Part 3 of 4
Points: 0.25 of 1
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Suppose 212 subjects are treated with a drug that is used to treat pain and 53 of them developed nausea. Use a 0.10 significance level to test the claim that more than $20 \%$ of users develop nausea.
$\checkmark$ A.
\[
\begin{array}{l}
H_{0}: p=0.20 \\
H_{1}: p> 0.20
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: p=0.20 \\
H_{1}: p< 0.20
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: p=0.20 \\
H_{1}: p \neq 0.20
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: p> 0.20 \\
H_{1}: p=0.20
\end{array}
\]

Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is $\mathbf{1 . 8 2}$.
(Round to two decimal places as needed.)
Identify the P-value for this hypothesis test.
The P-value for this hypothesis test is $\square$
(Round to three decimal places as needed.)

Step 1 ：State the null hypothesis and the alternative hypothesis. The null hypothesis is \(H_{0}: p=0.20\) and the alternative hypothesis is \(H_{1}: p>0.20\).

Step 2 ：The test statistic for this hypothesis test is 1.82.

Step 3 ：Use the test statistic to find the P-value. The P-value is the probability that we would observe a test statistic as extreme as the one we have, or more extreme, if the null hypothesis were true.

Step 4 ：Since we are testing the claim that more than 20% of users develop nausea, we are looking for the probability that we would observe a test statistic of 1.82 or greater, given that the true proportion is 20%.

Step 5 ：This is a one-tailed test, so we will use the cumulative distribution function (CDF) of the standard normal distribution to find the P-value. The CDF gives the probability that a random variable drawn from the given distribution is less than or equal to a given value, so to find the P-value, we need to subtract the value of the CDF at the test statistic from 1.

Step 6 ：The P-value for this hypothesis test is 0.034.

Step 7 ：Final Answer: The P-value for this hypothesis test is \(\boxed{0.034}\).