Question

suppose 241 subjects are treated with a drug that is used to treat pain and 54 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>0.20 )
b. ( h_0: p>0.20 )
( h_1: p = 0.20 )
c. ( h_0: overline{p}=0.20 )
( h_1: overline{p}
eq0.20 )
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
(round to two decimal places as needed.)

identify the p - value for this hypothesis test.

the p - value for this hypothesis test is

Explanation:

Step1: Identify 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 \(H_1: p>0.20\). This matches option A.

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

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 54\) (number of successes - those who developed nausea) and \(n=241\) (sample size). So \(\hat{p}=\frac{54}{241}\approx0.224\)

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 \(p = 0.20\), \(\hat{p}=0.224\), and \(n = 241\)

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Step4: Calculate the P - value

Since this is a right - tailed test (\(H_1: p>0.20\)), the P - value is \(P(Z>z)\)
Using a standard normal table or calculator, \(P(Z > 0.93)=1 - P(Z\leq0.93)\)
From the standard normal table, \(P(Z\leq0.93)=0.8238\)
So \(P - value=1 - 0.8238 = 0.1762\)

Answer:

• Hypotheses: A. \(H_0: p = 0.20\), \(H_1: p>0.20\)
• Test statistic: \(0.93\)
• P - value: \(0.18\) (rounded to two decimal places)