Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

We begin by calculating the sample proportion for each group. **For juniors:** - Sample size: \( n_j = 70 \) - Number watching TV: \( x_j = 56 \) - Sample proportion: \[ \hat{p}_j = \frac{56}{70} = 0.8. \] - Standard error: \[ SE_j = \sqrt{\frac{\hat{p}_j (1-\hat{p}_j)}{n_j}} = \sqrt{\frac{0.8(0.2)}{70}} = \sqrt{\frac{0.16}{70}} \approx 0.0478. \] - The margin of error for a \(95\%\) confidence interval (using \(Z = 1.96\)) is: \[ ME_j = 1.96 \times 0.0478 \approx 0.0937. \] - Thus, the \(95\%\) confidence interval for juniors is: \[ (0.8 - 0.0937,\; 0.8 + 0.0937) \approx (0.706, \;0.894). \] **For seniors:** - Sample size: \( n_s = 85 \) - Number watching TV: \( x_s = 47 \) - Sample proportion: \[ \hat{p}_s = \frac{47}{85} \approx 0.553. \] - Standard error: \[ SE_s = \sqrt{\frac{\hat{p}_s (1-\hat{p}_s)}{n_s}} = \sqrt{\frac{0.553(0.447)}{85}} \approx \sqrt{\frac{0.247}{85}} \approx \sqrt{0.00291} \approx 0.0539. \] - Margin of error: \[ ME_s = 1.96 \times 0.0539 \approx 0.1057. \] - The \(95\%\) confidence interval for seniors is: \[ (0.553 - 0.1057,\; 0.553 + 0.1057) \approx (0.447, \;0.659). \] **Interpreting the intervals:** The confidence interval for juniors is approximately \((0.706, 0.894)\) and for seniors is approximately \((0.447, 0.659)\). Notice that the entire interval for juniors is above that for seniors (since \(0.706 > 0.659\)); hence, the confidence intervals do not overlap. This leads to the conclusion that the proportion of juniors who watch television regularly is higher than the proportion of seniors. Among the given choices, only the statement that uses the intervals \((0.706,0.894)\) and \((0.447,0.659)\) and concludes that the junior interval is higher (non-overlapping) is correct. The correct answer is: The \(95\%\) confidence interval for juniors is \((0.706, 0.894)\), and the \(95\%\) confidence interval for seniors is \((0.447, 0.659)\). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.