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.

Testing Effectiveness of Nicotine Patches In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment and 32 were not smoking one year after the treatment (based on data from “High-Dose Nicotine Patch Therapy,” by Dale et al., Journal of the American Medical Association, Vol. 274, No. 17). Use a 0.05 significance level to test the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. Do these results suggest that the nicotine patch therapy is not effective?

It is given that 39 smokers who haveundergone the nicotine patch therapy aresmoking after one year of the treatment, while 32 arenot smoking one year after the treatment. It is claimed that most of the smokers who takethe treatment have started smoking after one year of the treatment.

The null hypothesis is written as follows.

The proportion of smokers who start smoking after one year of the treatment is equal to 0.5.

\({H_0}:p = 0.5\).

The alternative hypothesis is written as follows.

The proportion of smokers who start smoking after one year of the treatment is more than 0.5.

\[{H_1}:p > 0.5\].

The test is right-tailed.

The sample size is calculated below.

\(\begin{array}{c}n = 39 + 32\\ = 71\end{array}\).

The sample proportion of smokers who start smoking after one year of the treatment is computed below.

\[\begin{array}{c}\hat p = \frac{{39}}{{71}}\\ = 0.549\end{array}\].

The population proportion of smokers who start smoking after one year of the treatment is equal to 0.5.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.549 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{71}}} }}\\ = 0.826\end{array}\).

Thus, z=0.826

Referring to the standard normal 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 table, the p-value for the test statistic value of 0.826 is equal to 0.2044.

Asthe p-value is less than 0.05, the null hypothesis is not rejected.

There is not enough evidence to support the claim that the proportion of smokers who start smoking after one year of the treatment is greater than 50%.

Although the claim is rejected, it can be seen that the sample percentage of smokers who start smoking after one year of the treatment is equal to 54.9% (greater than 50%).

Thus, it can be said that nicotine patch therapy is not very effective in helping smokers quit smoking.