8.1.2.2 - Minitab: Hypothesis Tests for One Proportion

8.1.2.2 - Minitab: Hypothesis Tests for One Proportion

A hypothesis test for one proportion can be conducted in Minitab. This can be done using raw data or summarized data.

  • If you have a data file with every individual's observation, then you have raw data.
  • If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data.

The next two pages will show you how to use Minitab to conduct this analysis using either raw data or summarized data.

Note that the default method for constructing the sampling distribution in Minitab is to use the exact method.  If \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) then you will need to change this to the normal approximation method.  This must be done manually. Minitab will use the method that you select, it will not check assumptions for you!

8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw Data

8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw Data

If you have data in a Minitab worksheet, then you have what we call "raw data."  This is in contrast to "summarized data" which you'll see on the next page.

In order to use the normal approximation method both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\). Before we can conduct our hypothesis test we must check this assumption to determine if the normal approximation method or exact method should be used. This must be checked manually. Minitab will not check assumptions for you.

In the example below, we want to know if there is evidence that the proportion of students who are male is different from 0.50.

\(n=226\) and \(p_0=0.50\)

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\)

Both \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) so we can use the normal approximation method. 

Minitab®  – Conducting a One Sample Proportion z Test: Raw Data

Research question: Is the proportion of students who are male different from 0.50?

  1. Open Minitab file:
  2. In Minitab, select Stat > Basic Statistics > 1 Proportion
  3. Select One or more samples, each in a column from the dropdown
  4. Double-click the variable Biological Sex to insert it into the box
  5. Check the box next to Perform hypothesis test and enter 0.50 in the Hypothesized proportion box
  6. Select Options
  7. Use the default Alternative hypothesis setting of Proportion ≠ hypothesized proportion value 
  8. Use the default Confidence level of 95
  9. Select Normal approximation method
  10. Click OK and OK

The result should be the following output:

Method

Event: Biological Sex = Male
p: proportion where Biological Sex = Male
Normal approximation is used for this analysis.

Descriptive Statistics
N Event Sample p 95% CI for p
226 99 0.438053 (0.373368, 0.502738)
Test
Null hypothesis H 0: p = 0.5
Alternative hypothesis H 1: p ≠ 0.5
Z-Value P-Value
-1.86 0.063

Summary of Results

We could summarize these results using the five-step hypothesis testing procedure:

1. Check assumptions and write hypotheses

\(np_0 = 226(0.50)=113\) and \(n(1-p_0) = 226(1-0.50)=113\) therefore the normal approximation method will be used.

 \(H_0\colon p = 0.50\)

 \(H_a\colon p \ne 0.50\)

2. Calculate the test statistic

From the Minitab output, \(z\) = -1.86

3. Determine the p-value

From the Minitab output, \(p\) = 0.0625

4. Make a decision

\(p > \alpha\), fail to reject the null hypothesis

5. State a "real world" conclusion

There is NOT enough evidence that the proportion of all students in the population who are male is different from 0.50.

8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary Data

8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary Data

Example: Overweight

The following example uses a scenario in which we want to know if the proportion of college women who think they are overweight is less than 40%. We collect data from a random sample of 129 college women and 37 said that they think they are overweight.

First, we should check assumptions to determine if the normal approximation method or exact method should be used:

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

Minitab®  – Performing a One Proportion z Test with Summarized Data

To perform a one sample proportion z test with summarized data in Minitab:

  1. In Minitab, select Stat > Basic Statistics > 1 Proportion
  2. Select Summarized data from the dropdown
  3. For number of events, add 37 and for number of trials add 129.
  4. Check the box next to Perform hypothesis test and enter 0.40 in the Hypothesized proportion box
  5. Select Options
  6. Use the default Alternative hypothesis setting of Proportion < hypothesized proportion value 
  7. Use the default Confidence level of 95
  8. Select Normal approximation method
  9. Click OK and OK

The result should be the following output:

Method

Event: Event proportion
Normal approximation is used for this analysis.

Descriptive Statistics
N Event Sample p 95% Upper Bound for p
129 37 0.286822 0.352321
Test
Null hypothesis H 0: p = 0.4
Alternative hypothesis H 1: p < 0.4
Z-Value P-Value
-2.62 0.004

Summary of Results

We could summarize these results using the five-step hypothesis testing procedure:

1. Check assumptions and write hypotheses

\(np_0=129(0.40)=51.6\) and \(n(1-p_0)=129(1-0.40)=77.4\) both values are at least 10 so we can use the normal approximation method.

\(H_0\colon p = 0.40\)

\(H_a\colon p < 0.40\)

2. Calculate the test statistic

From output, \(z\) = -2.62

3. Determine the p-value

From output, \(p\) = 0.004

4. Make a decision

\(p \leq \alpha\), reject the null hypothesis

5. State a "real world" conclusion

There is evidence that the proportion of women in the population who think they are overweight is less than 40%.

8.1.2.2.2.1 - Minitab Example: Normal Approx. Method

8.1.2.2.2.1 - Minitab Example: Normal Approx. Method

Example: Gym membership

Research question: Are less than 50% of all individuals with a membership at one gym female?

A simple random sample of 60 individuals with a membership at one gym was collected. Each individual's biological sex was recorded. There were 24 females. 

First we have to check the assumptions:

 np = 60 (0.50) = 30

 n(1-p) = 60(1-0.50) = 30

The assumptions are met to use the normal approximation method.

To perform a one sample proportion z test with summarized data in Minitab:

  1. In Minitab, select Stat > Basic Statistics > 1 Proportion
  2. Select Summarized data from the dropdown
  3. For number of events, add 24 and for number of trials add 60.
  4. Check the box next to Perform hypothesis test and enter 0.50 in the Hypothesized proportion box
  5. Select Options
  6. Use the default Alternative hypothesis setting of Proportion < hypothesized proportion value 
  7. Use the default Confidence level of 95
  8. Select Normal approximation method
  9. Click OK and OK

The result should be the following output:

Method

Event: Event proportion
Normal approximation is used for this analysis.

Descriptive Statistics
N Event Sample p 95% Upper Bound for p
60 24 0.400000 0.504030
Test
Null hypothesis H 0: p = 0.5
Alternative hypothesis H 1: p < 0.5
Z-Value P-Value
-1.55 0.061

We could summarize these results using the five-step hypothesis testing procedure:

1. Check assumptions and write hypotheses

\(np_0=60(0.50)=30\) and \(n(1-p_0)=60(1-0.50)=30\) both values are at least 10 so we can use the normal approximation method.

\(H_0\colon p = 0.50\)

\(H_a\colon p < 0.50\)

2. Calculate the test statistic

From output, \(z\) = -1.55

3. Determine the p-value

From output, \(p\) = 0.061

4. Make a decision

\(p \geq \alpha\), fail to reject the null hypothesis

5. State a "real world" conclusion

There is not enough evidence to support the alternative that the proportion of women memberships at this gym is less than 50%.

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