Question

suppose 236 subjects are treated with a drug that is used to treat pain and 54 of them develop
significance level to test the claim that more than 20% of users develop nausea.

identify the test statistic for this hypothesis test.

the test statistic for the hypothesis test is
(round to two decimal places as needed.)

identify the p - value for the hypothesis test.

the p - value for the hypothesis test is
(round to three decimal places as needed.)

identify the conclusion for this hypothesis test.

a. fail to reject h0. there is not sufficient evidence to warrant support of the claim that more than 20%

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 54$ (number of successes) and $n=236$ (sample size). So, $\hat{p}=\frac{54}{236}\approx0.2288$. The hypothesized proportion $p_0 = 0.20$.

Step2: Calculate the test - statistic

The formula for the test - statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$.
Substitute the values:

$$
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$$

Step3: Calculate the P - value

Since this is a right - tailed test ($H_1:p>0.20$), the P - value is $P(Z>z)$. Using the standard normal distribution table or a calculator, $P(Z > 1.11)=1 - P(Z\leq1.11)$.
From the standard normal table, $P(Z\leq1.11)=0.8665$. So, $P(Z>1.11)=1 - 0.8665 = 0.1335\approx0.134$

Answer:

The test statistic for the hypothesis test is $1.11$.
The P - value for the hypothesis test is $0.134$.
If we assume a significance level (say $\alpha = 0.05$), since $P - value=0.134>\alpha = 0.05$, we fail to reject $H_0$. So the conclusion is A. Fail to reject $H_0$. There is not sufficient evidence to warrant support of the claim that more than 20% develop nausea.