorials
• Pre Algebra & Algebra
• Exponential Decay
• Worksheets By Grade
• Ph.D., Mathematics, Purdue University
• M.S., Mathematics, Purdue University
• B.A., Mathematics, Physics, and Chemistry, Anderson University

Hypothesis tests are one of the major topics in the area of inferential statistics. There are multiple steps to conduct a hypothesis test and many of these require statistical calculations. Statistical software, such as Excel, can be used to perform hypothesis tests. We will see how the Excel function Z.TEST tests hypotheses about an unknown population mean.

Conditions and Assumptions

We begin by stating the assumptions and conditions for this type of hypothesis test. For inference about the mean we must have the following simple conditions:

• The sample is a simple random sample .
• The sample is small in size relative to the population . Typically this means that the population size is more than 20 times the size of the sample.
• The variable being studied is normally distributed.
• The population standard deviation is known.
• The population mean is unknown.

All of these conditions are unlikely to be met in practice. However, these simple conditions and the corresponding hypothesis test are sometimes encountered early in a statistics class. After learning the process of a hypothesis test, these conditions are relaxed in order to work in a more realistic setting.

Structure of the Hypothesis Test

The particular hypothesis test we consider has the following form:

• State the null and alternative hypotheses .
• Calculate the test statistic, which is a z -score.
• Calculate the p-value by using the normal distribution. In this case the p-value is the probability of obtaining at least as extreme as the observed test statistic, assuming the null hypothesis is true.
• Compare the p-value with the level of significance to determine whether to reject or fail to reject the null hypothesis.

We see that steps two and three are computationally intensive compared two steps one and four. The Z.TEST function will perform these calculations for us.

Z.TEST Function

The Z.TEST function does all of the calculations from steps two and three above. It does a majority of the number crunching for our test and returns a p-value. There are three arguments to enter into the function, each of which is separated by a comma. The following explains the three types of arguments for this function.

• The first argument for this function is an array of sample data. We must enter a range of cells that corresponds to the location of the sample data in our spreadsheet.
• The second argument is the value of μ that we are testing in our hypotheses. So if our null hypothesis is H 0 : μ = 5, then we would enter a 5 for the second argument.
• The third argument is the value of the known population standard deviation. Excel treats this as an optional argument

Notes and Warnings

There are a few things that should be noted about this function:

• The p-value that is output from the function is one-sided. If we are conducting a two-sided test, then this value must be doubled.
• The one-sided p-value output from the function assumes that the sample mean is greater than the value of μ we are testing against. If the sample mean is less than the value of the second argument, then we must subtract the output of the function from 1 to get the true p-value of our test.
• The final argument for the population standard deviation is optional. If this is not entered, then this value is automatically replaced in Excel’s calculations by the sample standard deviation. When this is done, theoretically a t-test should be used instead.

We suppose that the following data are from a simple random sample of a normally distributed population of unknown mean and standard deviation of 3:

1, 2, 3, 3, 4, 4, 8, 10, 12

With a 10% level of significance we wish to test the hypothesis that the sample data are from a population with mean greater than 5. More formally, we have the following hypotheses:

• H 0 : μ= 5
• H a : μ > 5

We use Z.TEST in Excel to find the p-value for this hypothesis test.

• Enter the data into a column in Excel. Suppose this is from cell A1 to A9
• Into another cell enter =Z.TEST(A1:A9,5,3)
• The result is 0.41207.
• Since our p-value exceeds 10%, we fail to reject the null hypothesis.

The Z.TEST function can be used for lower tailed tests and two tailed tests as well. However the result is not as automatic as it was in this case. Please see here for other examples of using this function.

• What Is a P-Value?
• Example of Two Sample T Test and Confidence Interval
• Hypothesis Test Example
• Hypothesis Test for the Difference of Two Population Proportions
• An Example of a Hypothesis Test
• Functions with the T-Distribution in Excel
• The Runs Test for Random Sequences
• How to Conduct a Hypothesis Test
• How to Use the NORM.INV Function in Excel
• Chi-Square Goodness of Fit Test
• What Is the Difference Between Alpha and P-Values?
• What Level of Alpha Determines Statistical Significance?
• How to Find Degrees of Freedom in Statistics
• Robustness in Statistics
• How to Use the STDEV.S Function in Excel
• Example of a Chi-Square Goodness of Fit Test