Answer to Question #341737 in Statistics and Probability for Velle

Question

In a study to estimate the proportion of smokers among the residents in a certain city and its suburbs, it is found that 24 of 104 urban residents are smokers, while 34 of 108 suburban residents are smokers.

Answer 1

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{24+34}{104+108}=\dfrac{29}{106}$$

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

$H_0:p_1=p_2$

$H_a:p_1\not=p_2$

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:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$

$$=\dfrac{\dfrac{24}{104}-\dfrac{34}{108}}{\sqrt{\dfrac{29}{106}(1-\dfrac{29}{106})(1/104+1/108)}}$$

$$=-1.380342$$

Since it is observed that $|z| = 1.380342 \le 1.96=z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-1.380342)=0.167481,$ and since $p=0.167481>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p_1$ is different than $p_2,$ at the $\alpha = 0.05$ significance level.