**Constructing a Confidence Interval for the Difference between Two Population Proportions**

In order to determine if a new instructional technology improves students' scores, a professor wants to know if a larger percentage of students using the instructional technology passed the class than the percentage of students who did not use the new technology. Records show that 45 out of 50 randomly selected students who were in classes that used the instructional technology passed the class and 38 out of 51 randomly selected students who were in classes that did not use the instructional technology passed the class. Construct a 95%

confidence interval for the true difference between the proportion of students using the technology who passed and the proportion of students not using the technology who passed.

Solution

We are going to show how to construct the confidence interval first without a TI-83/84 Plus calculator and then with one.

Step 1: Find the point estimate.

First, we'll let Population 1 be those students who used the new technology and Population 2 be those students who did not. Next, we need to calculate the sample proportions. The sample proportion for Sample 1 (using instructional technology) is calculated as follows.

pˆ1=x1n1=4550=0.9

The sample proportion for Sample 2 (without the instructional technology) is found as follows.

pˆ2=x2n2=3851≈0.745098

Now that we have the sample proportions, we can calculate the point estimate.

pˆ1−pˆ2=0.9−0.745098=0.154902

Step 2: Find the margin of error.

Notice that the samples are indeed independent of one another. Because they are two separate groups of students, they are not connected in any way. We can assume that the other necessary conditions are met to allow us to use the standard normal distribution to calculate the margin of error. The level of confidence is c=0.95

, so the critical value is zα2/=z0.052/=z0.025=1.96

. Substituting the values into the formula gives us the following.

E=zα2/pˆ1(1−pˆ1)n1+pˆ2(1−pˆ2)n2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=1.960.9(1−0.9)50+0.745098(1−0.745098)51⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≈0.145675

Step 3: Subtract the margin of error from and add the margin of error to the point estimate.

Subtracting the margin of error from the point estimate and then adding the margin of error to the point estimate gives us the following endpoints of the confidence interval.

Lower endpoint: (pˆ1−pˆ2)−E=0.154902−0.145675≈0.009Upper endpoint: (pˆ1−pˆ2)+E=0.154902+0.145675≈0.301

Thus, the 95%

confidence interval for the difference between the two population proportions ranges from 0.009 to 0.301

. The confidence interval can be written mathematically using either inequality symbols or interval notation, as shown below.

0.009<p1−p2<0.301

or

(0.009,0.301)

Therefore, we are 95%

confident that the percentage of students who passed the class is between 0.9% and 30.1% higher for the population of students who used the new instructional technology (Population 1) than for the population of students who did not use the technology (Population 2). Thus, with 95%

confidence, the professor can conclude that the new instructional technology improves students' scores.

To calculate the confidence interval for the difference between two proportions on the calculator, we don't need to find the individual sample proportions; we just need to enter the number of successes and the sample size for each sample, as well as the level of confidence. Press STAT , scroll to TESTS, and then choose option B:2-PropZInt. x1 is the number of successes from the first sample and n1 is the first sample's size. Similarly, x2 is the number of successes from the second sample and n2 is the second sample's size. As usual, C-Level is the confidence level, which must be entered as a decimal. The data should be entered as shown in the first screenshot below. After you select Calculate and press ENTER , the results will be displayed on the screen as shown in the second screenshot below.

2-PropZInt data entry screen with x_1 equal to 45, n_1 equal to 50, x_2 equal to 38, n_2 equal to 51, and C-Level equal to .95. 2-PropZInt results screen shows ( .00923 , .30057 ), p hat_1 equal to .9 , p hat_2 equal to .7450980392, n_1 equal to 50, and n_2 equal to 51.

Notice that the calculator gives the same interval but with more decimal places. The interpretation of the confidence interval is still the same. The proportion of students passing the class was higher for the population of students who used the new instructional technology than for the population of students who did not use the technology.