Answer to Question #207370 in Statistics and Probability for Hafiz

Question

Given that two machines A and B produced 200 and 300 items respectively. From these samples, machine A produced 18 defective items and machine B produced 15 defective items. Construct a 95% confidence interval for the difference in the proportion of defective items produced by machine A and B. How do you interpret the interval?

Answer 1

The sample proportion 1 is computed as follows, based on the sample size $n_1=200$

and the number of defective items $x_1=18$:

$$\hat{p_1}=\dfrac{x_1}{n_1}=\dfrac{18}{200}=0.09$$

The sample proportion 2 is computed as follows, based on the sample size $n_2=300$

and the number of defective items $x_2=15$:

$$\hat{p_2}=\dfrac{x_2}{n_2}=\dfrac{15}{300}=0.05$$

The critical value for $\alpha=0.95$ is $z_c=z_{1-\alpha/2}=1.96.$ The corresponding confidence interval is computed as shown below:

$$CI=\bigg(\hat{p_1}-\hat{p_2}-z_c\sqrt{\dfrac{\hat{p_1}(1-\hat{p_1})}{n_1}+\dfrac{\hat{p_2}(1-\hat{p_2})}{n_2}},$$

$$\hat{p_1}-\hat{p_2}+z_c\sqrt{\dfrac{\hat{p_1}(1-\hat{p_1})}{n_1}+\dfrac{\hat{p_2}(1-\hat{p_2})}{n_2}}\bigg)$$

$$=\big(0.09-0.05-1.96\sqrt{\dfrac{0.09(1-0.09)}{200}+\dfrac{0.05(1-0.05)}{300}},$$

$$0.09-0.05+1.96\sqrt{\dfrac{0.09(1-0.09)}{200}+\dfrac{0.05(1-0.05)}{300}}\big)$$

$$=(-0.0067, 0.0867)$$

Therefore, based on the data provided, the 95% confidence interval for the difference between the population proportions $p_1-p_2$ is $-0.0067<p_1-p_2<0.0867,$ which indicates that we are 95% confident that the true difference between population proportions is contained by the interval $(-0.0067, 0.0867).$