Question

a random sample of 250 bolts from machine a contained 17 defective bolts, while an independently - chosen, random sample of 200 bolts from machine b contained 18 defective bolts. let $p_1$ be the proportion of the population of all bolts from machine a that are defective, and let $p_2$ be the proportion of the population of all bolts from machine b that are defective. find a 90% confidence interval for $p_1 - p_2$. then find the lower limit and upper limit of the 90% confidence interval. carry your intermediate computations to at least three decimal places. round your responses to at least three decimal places. (if necessary, consult a list of formulas.) lower limit: upper limit:

Explanation:

Step1: Calculate sample proportions

$p_1=\frac{17}{250}=0.068$, $p_2 = \frac{18}{200}=0.09$

Step2: Calculate the standard error

$SE=\sqrt{\frac{p_1(1 - p_1)}{n_1}+\frac{p_2(1 - p_2)}{n_2}}=\sqrt{\frac{0.068\times(1 - 0.068)}{250}+\frac{0.09\times(1 - 0.09)}{200}}\approx\sqrt{\frac{0.068\times0.932}{250}+\frac{0.09\times0.91}{200}}\approx\sqrt{\frac{0.063376}{250}+\frac{0.0819}{200}}\approx\sqrt{0.000253504 + 0.0004095}\approx\sqrt{0.000663004}\approx0.0257$

Step3: Find the z - value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90 = 0.10$, and $\alpha/2=0.05$. The z - value $z_{\alpha/2}=z_{0.05}\approx1.645$

Step4: Calculate the confidence interval

The confidence interval for $p_1 - p_2$ is $(p_1 - p_2)\pm z_{\alpha/2}\times SE$.
$(p_1 - p_2)=0.068 - 0.09=- 0.022$
Lower limit: $(p_1 - p_2)-z_{\alpha/2}\times SE=-0.022-1.645\times0.0257=-0.022 - 0.04237=-0.064$
Upper limit: $(p_1 - p_2)+z_{\alpha/2}\times SE=-0.022 + 1.645\times0.0257=-0.022+0.04237 = 0.020$

Answer:

Lower limit: -0.064
Upper limit: 0.020