Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5754 patients treated with this drug, 157 developed the adverse reaction of nausea. Use a 0.10 significance level to test the claim that \( 3 \% \) of users develop nausea. Does nausea appear to be a problematic adverse reaction? Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. \( H_{0}: p=0.03 \) \[ H_{1}: p \neq 0.03 \] B. \( H_{0}: p=0.03 \) \( H_{1}: p<0.03 \) C. \( H_{0}: p \neq 0.03 \) \( H_{1}: p=0.03 \) D. \[ \begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p>0.03 \end{array} \] Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is \( \square \) \( \square \). (Round to two decimal places as needed.)

To address the test of whether nausea appears to be a problematic reaction, consider the hypotheses: The correct answer is: A. \( H_{0}: p=0.03 \) \[ H_{1}: p \neq 0.03 \] Now, let's calculate the test statistic. You can find the test statistic using the formula for the proportion: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Where: - \( \hat{p} \) is the sample proportion (157/5754) - \( p_0 \) is the claimed proportion (0.03) - \( n \) is the total number of patients (5754) Calculating \( \hat{p} \): \[ \hat{p} = \frac{157}{5754} \approx 0.0273 \] Now plug in the values: \[ z = \frac{0.0273 - 0.03}{\sqrt{\frac{0.03(1 - 0.03)}{5754}}} \approx \frac{-0.0027}{\sqrt{\frac{0.03 \cdot 0.97}{5754}}} \approx \frac{-0.0027}{0.0036} \approx -0.75 \] Thus, rounding to two decimal places, the test statistic for this hypothesis test is \( -0.75 \). Now you have both the hypotheses and the test statistic to evaluate whether nausea is a problematic adverse reaction!