Several years ago, $34 \%$ of parents who had children in grades $\mathrm{K}-12$ were satisfied with the quality of education the students receive. $A$ recent poll asked 1,155 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,155 surveyed, 456 indicated that they were satisfied. Construct a $99 \%$ confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed.
What are the null and alternative hypotheses?
$\mathrm{H}_{0}: \mathrm{p} \nabla \square$ versus $\mathrm{H}_{1}: \mathrm{p}$
(Round to two decimal places as needed.)
Answer
\(\boxed{\text{Final Answer: The null hypothesis is } H_{0}: p = 0.34 \text{ and the alternative hypothesis is } H_{1}: p \neq 0.34. \text{ The 99% confidence interval for the proportion of parents who are satisfied with the quality of education is approximately } (0.358, 0.432). \text{ Since the proportion from several years ago (0.34) does not fall within this interval, we reject the null hypothesis. There is evidence to suggest that parents' attitudes toward the quality of education have changed.}}\)
Step 1 ：Define the null and alternative hypotheses. The null hypothesis \(H_{0}\) is that the proportion of parents who are satisfied with the quality of education has not changed, i.e., it is still 34%. The alternative hypothesis \(H_{1}\) is that the proportion has changed, i.e., it is not 34% anymore.
Step 2 ：Calculate the sample proportion, which is the number of parents who indicated that they were satisfied divided by the total number of parents surveyed. In this case, the sample size \(n\) is 1155 and the number of parents who were satisfied \(x\) is 456. So, the sample proportion \(\hat{p}\) is \(\frac{x}{n} = \frac{456}{1155} = 0.395\).
Step 3 ：Construct a 99% confidence interval for the proportion. The formula for a confidence interval for a proportion is given by \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(Z\) is the Z-score corresponding to the desired level of confidence. For a 99% confidence level, the Z-score is approximately 2.576.
Step 4 ：Calculate the standard error \(SE\) using the formula \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.395(1-0.395)}{1155}} = 0.0144\).
Step 5 ：Calculate the lower and upper bounds of the confidence interval using the formula \(\hat{p} \pm Z \times SE\). The lower bound is \(0.395 - 2.576 \times 0.0144 = 0.358\) and the upper bound is \(0.395 + 2.576 \times 0.0144 = 0.432\). So, the 99% confidence interval is approximately \((0.358, 0.432)\).
Step 6 ：Compare the proportion from several years ago (34% or 0.34) with the confidence interval. Since 0.34 does not fall within the interval \((0.358, 0.432)\), we reject the null hypothesis \(H_{0}\).
Step 7 ：\(\boxed{\text{Final Answer: The null hypothesis is } H_{0}: p = 0.34 \text{ and the alternative hypothesis is } H_{1}: p \neq 0.34. \text{ The 99% confidence interval for the proportion of parents who are satisfied with the quality of education is approximately } (0.358, 0.432). \text{ Since the proportion from several years ago (0.34) does not fall within this interval, we reject the null hypothesis. There is evidence to suggest that parents' attitudes toward the quality of education have changed.}}\)
