AP Statistics 6.11 Carrying Out a Test for the Difference of Two Population Proportions- FRQs - Exam Style Questions

State:
We will perform a two-sample z-test for a difference in proportions. Let \(p_{younger}\) be the true proportion of members ages \(18\) to \(55\) interested in online classes, and \(p_{older}\) be the true proportion for members ages \(56\) and older. The significance level is \(\alpha=0.05\).
The hypotheses are:
\(H_0: p_{younger} – p_{older} = 0\)
\(H_a: p_{younger} – p_{older} \ne 0\)

Plan:
We check the conditions for inference.
1. Random: The data come from two independent random samples.
2. Independence (10% condition): \(170\) and \(230\) are likely less than \(10\%\) of all members in their respective age groups at a “large exercise center.”
3. Normality (Large Counts):
– \(\hat{p}_{younger} = \frac{51}{170} = 0.3\). Successes = \(51\), Failures = \(119\). Both are \(\ge 10\).
– \(\hat{p}_{older} = \frac{79}{230} \approx 0.343\). Successes = \(79\), Failures = \(151\). Both are \(\ge 10\).
All conditions are met.

Do:
First, calculate the pooled proportion, \(\hat{p}_c\):
\(\hat{p}_c = \frac{51+79}{170+230} = \frac{130}{400} = 0.325\)

Next, calculate the z-test statistic:
\(z = \frac{(\hat{p}_{younger} – \hat{p}_{older}) – 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)(\frac{1}{n_{younger}} + \frac{1}{n_{older}})}}\)
\(z = \frac{(0.3 – 0.3435)}{\sqrt{(0.325)(0.675)(\frac{1}{170} + \frac{1}{230})}} \approx -0.918\)

Finally, find the two-sided p-value:
p-value = \(2 \times P(Z < -0.918) \approx 0.359\)

Conclude:
Since the p-value (\(0.359\)) is greater than the significance level (\(\alpha = 0.05\)), we fail to reject the null hypothesis.

There is not convincing statistical evidence to conclude that there is a difference in the proportion of all younger members and all older members at this exercise center who would be interested in taking online fitness classes.