Testing Claims About Proportions. In Exercises 7–22, 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.

Clinical Trials of OxyContin OxyContin (oxycodone) is a drug used to treat pain, butit is well known for its addictiveness and danger. In a clinical trial, among subjects treatedwith OxyContin, 52 developed nausea and 175 did not develop nausea. Among other subjectsgiven placebos, 5 developed nausea and 40 did not develop nausea (based on data from PurduePharma L.P.). Use a 0.05 significance level to test for a difference between the rates of nauseafor those treated with OxyContin and those given a placebo.

a. Use a hypothesis test.

b. Use an appropriate confidence interval.

c. Does nausea appear to be an adverse reaction resulting from OxyContin?

In a clinical trial, 52 subjects developed nausea, and 175 did not develop nausea among the subjects who were treated with OxyContin.

Also, among the subjects who were given placebos, fivedeveloped nausea, and 40 did not develop nausea.

Null hypothesis:There is no difference between the rates of nausea for subjectswho were treated with OxyContin and subjectswho were given a placebo.

\({H_0}:{p_1} = {p_2}\)

Alternate hypothesis:There is a significant difference between the rates of nausea for subjectswho were treated with OxyContin and subjectswho were given a placebo.

\({\rm{ }}{H_1}:{p_1} \ne {p_2}\)

Let\({n_1}\)and\({n_2}\)be the number ofsubjectstreated with OxyContin and the number of subjectsgiven a placebo, respectively.

The sample size\(\left( {{n_1}} \right)\)is computed below:

\(\begin{array}{c}{n_1} = 52 + 175\\ = 227\end{array}\)

The sample size\(\left( {{n_2}} \right)\)is computed below:

\(\begin{array}{c}{n_2} = 5 + 40\\ = 45\end{array}\)

Assume that\({x_1}\)and\({x_2}\)are the number ofsubjects who developed nausea and were given OxyContin and placebo, respectively.

Let \({\hat p_1}\)be the sample proportionof subjects who developed nausea and were treated with OxyContin.

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{52}}{{227}}\\ = 0.229\end{array}\)

\(\begin{array}{c}{{\hat q}_1} = 1 - {{\hat p}_1}\\ = 1 - 0.229\\ = 0.771\end{array}\)

Let \({\hat p_2}\)be the sample proportion ofsubjects who developed nausea and were given a placebo.

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{5}{{45}}\\ = 0.111\end{array}\)

\(\begin{array}{c}{{\hat q}_2} = 1 - {{\hat p}_2}\\ = 1 - 0.111\\ = 0.889\end{array}\)

The value of the pooled sample proportion is equal to:

\(\begin{array}{c}\bar p = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{52 + 5}}{{227 + 45}}\\ = 0.210\end{array}\)

Hence,

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.210\\ = 0.790\end{array}\)

The test statistic is equal to:

\(\begin{array}{c}z = \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} }}\\ = \frac{{\left( {0.229 - 0.111} \right) - 0}}{{\sqrt {\frac{{\left( {0.210} \right)\left( {0.790} \right)}}{{227}} + \frac{{\left( {0.210} \right)\left( {0.790} \right)}}{{45}}} }}\\ = 1.776\end{array}\)

Referring to the standard normal distribution table, the critical values of z corresponding to\(\alpha = 0.05\)for a two-tailed test are -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0757.

Here, the value of the test statistic lies between the two critical values, and the p-value is greater than 0.05.

Therefore, the null hypothesis is failed to reject.

a.

There is not sufficient evidence to concludethat there is a significant difference between the rates of nausea for subjectswho were treated with OxyContin and subjectswho were given a placebo.

If the level of significance for a two-tailed test is equal to 0.05, then the corresponding confidence level to construct the confidence interval is equal to 95%.

The expression for computing the confidence interval is given below:

\(\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\)

E is the margin of error and has the following formula:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ = 1.96 \times \sqrt {\frac{{\left( {0.229} \right)\left( {0.771} \right)}}{{227}} + \frac{{\left( {0.111} \right)\left( {0.889} \right)}}{{45}}} \\ = 0.1069\end{array}\)

b.

Substituting the required values, the following confidence interval is obtained:

\(\begin{array}{c}\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E < {p_1} - {p_2} < \left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E\\(0.229 - 0.111) - 0.1069 < {p_1} - {p_2} < (0.229 - 0.111) + 0.1069\\0.011 < {p_1} - {p_2} < 0.225\end{array}\)

Thus, the 95% confidence interval is equal to (0.011, 0.225).

This confidenceinterval does not contain zero,which means that there is a significant difference between the two proportions of subjects who developed nausea.

Therefore, there is sufficient evidence to conclude thatthere is a significant difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

c.

The sample results show that 22.9% of the subjects who were treated with OxyContin developed nausea.

Since the proportion is high, it can be said that nausea is an adverse reaction of OxyContin.