Answer to Question #259456 in Statistics and Probability for izi

Question

A manufacturer turns out a product item that is labeled either “defective” or “not defective.” In order to estimate the proportion defective, a random sample of 100 items is taken from production, and 10 are found to be defective. Following implementation of a quality improvement program, the experiment is conducted again. A new sample of 100 is taken, and this time only 6 are found to be defective. (a) Give a 95% confidence interval on p_{1}-p_{2},

p1

​−p2

​, where p_1

p1

​ is the population proportion defective before improvement and p_2

p2

​ is the proportion defective after improvement. (b) Is there information in the confidence interval found in (a) that would suggest that p_{1}>p_{2} ?

p1

​>p2

​? Explain.

Answer 1

Let us make a summary of the information given about the samples.

Sample 1(before improvement)

$n_1=100$

$x_1=10$

sample 2(after improvement)

$n_2=100$

$x_2=6$

Point estimate for proportion in sample 1 is given as,

$\hat{p}_1=x_1/n_1=10/100=0.1$

For sample 2,

$\hat{p}_2=x_2/n_2=6/100=0.06$

Define $\hat{q}_1=1-\hat{p}_1\space and\space \hat{q}_2=1-\hat{p}_2$

a.

A 95% confidence interval for the difference in proportions$(p_1-p_2)$ is given as,

$C.I=(\hat{p}_1-\hat{p}_2)\pm Z_{\alpha/2}*\sqrt{(\hat{p}_1*\hat{q}_1)/n_1+(\hat{p}_2*\hat{q}_2)/n_2}$ , where $Z_{\alpha/2}=Z_{0.05/2}=Z_{0.025}=1.96$ is the value in the standard normal distribution that leaves an area of 0.025 to the right.

Now,

$C.I=(0.1-0.06)\pm 1.96*\sqrt{(0.09/100)+(0.0564/100)}$

$C.I=0.04\pm 1.96*\sqrt{0.001464}$

$C.I=0.04\pm 1.96*0.03826225$

$C.I=0.04\pm0.07499401$

Therefore, a 95% confidence interval for $(p_1-p_2)$ is given as, $(-0.035,\space 0.115)$

b.

No.

If $p_1\gt p_2$ then we would expect that this interval would not include 0. Since this confidence interval includes zero, there is no sufficient evidence to show that $p_1\gt p_2$