Consider a drug that is used to help prevent blood clots in certain patients. In clinical trials, among 5995 patients treated with this drug. 149 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. \[ \begin{array}{r} H_{0}: p \neq 0.03 \\ H_{1}: p=0.03 \end{array} \] B. \[ \begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p \neq 0.03 \end{array} \] C. \[ \begin{array}{l} H_{0}: p=0.03 \\ H_{1}: p>0.03 \end{array} \] D. \[ \begin{array}{l} H_{0}: p=0.63 \\ H_{1}: p<0.03 \end{array} \] Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is -2.34 . (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.)
Solution
Step 1 ：Identify the null and alternative hypotheses for this test. The correct answer is: \n \[ \begin{array}{l} H_{0}: p=0.03 \ H_{1}: p \neq 0.03 \end{array} \]
Step 2 ：Identify the test statistic for this hypothesis test. The test statistic for this hypothesis test is -2.34.
Step 3 ：To calculate the P-value, we need to use the cumulative distribution function (CDF) of the normal distribution. The CDF gives us the probability that a random variable is less than or equal to a certain value. Since our test statistic is negative, we want to find the probability that a random variable from the standard normal distribution is less than or equal to -2.34. This will give us the P-value for our hypothesis test.
Step 4 ：The P-value for this hypothesis test is 0.00964186994535833.
Step 5 ：Final Answer: The null and alternative hypotheses for this test are: \n \[ \begin{array}{l} H_{0}: p=0.03 \ H_{1}: p \neq 0.03 \end{array} \] \n The test statistic for this hypothesis test is -2.34. \n The P-value for this hypothesis test is \(\boxed{0.010}\).
