Suppose 227 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.10 significance level to test the claim that more than \( 20 \% \) of users develop nausea. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. \( H_{0}: p=0.20 \) \( H_{1}: p>0.20 \) B. \( H_{0}: p>0.20 \) \( H_{1}: p=0.20 \) C. \( H_{0}: p=0.20 \) \( H_{1}: p<0.20 \) D. \( H_{0}: p=0.20 \) \( H_{1}: p \neq 0.20 \) Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is (Round to two decimal places as needed.)

The correct answer for the null and alternative hypotheses is: A. \( H_{0}: p=0.20 \) \( H_{1}: p>0.20 \) Now, let's get to the test statistic! To calculate it, we first need the sample proportion of subjects who developed nausea. This is given by: \[ \hat{p} = \frac{\text{number of subjects who developed nausea}}{\text{total number of subjects}} = \frac{54}{227} \approx 0.237 \] Next, we use the formula for the test statistic for a proportion, which is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Here, \( p_0 = 0.20 \), \( n = 227 \). Calculating the standard error: \[ \text{Standard error} = \sqrt{\frac{0.20(1-0.20)}{227}} \approx \sqrt{\frac{0.20 \times 0.80}{227}} \approx 0.034 \] Now plug it into the z formula: \[ z = \frac{0.237 - 0.20}{0.034} \approx \frac{0.037}{0.034} \approx 1.09 \] So, rounding to two decimal places, the test statistic for this hypothesis test is approximately \( z \approx 1.09 \).