AP Statistics 6.3 Justifying a Claim Based on a Confidence Interval for a Population Proportion- FRQs - Exam Style Questions

(a)
State: We will construct a one-sample z-interval for a population proportion.
Let \( p \) = the true proportion of all teenagers in the United States who would respond that they use a video streaming service every day.

Plan: We verify the conditions:
• Random: The sample was randomly selected.
• 10% Condition: \( 920 \) is less than 10% of all teenagers in the United States.
• Large Counts: \( n\hat{p} = 920 \times 0.59 = 542.8 \geq 10 \) and \( n(1-\hat{p}) = 920 \times 0.41 = 377.2 \geq 10 \)

Do: The \( 95\% \) confidence interval is:
\( \hat{p} \pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = 0.59 \pm 1.96\sqrt{\frac{0.59 \times 0.41}{920}} \)
\( = 0.59 \pm 1.96\sqrt{0.000263} \approx 0.59 \pm 1.96 \times 0.0162 \approx 0.59 \pm 0.0318 \)
The interval is \( (0.5582, 0.6218) \)

Conclude: We are \( 95\% \) confident that the true proportion of all teenagers in the United States who would respond that they use a video streaming service every day is between \( 0.558 \) and \( 0.622 \).

(b)
Yes, the sample data provide convincing statistical evidence that the proportion is not \( 0.5 \).
Justification: The \( 95\% \) confidence interval \( (0.558, 0.622) \) does not contain \( 0.5 \). Since \( 0.5 \) is not a plausible value for the population proportion based on this interval, we have convincing evidence that the true proportion is different from \( 0.5 \).