y Statistics
Aja Andrews
$11 / 21 / 2310: 52 \mathrm{PM}$
ver $8-2 \& 8-3$ )
Question 7 of 13
This quiz: 13 point(s) possible
This question: 1 point(s) possible
Submit quiz
Suppose 233 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
The test statistic for this hypothesis test is $\square$
(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.)
Identify the conclusion for this hypothesis test.
A. Reject $\mathrm{H}_{0}$. There is sufficient evidence to warrant support of the claim that more than $20 \%$ of users develop nausea
B. Reject $\mathrm{H}_{0}$. There is not sufficient evidence to warrant support of the claim that more than $20 \%$ of users develop nausea
C. Fail to reject $\mathrm{H}_{0}$. There is not sufficient evidence to warrant support of the claim that more than $20 \%$ of users develop nausea
D. Fail to reject $\mathrm{H}_{0}$. There is sufficient evidence to warrant support of the claim that more than $20 \%$ of users develop nausea.
(1) Time Remaining: 0106.28
Next
Answer
\(\boxed{\text{Since the P-value (0.248) is greater than the significance level (0.05), we cannot reject the null hypothesis. Therefore, there is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea.}}\)
Step 1 ：State the null hypothesis H0: \(p = 0.20\) and the alternative hypothesis Ha: \(p > 0.20\)
Step 2 ：Plan to use the one-sample z-test for a proportion
Step 3 ：Calculate the pooled sample proportion (p) and the standard error (SE) using the formulae \(p = \frac{51}{233} = 0.219\) and \(SE = \sqrt{\frac{p * (1 - p)}{n}} = \sqrt{\frac{0.219 * (1 - 0.219)}{233}} = 0.028\)
Step 4 ：Compute the z-score test statistic (z) using the formula \(z = \frac{p - P}{SE} = \frac{0.219 - 0.20}{0.028} = 0.68\)
Step 5 ：Find the P-value associated with the observed z-score (0.68) in the Normal Distribution, which is 0.248
Step 6 ：\(\boxed{\text{Since the P-value (0.248) is greater than the significance level (0.05), we cannot reject the null hypothesis. Therefore, there is not sufficient evidence to warrant support of the claim that more than 20% of users develop nausea.}}\)
