Question

A garment factory distributes two brands of jeans. If it is found that 75 out of 250 customers prefer brand A and that 30 out of 150 prefer brand B, can we conclude at 0.05 level of significance that brand A outsells brand B?

Answer 1

a.

$H_0: p_1\le p_2$

$H_a:p_1>p_2$

b.The significance level is $\alpha = 0.05.$

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

d. The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{75+30}{250+150}=0.2625$$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})(1/n_1+1/n_2)}}$$ $$=\dfrac{75/250+30/105}{\sqrt{0.2625(1-0.2625)(1/250+1/150)}}=2.2006$$

e. Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.6449.$

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

g. Since it is observed that $z = 2.2006 >1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.2006)=0.013882,$ and since $p=0.013882<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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