Suppose 216 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.10 significance level to test the claim that more than $20 \%$ of users develop nausea.
Identify the test statistic for this hypothesis test.
The test statistic for this hypothesis test is
(Round to two decimal places as needed.)

Step 1 ：We are given that 216 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. We are asked to use a 0.10 significance level to test the claim that more than $20 \%$ of users develop nausea.

Step 2 ：We are asked to identify the test statistic for this hypothesis test.

Step 3 ：The test statistic for a hypothesis test for a proportion is given by the formula: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\] where: \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size.

Step 4 ：In this case, \(\hat{p} = \frac{51}{216}\), \(p_0 = 0.20\), and \(n = 216\).

Step 5 ：We can plug these values into the formula to find the test statistic.

Step 6 ：\[Z = \frac{0.2361111111111111 - 0.2}{\sqrt{\frac{0.2(1-0.2)}{216}}}\]

Step 7 ：Calculating the above expression, we get \(Z = 1.326806944007554\)

Step 8 ：Rounding to two decimal places as needed, we get the final answer.

Step 9 ：Final Answer: The test statistic for this hypothesis test is \(\boxed{1.33}\).