Question

b) In a sample of 600 men from a city 450 are found to be smokers. In another sample of 900 men from another city 450 are smokers. Do the data indicate that the cities are significantly different with respect to the habit of smoking among men? ( = 5%)

Answer 1

For sample 1, we have that the sample proportion is $\hat{p}_1=\dfrac{X_1}{N_1}=\dfrac{450}{600}=0.75.$

For sample 1, we have that the sample proportion is $\hat{p}_2=\dfrac{X_2}{N_2}=\dfrac{450}{900}=0.5.$

The value of the pooled proportion is computed as 

$$\dfrac{X_1+X_2}{N_1+N_2}=\dfrac{450+450}{600+900}=0.6$$

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:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(\dfrac{1}{N_1}+\dfrac{1}{N_2})}}$$

$$=\dfrac{0.75-0.5}{\sqrt{0.6(1-0.6)(\dfrac{1}{600}+\dfrac{1}{900})}}\approx9.682$$

Since it is observed that $|z|=9.682>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z>9.682)=0,$ 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.