Question

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

OxyContin The drug OxyContin (oxycodone) is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 227 subjects were treated with OxyContin and 52 of them developed nausea (based on data from Purdue Pharma L.P.). Use a 0.05 significance level to test the claim that more than 20% of OxyContin users develop nausea. Does the rate of nausea appear to be too high?

Short answer

Nullhypothesis: The proportion of subjects who developed nausea is equal to 20%.

Alternativehypothesis: The proportion of subjects who developed nausea is more than 20%.

Teststatistic: 1.092

Criticalvalue: 1.645

P-value: 0.1374

The null hypothesis is failed to reject.

There is not enough evidence to support the claim that the proportion of subjects who developed nausea is more than 20%.

Although the null hypothesis is failed to reject, the sample proportion of subjects who developed nausea equal to 0.229 seems to be high.

Step-by-step solution

Given information

Out of 227 subjects, the number of subjects who developed nausea is equal to 52.

Hypotheses

The null hypothesis is written as follows:

The proportion of subjects who developed nausea is equal to 20%.

$H_0:p=0.20$

The alternative hypothesis is written as follows:

The proportion of subjects who developed nausea is more than 20%.

$H_1:p>0.20$

The test is right-tailed.

Sample size, sample proportion, and population proportion

The sample size is equal to n=227.

The sample proportion of subjects who developed nausea isas follows:

$$\begin{gathered} \hat{p}=\frac{52}{227} \\ =0.229 \end{gathered}$$

The population proportion of subjects who developed nausea is equal to 0.20.

Test statistic

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.229-0.20}{\sqrt{\frac{0.20\left(1-0.20\right)}{227}}} \\ =1.092 \end{gathered}$$

Thus, z=1.092.

Critical value and p-value

Referring to the standard normal distribution table, the critical value of z at $\alpha =0.05$ for a right-tailed test is equal to 1.645.

Referring to the standard normal distribution table, the p-value for the test statistic value of 1.092 is equal to 0.1374.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

Conclusion of the test

There is not enough evidence to support the claim that the proportion of subjects who developed nausea is more than 20%.

Although the null hypothesis is failed to reject, the sample proportion of subjects who developed nausea equal to 0.229 appears to be quite high.