Question

suppose 232 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. use a 0.05 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 ( square ).
(round to two decimal places as needed.)

Explanation:

Step1: Determine the null and alternative hypotheses

The claim is that more than 20% of users develop nausea. The null hypothesis \(H_0\) is a statement of equality, so \(H_0: p = 0.20\). The alternative hypothesis \(H_1\) is the claim we are testing, so \(H_1: p>0.20\) (since we are testing for "more than").

Step2: Calculate the sample proportion \(\hat{p}\)

The sample size \(n = 232\) and the number of successes (those who developed nausea) \(x = 51\). The sample proportion \(\hat{p}=\frac{x}{n}=\frac{51}{232}\approx0.22\)

Step3: Calculate the test - statistic \(z\)

The formula for the test - statistic in a one - sample proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.20\), \(\hat{p}=0.22\), and \(n = 232\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

The null hypothesis \(H_0: p = 0.20\) and the alternative hypothesis \(H_1: p>0.20\) (corresponding to option B). The test - statistic \(z\approx0.76\)