Question

A random sample of 350 bolts from machine A contained 31 defective bolts, while an independently chosen, random sample of 375 balts from machine B contained 30 defective bolts. Let P, be the proportion of the population of all bolts from machine A that are defective, and let py be the proportion of the population of all bolts from machine B that are defective. Find a 95% confidence interval for P-P2. 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? D X ?

Answer 1

This is question for the difference of two binomial distributions. We need to find the interval for the difference of the population proportions. Since both n > 30 we can use the normal approximation.

Let 'x_i  be the no. of defective bolts in machine 'i'  and n_i be the total machines sampled

21 Pi ni
  
12 p2 12

21+22 ni+n2

(1- $1596223911655_blob.png$ )% CI for population proportion

(p1 - P2+Za/2V(1 - )(1/nı+1/n2))

$1596223927425_blob.png$ =1 - 0.95 = 0.05

Therefore the critical value at

2012

= Z_0.025

= 1.96 .............using normal percentage tables with p = 0.025

Margin of error = Critical value * SE

=
2012
*
Vp(1-)(1/n1+1/n2)

	Machine A	Mchine B
x	31	30
n	350	375
Sample $\hat{p_i}$	0.0886	0.08
Sample $\hat{p_i}$ difference	0.0086
pooled $\hat{p}$	0.0841
SE	0.0206
C.V.	1.9600
MOE	0.0404
Lower L	-0.0319
Upper L	0.0490

The experts are to solve one question compulsorily.