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.

Eliquis The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol-Myers Squibb Co.). Use a 0.05 significance level to test the claim that 3% of Eliquis users develop nausea. Does nausea appear to be a problematic adverse reaction?

Among 5,924 patients treated with Eliquis, 153 developed the adverse reaction of nausea.

The null hypothesis is written as follows:

The proportion of patients who suffered from the adverse reaction is equal to 3%.

\({H_0}:p = 0.03\)

The alternative hypothesis is written as follows:

The proportion of patients who suffered from the adverse reaction is not equal to 3%.

\({H_1}:p \ne 0.03\)

The test is two-tailed.

The sample proportion of patients who suffered from the adverse reaction isas follows:

\[\begin{array}{c}\hat p = \frac{{{\rm{Number}}\;{\rm{of}}\;patients\;{\rm{who}}\;{\rm{developedadverse}}\;{\rm{reaction}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{patients}}}}\\ = \frac{{153}}{{5924}}\\ = 0.0258\end{array}\]

The population proportion of patients who developed an adverse reaction is equal to p=0.03.

The sample size (n) is equal to 5924.

The value of the test statistic is computed below:

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.0258 - 0.03}}{{\sqrt {\frac{{0.03\left( {1 - 0.03} \right)}}{{5924}}} }}\\ = - 1.895\end{array}\)

Thus, z=-1.895.

Referring to the standard normal distribution table, the critical value of z at\(\alpha = 0.05\)for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the two-tailed test using absolute test statistic (1.895) is equal to 0.0581.

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

There is not enough evidence to reject the claim that the proportion of patients who developed an adverse reaction is equal to 0.03.

Only 2.6% of the subjects developed nausea as an adverse reaction of even less than 3%. Thus, it can be said that nausea does not appear to be a serious problem.