Question

A random sample of 250 bolts from machine A contained 17 defective bolts, while an independently chosen, random sample of 225 bolts from machine B contained 24 defective bolts. Let p, be the proportion of the population of all bolts from machine A that are defective, and let p, be the proportion of the population of all bolts from machine B that are defective. Find a 95% confidence interval for p-p.. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval? x Ś ?

Please answer neatly and correctly!

Answer 1

Solution :

Given that,

n₁ = 250

x₁ = 17

$\hat p$₁ = x₁ / n₁ = 0.068

n₂ =225

x₂ = 24

$\hat p$₂ = x₂ / n₂ = 0.107

1) Point estimate of difference between two proportions

= $\hat p$ ₁ -  $\hat p$₂

= 0.068- 0.107

= -0.039

2)

$\therefore$$Z_{\alpha/2}$ = 1.96 (use z table)

Margin of error =  $Z_{\alpha/2}$ * $\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

=  

= $1.96 * \sqrt{\frac{0.068(1-0.068)}{250}+\frac{0.107(1-0.107)}{225}}$

= 0.051

3) Required interval is

Point estimate  $\pm$ Margin of error

-0.039$\pm$  0.051

(-0.039- 0.051 , -0.039+  0.051)

(-0.09 , 0.012)

Lower limit = -0.09

Upper limit = 0.012