Answer to Question #288504 in Statistics and Probability for Maryann nganga

Question

A company has two manufacturing plants I and II.in a quality control inspection operation,a random sample of 50 items produced by plant I gave 8 defective while a sample of 50 from plant II gave 12 defective construct confidence interval for the difference in the proportion of items produced by the two plants..

Answer 1

For plant 1,

$n_1=50\\ x_1=8\\ \hat{p_1}={x_1\over n_1}={8\over50}={4\over25}=0.16$

For plant 2,

$n_2=50\\ x_2=12\\ \hat{p_2}={x_2\over n_2}={12\over50}={6\over25}=0.24$

$\alpha=0.05\\ \hat{q_1}=1-\hat{p_1}=1-0.16=0.84\\ \hat{q_2}=1-\hat{p_2}=1-0.24=0.76$

A 95% confidence interval for the difference in  proportion of items is given by,

$C.I=(\hat{p_1}-\hat{p_2})\pm Z_{\alpha\over2}\sqrt{{\hat{p_1}(\hat{q_1})\over n_1}+{p_2(\hat{q_2})\over n_2}}$

Where,

$Z_{\alpha\over2}=Z_{0.05\over2}=Z_{0.025}=1.96$

Therefore,

$C.I=(0.16-0.24)\pm1.96\sqrt{{0.16\times0.84\over50}+{0.24\times0.76\over50}}\\ C.I=-0.08\pm1.96\sqrt{0.006136}\\ C.I=-0.08\pm0.153531943$

Therefore, a 95% confidence interval for the difference in the proportions of items produced by the two plants is,

$(-0.2335, 0.0735)$