A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The manufacturer of the drug claims that more than $91 \%$ of patients taking the drug are healed within 8 weeks. In clinical trials, 210 of 228 patients suffering from acid reflux disease were healed after 8 weeks. Test the manufacturer's claim at the $\alpha=0.01$ level of significance.
Click here to view the standard normal distribution table (page 1).
Click here to view the standard normal distribution table (page 2).
Because $n p_{0}\left(1-p_{0}\right)=18.7> 10$, the sample size is less than $5 \%$ of the population size, and the sample (can be reasonably assumed to be random, $\quad \hat{*}$ the requirements for testing the hypothesis are $\$$ satisfied.
(Round to one decimal place as needed.)
What are the null and alternative hypotheses?
\[
\mathrm{H}_{0}: \mathrm{p}=.91 \text { versus } \mathrm{H}_{1}: \mathrm{p}> .91
\]
(Type integers or decimals. Do not round.)
Determine the test statistic, $z_{0}$.
$\mathrm{z}_{0}=.58$ (Round to two decimal places as needed.)
Determine the critic్fal value(s). Select the correct choice below and fill in the answer box to complete your choice.
(Round to two decimal places as needed.)
A. $\pm z_{\alpha / 2}= \pm$
B. $z_{\alpha}=$
Answer
The final answer is that the critical value is \(\boxed{2.33}\).
Step 1 ：The null and alternative hypotheses are \(H_{0}: p=0.91\) versus \(H_{1}: p>0.91\).
Step 2 ：The test statistic, \(z_{0}\), is determined to be 0.58.
Step 3 ：The critical value is determined by finding the z-score that corresponds to the area to the right of the standard normal distribution curve that equals 0.01. This is a one-tailed test, so we use 1 - 0.01 = 0.99 as the input to the inverse of the cumulative distribution function.
Step 4 ：The critical value at the significance level of 0.01 is approximately 2.33. This is the z-score that corresponds to the area to the right of the standard normal distribution curve that equals 0.01.
Step 5 ：The final answer is that the critical value is \(\boxed{2.33}\).
