Answer to Question #273791 in Statistics and Probability for kyla

Question

In a sample of 150 ilocanos, 65% wished that they were rich. In a sample of 260 Visayas, 60% wished that they were rich. At 𝛼 = 0.05, is there a difference in the proportions? Find the 95% confidence interval for the difference of two proportions

Answer 1

a) The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{0.65(150)+0.6(260)}{150+260}=0.6183$$

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:

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

$$=\dfrac{0.65-0.6}{\sqrt{0.6183(1-0.6183)(1/150+1/260)}}$$

$$=1.0038$$

Since it is observed that $|z| = 1.0038 \le1.96= z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p =2P(Z>1.0038)= 0.315475,$ and since $p= 0.315475>0.05=\alpha,$ 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.

b) The critical value for $\alpha = 0.05$ is $z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\hat{p}_1-\hat{p}_2-z_c\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}},$$

$$\hat{p}_1-\hat{p}_2+z_c\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}})$$

$$=(0.65-0.6-1.96\sqrt{\dfrac{0.65(1-0.65)}{150}+\dfrac{0.6(1-0.6)}{260}},$$

$$0.65-0.6+1.96\sqrt{\dfrac{0.65(1-0.65)}{150}+\dfrac{0.6(1-0.6)}{260}})$$

$$\approx(-0.047, 0.147)$$

Therefore, based on the data provided, the 95 % confidence interval for the difference between the population proportions $p_1 - p_2$ is $-0.047 < p_1 - p_2 < 0.147,$ which indicates that we are 95% confident that the true difference between population proportions is contained by the interval $(-0.047, 0.147).$