Answer to Question #230727 in Statistics and Probability for Solomon

Question

There are 1000 students in the school of Engeering of 20,000 in the whole University of Zambia. In the study 200 students were found to be smokers in the school and 1000 in the whole university. Is there a statistical significant difference between the proportions of smokers in the school and the university

Answer 1

Given

$$n_1=1000, x_1=200, \hat{p_1}=0.2,$$

$$n_2=20000, x_2=1000, \hat{p_2}=0.05.$$

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{x_1+x_2}{n_1+n_2}=\dfrac{200+1000}{1000+20000}=\dfrac{2}{35}\approx0.0571$$

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

$H_0: p_1=p_2$

$H_1:p_1\not=p_2$

This corresponds to a two-tailed test, and a z-test for two population proportions will be used.

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 $R=\{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p_1}-\hat{p_2}}{\sqrt{\bar{p}(1-\bar{p})(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{0.2-0.05}{\sqrt{\dfrac{2}{35}(1-\dfrac{2}{35})(\dfrac{1}{1000}+\dfrac{1}{20000})}}$$

$$\approx19.9431$$

Since it is observed that $|z|=19.9431>z_c=1.96,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z>19.9431)\approx0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p_1$ is different than $p_2,$ at the $\alpha=0.05$ significance level.

Therefore there is a statistical significant difference between the proportions of smokers in the school and the university.