Question

suppose 220 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
eq 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<0.20 ) identify the test statistic for this hypothesis test. the test statistic for this hypothesis test is 1.69 (round to two decimal places as needed) identify the p - value for this hypothesis test. the p - value for this hypothesis test is (round to three decimal places as needed)

Explanation:

Step1: Determine the type of test

This is a right - tailed test (since \(H_1:p > 0.20\)). The formula for the \(z\) - test statistic in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(\hat{p}=\frac{x}{n}\), \(x = 54\), \(n=220\), and \(p = 0.20\). We are given \(z = 1.69\).
The \(P\) - value for a right - tailed \(z\) - test is \(P(Z>z)\).

Step2: Calculate the \(P\) - value

Using the standard normal distribution table or a calculator with a normal distribution function. For a \(z\) - score of \(z = 1.69\), the \(P\) - value is \(P(Z>1.69)=1 - P(Z\leq1.69)\).
From the standard normal table, \(P(Z\leq1.69)=0.9545\).
So, \(P(Z > 1.69)=1-0.9545 = 0.0455\approx0.046\)

Answer:

The \(P\) - value for this hypothesis test is \(0.046\)