Question

suppose 209 subjects are treated with a drug that is used to treat pain and 50 of them developed nausea. use a 0.05 significance level to test the claim that more than 20% of users develop nausea. identify the conclusion for this hypothesis test. a. fail to reject ( h_0 ). there is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea. b. fail to reject ( h_0 ). there is sufficient evidence to warrant support of the claim that more than 20% of users develop nausea. c. reject ( h_0 ). there is sufficient evidence to warrant support of the claim that more than 20% of users develop nausea. d. reject ( h_0 ). there is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea.

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 50$ (number of successes) and $n=209$ (sample size). So, $\hat{p}=\frac{50}{209}\approx0.239$.

Step2: State hypotheses

The null hypothesis $H_0:p = 0.20$ and the alternative hypothesis $H_1:p>0.20$ (right - tailed test).

Step3: Calculate test statistic

The test statistic for a proportion is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$. Substituting $p = 0.20$, $\hat{p}=0.239$, and $n = 209$:

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Step4: Find critical value and p - value

For a significance level $\alpha = 0.05$ in a right - tailed test, the critical value $z_{\alpha}=1.645$. The p - value for $z = 1.41$ (using standard normal table) is $P(Z>1.41)=1 - P(Z\leq1.41)=1 - 0.9207=0.0793$.

Step5: Make decision

Since $z = 1.410.05$), we fail to reject $H_0$.

Answer:

A. Fail to reject $H_0$. There is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea.