# 9.1.2.2 - Minitab Express: Difference Between 2 Independent Proportions

When conducting a hypothesis test comparing the proportions of two independent proportions in Minitab Express 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 Express Note!

When conducting a hypothesis test for a difference between two independent proportions in Minitab Express 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.

## MinitabExpress – 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 Express file:
2. On a PC: In the menu bar select STATISTICS > Two Samples > Proportions
3. On a Mac: In the menu bar select Statistics > 2-Sample Inference > Proportions
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

This should result in the following output:

2-Sample Proportions: Try Weed by Biological Sex
 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)
Null hypothesis $$H_0$$: $$p_1-p_2=0$$ $$H_1$$: $$p_1-p_2\neq0$$
Method Z-Value P-Value
Fisher's exact   0.0072
Normal approximation -2.77 0.0057

The pooled estimate of the proportion (0.522124) is used for the tests.

Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

## Five Step Hypothesis Testing Procedure Section

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, $$p=0.0057$$

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.

## MinitabExpress – 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. Open Minitab Express without data
2. On a PC: In the menu bar select STATISTICS > Two Samples > Proportions
3. On a Mac: In the menu bar select Statistics > 2-Sample Inference > Proportions
4. Change Both samples are in one column to Summarized data
5. For Sample 1 next to Number of events enter 92 and next to Number of trials enter 120
6. For Sample 2 next to Number of events enter 23 and next to Number of trials enter 160
7. Under the Options tab change the Test method to Use the pooled estimate of the proportion
8. Click OK

This should result in the following output:

2-Sample Proportions
 $$p_1$$: 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)
Null hypothesis $$H_0$$: $$p_1-p_2=0$$ $$H_1$$: $$p_1-p_2\neq0$$
Method Z-Value P-Value
Fisher's exact   <0.0001
Normal approximation 10.49 <0.0001

The pooled estimate of the proportion (0.410714) is used for the tests.

Video Walkthrough

Select your operating system below to see a step-by-step guide for this example.

## Five Step Hypothesis Testing Procedure Section

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, $$p<.0001$$

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.