Answer to Question #273795 in Statistics and Probability for kyla

Question

In a sample of 240 store customers, 72 use visa card. In another sample of 190, 76 used a mastercard. At 𝛼 = 0.10, is there a difference in the proportion of people who use each type of credit card?

Answer 1

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{72+76}{240+190}=0.344186$$

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 $\alpha = 0.10,$ and the critical value for a two-tailed test is $z_c = 1.6449.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.6449\}.$

The z-statistic is computed as follows:

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

$$=\dfrac{72/240-76/190}{\sqrt{0.344186(1-0.344186)(1/240+1/190)}}$$

$$=-2.1675$$

Since it is observed that $|z| = 2.1675> 1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p =2P(Z<-2.1675)= 0.0302,$ and since $p= 0.03002<0.10=\alpha,$ 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 $\alpha = 0.10$ significance level.