AP Statistics 6.6 Concluding a Test for a Population Proportion- FRQs - Exam Style Questions

State: We want to perform a hypothesis test to determine if there is convincing evidence that the proportion of all students at Karen’s school who use the app for homework at least once per week is greater than \(0.22\). Let \(p\) be this true proportion. The hypotheses are: \[ H_0: p = 0.22 \] \[ H_a: p > 0.22 \] We will use a significance level of \(\alpha = 0.05\).

Plan: The appropriate inference procedure is a one-sample z-test for a population proportion. We need to check the conditions:

1. Random: Karen took a simple random sample of \(130\) students. Condition met.
2. \(10\%\) Condition (Independence): The school has more than \(2,000\) students. \(130 \le 0.10(2000) = 200\). Sampling without replacement is okay. Condition met.
3. Large Counts (Normality): Check \(np_0 \ge 10\) and \(n(1-p_0) \ge 10\).
   • \(130(0.22) = 28.6 \ge 10\)
   • \(130(1-0.22) = 130(0.78) = 101.4 \ge 10\)

   Condition met. The sampling distribution of \(\hat{p}\) is approximately normal.

Do: The sample proportion is \(\hat{p} = \frac{38}{130} \approx 0.2923\).
The standard error under \(H_0\) is: \[ \sigma_{\hat{p}} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.22(0.78)}{130}} \approx 0.0363 \] The test statistic is: \[ z = \frac{\hat{p} – p_0}{\sigma_{\hat{p}}} = \frac{0.2923 – 0.22}{0.0363} \approx 1.99 \] The p-value is the probability of observing a test statistic \( \ge 1.99 \), assuming \(H_0\) is true: \[ \text{p-value} = P(Z \ge 1.99) \approx 0.0233 \]

Conclude: Since the p-value (\(\approx 0.0233\)) is less than the significance level (\(\alpha = 0.05\)), we reject \(H_0\). There is convincing statistical evidence that the proportion of all students at Karen’s high school who use the app to help them with their homework at least once per week is greater than the national proportion of \(0.22\).