91影视

A sample of 198 patients has beengiven sustained care, and 82.8% of them have stopped smoking after one month. It is claimed that 80% of the patients stop smoking after they are given sustained care.

The null hypothesis is written as follows.

The proportion of patients who stop smoking when they are given sustained care is equal to 80%.

\({H_0}:p = 0.80\).

The alternative hypothesis is written as follows.

The proportion of patients who stop smoking when they are given sustained care is not equal to 80%.

\({H_1}:p \ne 0.80\).

The test is two-tailed.

The sample size is n=198.

The sample proportion of patients who stop smoking after they are given sustained care is given as follows.

\(\begin{array}{c}\hat p = 82.8\% \\ = \frac{{82.8}}{{100}}\\ = 0.828\end{array}\).

The sample proportion of patients who stop smoking after they are given sustained care is equal to 0.80.

The value of the test statistic is computed below.

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.828 - 0.80}}{{\sqrt {\frac{{0.80\left( {1 - 0.80} \right)}}{{198}}} }}\\ = 0.985\end{array}\)

Thus, z=0.985.

Referring to the standard normal table, the critical value of z at\(\alpha = 0.01\)for a two-tailed test is equal to 2.5758.

Referring to the standard normal table, the p-value for the test statistic value of 0.985 is equal to 0.3246.

Asthe p-value is greater than 0.01, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of patients who stop smoking after they are given sustained care is equal to 80%.

A significant percentage of people are able to quit smoking with the help of sustained care.

Thus, sustained care proves to be effective.