9.1.2.2 - Minitab: Difference Between 2 Independent Proportions

9.1.2.2 - Minitab: Difference Between 2 Independent Proportions

When conducting a hypothesis test comparing the proportions of two independent proportions in Minitab two p-values are provided. If \(np \ge 10\) and \(n(1-p) \ge 10\), use the p-value associated with the normal approximation method. If this assumption is not met, use the p-value associated with Fisher's exact method.

Minitab Note!

When conducting a hypothesis test for a difference between two independent proportions in Minitab you need to remember to change the "test method" to "use the pooled estimate of the proportion." This is because the null hypothesis is that the two proportions are equal.

Minitab®  – Testing Two Independent Proportions using Raw Data

Let's compare the proportion of females who have tried weed to the proportion of males who have tried weed. 

  1. Open Minitab file: class_survey.mpx
  2. Select Stat > Basic Statistics > 2 Proportions
  3. Select Both samples are in one column from the dropdown
  4. Double click the variable Try Weed in the box on the left to insert the variable into the Samples box
  5. Double click the variable Biological Sex in the box on the left to insert the variable into the Sample IDs box
  6. Under the Options tab change the Test method to Use the pooled estimate of the proportion
  7. Click OK and OK

This should result in the following output:

Method

Event: Try_Weed = Yes

\(p_1\): proportion where Try_Weed = Yes and Biological Sex = Female

\(p_2\): proportion where Try-Weed = Yes and Biological Sex = Male

Difference: \(p_1\)-\(p_2\)

Descriptive Statistics: Try Weed
Biological Sex N Event Sample p
Female 127 56 0.440945
Male 99 62 0.626263
Estimation for Difference
Difference 95% CI for Difference
-0.185318 (-0.313920, -0.056716)

CI based on normal approximation

Test
Null hypothesis \(H_0\): \(p_1-p_2=0\)
Alternative hypothesis \(H_1\): \(p_1-p_2\neq0\)
Method Z-Value P-Value
Normal approximation -2.77 0.006
Fisher's exact   0.007

The test based on the normal approximation uses the pooled estimate of the proportion (0.522124)

Five Step Hypothesis Testing Procedure

Step 1:
\(H_0 : p_f - p_m =0\)
\(H_a : p_f - p_m \neq 0\)

\(n_f p_f = 56\)
\(n_f (1-p_f) = 127-56 = 71\)
\(n_m p_m = 62\)
\(n_m (1-p_m) = 99-62 = 37\)

All \(np\) and \(n(1-p)\) are at least ten, therefore we can use the normal approximation method.

Step 2:
From output, \(z=-2.77\)

Step 3:
From output, \(p=0.005\)

Step 4:
\(p \leq \alpha\), reject the null hypothesis

Step 5:
There is evidence that in the population the proportion of females who have tried weed is different from the proportion of males who have tried weed.

Minitab®  – Testing Two Independent Proportions using Summarized Data

Let's compare the proportion of Penn State World Campus graduate students who have children to the proportion of Penn State University Park graduate students who have children.  In a representative sample there were 120 World Campus graduate students; 92 had children.  There were 160 University Park graduate students; 23 had children.

  1. In Minitab, select Stat > Basic Statistics > 2 Proportions
  2. Change Both samples are in one column to Summarized data in the dropdown
  3. For Sample 1 next to Number of events enter 92 and next to Number of trials enter 120
  4. For Sample 2 next to Number of events enter 23 and next to Number of trials enter 160
  5. Under the Options tab change the Test method to Use the pooled estimate of the proportion
  6. Click OK and OK

This should result in the following output:

Method

\(p_1\): proportion where Sample 1 = Event

\(p_2\): proportion where Sample 2 = Event

Difference: \(p_1\)-\(p_2\)

Descriptive Statistics
Sample N Event Sample p
Sample 1 120 92 0.766667
Sample 2 160 23 0.143750
Estimation for Difference
Difference 95% CI for Difference
0.622917 (0.529740, 0.716093)

CI based on normal approximation

Test
Null hypothesis \(H_0\): \(p_1-p_2=0\)
Alternative hypothesis \(H_1\): \(p_1-p_2\neq0\)
Method Z-Value P-Value
Normal approximation 10.49 0.000
Fisher's exact   0.000

The test based on the normal approximation uses the pooled estimate of the proportion (0.410714)

Five Step Hypothesis Testing Procedure

Step 1:
\(H_0 : p_W - p_U =0\)
\(H_a : p_W - p_U \neq 0\)

\(n_W p_W = 92\)
\(n_W (1-p_W) = 120-92 = 28\)
\(n_U p_U = 23\)
\(n_U (1-p_u) = 160-23 = 137\)

All \(np\) and \(n(1-p)\) are at least ten, therefore we can use the normal approximation method.

Step 2:
From output, \(z=10.49\)

Step 3:
From output, \(p=0.000\)

Step 4:
\(p \leq \alpha\), reject the null hypothesis

Step 5:
There is evidence that in the population the proportion of World Campus students who have children is different from the proportion of University Park students who have children.


9.1.2.2.1 - Example: Dating

9.1.2.2.1 - Example: Dating

Example: Dating

This example uses the course survey dataset:

A random sample of Penn State University Park undergraduate students were asked, "Would you date someone with a great personality even if you did not find them attractive?"  Let's compare the proportion of males and females who responded "yes" to determine if there is evidence of a difference.

1. Check any necessary assumptions and write hypotheses.

We are looking for a "difference," so this is a two-tailed test.

\(H_{0} : p_1 - p_2 =0\)
\( H_{a} :p_1 - p_2 \neq 0 \)

Method
Event: DatePerly = Yes
\(p_1\): proportion where DatePerly = Yes and Gender = Female
\(p_2\): proportion where DatePerly = Yes and Gender = Male
Difference: \(p_1-p_2\)
Descriptive Statistics: DatePerly
Gender N Event Sample p
Female 571 367 0.642732
Male 433 148 0.341801
Estimation for Difference
Difference 95% CI for Difference
0.300931 (0.241427, 0.360435)
Test
Null hypothesis \(H_0\): \(p_1-p_2=0\)
Alternative hypothesis \(H_1\): \(p_1-p_2\neq0\)
Method Z-Value P-Value
Fisher's exact   <0.0001
Normal approximation 9.45 <0.0001

The pooled estimate of the proportion (0.512948) is used for the tests.

\(n_f p_f = 367\)

\(n_f (1-p_f) = 571 - 367 = 204\)

\(n_m p_m = 148\)

\(n_m (1 - p_m) = 433 - 148 = 285\)

All of these counts are at least 10 so we will use the normal approximation method.

2. Calculate an appropriate test statistic.

From output, \(z=9.45\)

3. Determine the p-value associated with the test statistic.

From output, \(p<0.0001\)

4. Decide between the null and alternative hypotheses.

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

5. State a "real world" conclusion.

There is evidence that in the population of all Penn State University Park undergraduate students the proportion of men who would date someone with a great personality even if they did not find them attractive is different from the proportion of women who would date someone with a great personality even if they did not find them attractive.


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