Answer to Question #275914 in Statistics and Probability for sunshine

Question

Researchers suspect that 18% of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, with an enrollment of 300 students, a study found that 50 students smoked at least one pack of cigarettes a day. At a=0.05, test the claim that 18% of all high school students smoke at least one pack of cigarettes day.

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\geq 0.18$

$H_1:p<0.18$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this left-tailed test is $R = \{z: z < -1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$$

$$=\dfrac{50/300-0.18}{\sqrt{\dfrac{0.18(1-0.18)}{300}}}\approx-0.601113$$

Since it is observed that $z = -0.601113 \ge -1.6449=z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p=P(Z<-0.601113)$$=0.273882,$ and since $p = 0.273882>0.05=\alpha ,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion  $p$ is less than $0.18,$ at the $\alpha = 0.05$  significance level.

Therefore, there is enough evidence to claim that at least 18% of all high school students smoke at least one pack of cigarettes day at the $\alpha = 0.05$  significance level.