Question

A researcher wishes to determine the proportion of college students who smoke. He determined that from a sample of 150 college students, 35admitted that they smoke.Is it valid to infer that 23% of college students smoke? Use a = 0.05

Answer 1

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

$H_0:p=0.23$

$H_a:p\not=0.23$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{35}{150}-0.23}{\sqrt{\dfrac{0.23(1-0.23)}{150}}}\approx0.0970$$

The p-value is $p =2P(Z>0.0970)= 0.922726$

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\}.$

Since it is observed that $z = 0.0970 \le1.96= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p =0.922726,$ and since $p = 0.922726 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than 0.23, at the $\alpha = 0.05$ significance level.