Several years ago, $34 \%$ of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,155 parents who have children in grades $\mathrm{K}-12$ 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?
\[
H_{0}: p=0.34 \text { versus } H_{1}: p \neq 0.34
\]
(Round to two decimal places as needed.)
Use technology to find the $99 \%$ confidence interval.
The lower bound is
The upper bound is
(Round to two decimal places as needed.)

Step 1 ：Define the null and alternative hypotheses as follows: \(H_{0}: p=0.34\) versus \(H_{1}: p \neq 0.34\)

Step 2 ：Calculate the sample proportion (p̂) which is the number of parents who indicated they were satisfied divided by the total number of parents surveyed. In this case, \(p̂ = \frac{456}{1155} = 0.395\)

Step 3 ：Calculate the standard error (SE) which is the square root of \(\frac{p̂*(1-p̂)}{n}\). In this case, \(SE = \sqrt{\frac{0.395*(1-0.395)}{1155}} = 0.0144\)

Step 4 ：Calculate the confidence interval using the formula \(p̂ ± Z*SE\), where Z is the Z-score for a 99% confidence interval (which is approximately 2.576). In this case, the lower bound of the confidence interval is \(0.395 - 2.576*0.0144 = 0.358\) and the upper bound is \(0.395 + 2.576*0.0144 = 0.432\)

Step 5 ：\(\boxed{\text{Final Answer: The 99% confidence interval for the proportion of parents who are satisfied with the quality of education their children receive is approximately } [0.36, 0.43]}\)