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.

h₁: 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
(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% of users develop nausea.

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 50$ (number of successes) and $n=209$ (sample size).
$\hat{p}=\frac{50}{209}\approx0.2392$

Step2: Calculate the test statistic

The formula for the test statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}$, where $p = 0.20$ (hypothesized proportion).
Substitute the values:

$$
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$$

Step3: Calculate the P - value

Since this is a right - tailed test (claim is $p>0.20$), the P - value is $P(Z>z)$.
Using a standard normal table or calculator, $P(Z > 1.41)=1 - P(Z\leq1.41)$.
From the standard normal table, $P(Z\leq1.41)=0.9207$.
So, $P(Z>1.41)=1 - 0.9207 = 0.0793\approx0.079$

Answer:

The test statistic for this hypothesis test is $1.41$.
The P - value for this hypothesis test is $0.079$.
For the conclusion:
Since the P - value ($0.079$) $>0.05$ (significance level), we fail to reject $H_0$.
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.