9.1.2  Hypothesis Testing
9.1.2  Hypothesis TestingHere we will walk through the fivestep hypothesis testing procedure for comparing the proportions from two independent groups. In order to use the normal approximation method there must be at least 10 "successes" and at least 10 "failures" in each group. In other words, \(n p \geq 10\) and \(n (1p) \geq 10\) for both groups.
If this assumption is not met you should use Fisher's exact method in Minitab or bootstrapping methods in StatKey.
9.1.2.1  Normal Approximation Method Formulas
9.1.2.1  Normal Approximation Method Formulas1. Check any necessary assumptions and write null and alternative hypotheses.
To use the normal approximation method a minimum of 10 successes and 10 failures in each group are necessary (i.e., \(n p \geq 10\) and \(n (1p) \geq 10\)).
The two groups that are being compared must be unpaired and unrelated (i.e., independent).
Below are the possible null and alternative hypothesis pairs:
Research Question  Are the proportions of group 1 and group 2 different?  Is the proportion of group 1 greater than the proportion of group 2?  Is the proportion of group 1 less than the proportion of group 2? 

Null Hypothesis, \(H_{0}\)  \(p_1  p_2=0\)  \(p_1  p_2=0\)  \(p_1  p_2=0\) 
Alternative Hypothesis, \(H_{a}\)  \(p_1  p_2 \neq 0\)  \(p_1  p_2> 0\)  \(p_1  p_2<0\) 
Type of Hypothesis Test  Twotailed, nondirectional  Righttailed, directional  Lefttailed, directional 
The null hypothesis is that there is not a difference between the two proportions (i.e., \(p_1 = p_2\)). If the null hypothesis is true then the population proportions are equal. When computing the standard error for the difference between the two proportions a pooled proportion is used as opposed to the two proportions separately (i.e., unpooled). This pooled estimate will be symbolized by \(\widehat{p}\). This is similar to a weighted mean, but with two proportions.
 Pooled Estimate of \(p\)
 \(\widehat{p}=\dfrac{\widehat{p}_1n_1+\widehat{p}_2n_2}{n_1+n_2}\)
The standard error for the difference between two proportions is symbolized by \(SE_{0}\). The subscript 0 tells us that this standard error is computed under the null hypothesis (\(H_0: p_1p_2=0\)).
 Standard Error

\(SE_0={\sqrt{\dfrac{\widehat{p} (1\widehat{p})}{n_1}+\dfrac{\widehat{p}(1\widehat{p})}{n_2}}}=\sqrt{\widehat{p}(1\widehat{p})\left ( \dfrac{1}{n_1}+\dfrac{1}{n_2} \right )}\)
Note that the default in many statistical programs, including Minitab, is to estimate the two proportions separately (i.e., unpooled). In order to obtain results using the pooled estimate of the proportion you will need to change the method.
Also note that this standard error is different from the one that you used when constructing a confidence interval for \(p_1p_2\). While the hypothesis testing procedure is based on the null hypothesis that \(p_1p_2=0\), the confidence interval approach is not based on this premise. The hypothesis testing approach uses the pooled estimate of \(p\) while the confidence interval approach will use an unpooled method.
 Test Statistic for Two Independent Proportions
 \(z=\dfrac{\widehat{p}_1\widehat{p}_2}{SE_0}\)
The \(z\) test statistic found in Step 2 is used to determine the \(p\) value. The \(p\) value is the proportion of the \(z\) distribution (normal distribution with a mean of 0 and standard deviation of 1) that is more extreme than the test statistic in the direction of the alternative hypothesis.
If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.
Based on your decision in Step 4, write a conclusion in terms of the original research question.
9.1.2.1.1 – Example: Ice Cream
9.1.2.1.1 – Example: Ice CreamExample: Ice Cream
The Creamery wants to compare adults and children in terms of preference for eating their ice cream out of a cone. They take a representative sample of 500 customers (240 adults and 260 children) and ask if they prefer cones over bowls. They found that 124 adults preferred cones and 90 children preferred cones.
\(H_0: p_a  p_c = 0\)
\(H_a:p_a  p_c \ne 0\)
\(n_a \widehat p_a = 124\)
\(n_a (1\widehat p_a) = 240  124 = 116\)
\(n_c \widehat p_c = 90\)
\(n_c (1\widehat p_c) = 26090 = 170\)
These counts are all at least 10 so we can use the normal approximation method.
 Pooled Estimate of \(p\)
 \(\widehat{p}=\dfrac{\widehat{p}_1n_1+\widehat{p}_2n_2}{n_1+n_2}\)
\(\widehat{p}=\dfrac{124+90}{240+260}=\dfrac{214}{500}=0.428\)
 Standard Error of \(\widehat{p}\)
 \(SE_{0}={\sqrt{\frac{\widehat{p} (1\widehat{p})}{n_1}+\frac{\widehat{p}(1\widehat{p})}{n_2}}}=\sqrt{\widehat{p}(1\widehat{p})\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}\)
\(SE_{0}=\sqrt{0.428(10.428)\left ( \frac{1}{240}+\frac{1}{260} \right )}=0.04429\)
 Test Statistic for Two Independent Proportions
 \(z=\dfrac{\widehat{p}_1\widehat{p}_2}{SE_0}\)
\(z=\dfrac{\frac{124}{240}\frac{90}{260}}{0.04429}=3.850\)
This is a twotailed test. Our pvalue will be the area of the \(z\) distribution more extreme than \(3.850\).
\(p = 0.0000591 \times 2 = 0.0001182\)
\(p \le 0.05\)
Reject the null hypothesis
There is evidence that, in the population of Creamery customers, the proportion of adults who prefer cones is different from the proportion of children who prefer cones in the population.
9.1.2.1.2 – Example: Same Sex Marriage
9.1.2.1.2 – Example: Same Sex MarriageExample: Same Sex Marriage
A survey was given to a random sample of college students. They were asked whether they think same sex marriage should be legal. We're going to compare the students who identified as women and men in terms of whether or not they responded "yes" to this question. Of the 251 women in the sample, 185 said "yes." Of the 199 men in the sample, 107 said "yes."
For women, there were 185 who said "yes" and 66 who said "no." For men, there were 107 who said "yes" and 92 who said "no." There are at least 10 successes and failures in each group so the normal approximation method can be used.
 \(\widehat{p}_{w}=\dfrac{185}{251}\)
 \(\widehat{p}_{m}=\dfrac{107}{199}\)
This is a twotailed test because we are looking for a difference between women and men, we were not given a specific direction.
 \( H_{0} : p_{w} p_{m}=0 \)
 \( H_{a} : p_{w} p_{m}\neq 0 \)
\(\widehat{p}=\dfrac{185+107}{251+199}=\dfrac{292}{450}=0.6489\)
\(SE_0=\sqrt{\frac{292}{450}\left ( 1\frac{292}{450} \right )\left ( \frac{1}{251}+\frac{1}{199} \right )}=0.0453\)
\(z=\dfrac{\frac{185}{251}\frac{107}{199}}{0.0453}=4.400\)
Our test statistic is \(z=4.400\)
\(P(z>4.400)=0.0000054\), this is a twotailed test, so this value must be multiplied by two: \(0.0000054\times 2= 0.0000108\)
\(p<0.0001\)
\(p\leq0.05\), therefore we reject the null hypothesis.
There is evidence, in this population of students, that there is a difference between the proportion of women and men who think that same sex marriage should be legal.
9.1.2.2  Minitab: Difference Between 2 Independent Proportions
9.1.2.2  Minitab: Difference Between 2 Independent ProportionsWhen conducting a hypothesis test comparing the proportions of two independent proportions in Minitab two pvalues are provided. If \(np \ge 10\) and \(n(1p) \ge 10\), use the pvalue associated with the normal approximation method. If this assumption is not met, use the pvalue 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.
 Open Minitab file: class_survey.mpx
 Select Stat > Basic Statistics > 2 Proportions
 Select Both samples are in one column from the dropdown
 Double click the variable Try Weed in the box on the left to insert the variable into the Samples box
 Double click the variable Biological Sex in the box on the left to insert the variable into the Sample IDs box
 Under the Options tab change the Test method to Use the pooled estimate of the proportion
 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 TryWeed = 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  \(H_0\): \(p_1p_2=0\) 

Alternative hypothesis  \(H_1\): \(p_1p_2\neq0\) 
Method  ZValue  PValue 

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:
\(H_0 : p_f  p_m =0\)
\(H_a : p_f  p_m \neq 0\)
\(n_f p_f = 56\)
\(n_f (1p_f) = 12756 = 71\)
\(n_m p_m = 62\)
\(n_m (1p_m) = 9962 = 37\)
All \(np\) and \(n(1p)\) 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.
 In Minitab, select Stat > Basic Statistics > 2 Proportions
 Change Both samples are in one column to Summarized data in the dropdown
 For Sample 1 next to Number of events enter 92 and next to Number of trials enter 120
 For Sample 2 next to Number of events enter 23 and next to Number of trials enter 160
 Under the Options tab change the Test method to Use the pooled estimate of the proportion
 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  \(H_0\): \(p_1p_2=0\) 

Alternative hypothesis  \(H_1\): \(p_1p_2\neq0\) 
Method  ZValue  PValue 

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:
\(H_0 : p_W  p_U =0\)
\(H_a : p_W  p_U \neq 0\)
\(n_W p_W = 92\)
\(n_W (1p_W) = 12092 = 28\)
\(n_U p_U = 23\)
\(n_U (1p_u) = 16023 = 137\)
All \(np\) and \(n(1p)\) 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: DatingExample: 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.
We are looking for a "difference," so this is a twotailed test.
\(H_{0} : p_1  p_2 =0\)
\( H_{a} :p_1  p_2 \neq 0 \)
Event: DatePerly = Yes 
\(p_1\): proportion where DatePerly = Yes and Gender = Female 
\(p_2\): proportion where DatePerly = Yes and Gender = Male 
Difference: \(p_1p_2\) 
Gender  N  Event  Sample p 

Female  571  367  0.642732 
Male  433  148  0.341801 
Difference  95% CI for Difference 

0.300931  (0.241427, 0.360435) 
Null hypothesis  \(H_0\): \(p_1p_2=0\) 

Alternative hypothesis  \(H_1\): \(p_1p_2\neq0\) 
Method  ZValue  PValue 

Fisher's exact  <0.0001  
Normal approximation  9.45  <0.0001 
The pooled estimate of the proportion (0.512948) is used for the tests.
\(n_f p_f = 367\)
\(n_f (1p_f) = 571  367 = 204\)
\(n_m p_m = 148\)
\(n_m (1  p_m) = 433  148 = 285\)
All of these counts are at least 10 so we will use the normal approximation method.
From output, \(z=9.45\)
From output, \(p<0.0001\)
\(p \leq \alpha\), reject the null hypothesis
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.