Answer to Question #275469 in Statistics and Probability for Maxwell

Question

A Dell company manufactures computers. For quality control,two sets of laptops were tested. In the first group, 32 out of 500 were found to contain some sort of defect. In the second group,30 out of 500 were found to have a defect. Is the difference between the groups significant?(use a significance level of 0.05.

Answer 1

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{32+30}{500+500}=0.062$$

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:

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

$$=\dfrac{32/500-30/500}{\sqrt{0.062(1-0.062)(1/500+1/500)}}$$

$$=0.26226$$

Since it is observed that $|z| =0.26226<1.96= z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p =2P(Z>0.26226)= 0.793121,$ and since $p= 0.793121>0.05=\alpha,$ 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.