Question

Defective electronics: A team of designers was given the task of reducing the defect rate in the manufacture of a certain printed ole circuit board. The team decided to reconfigure the cooling system. A total of 985 boards were produced the week before the reconfiguration was implemented, and 260 of these were defective. A total of 842 boards were produced the week after reconfiguration, and 194 of these were defective. Part: 0/2 Part 1 of 2 (a) Construct a 99% confidence interval for the decrease in the defective rate after the reconfiguration. Use the TI-84 Plus calculator and round the answers to three decimal places. A 99% confidence interval for the decrease in the defective rate after the reconfiguration is <21-22<0.

Answer 1

Ans :

Random variable & Parameters :

x1 : Number of successes from group 1

x2 : Number of successes from group 2

p1: Proportion of successes in group 1

p2: Proportion of successes in group 2

n1 : number of trial in group 1

n2=number of trial in group 2

From the given information is summaries as :

X1 = 260

n1 : =985

X2= 194

n2 = 842

then

p1=260/985

p1 =0.2640

p2=194/842

p2 = 0.2304

Construct (1-α)100% confidence interval for proportion :

  
((p1 - p2) – z** p11-p1) 1)p2(1 – p2) < p1, p2 > n1 n2

(p1 - p2) +z** pi(1-p1)p2(1 - p2) + n1 n2 V

here $\alpha$ = 0.01

then value of z* = z_1-$\alpha$/2 = z_0.995 =2.576

Therfore 99% confidence interval for the decrease in the defective rate after the reconfiguration is

((0.2640 - 0.2304) - 2.576* 0.2640(1 – 0.2640) 0.2304(1 – 0.2304) + 985 842 <p1 - P2 >

(0.2640 -0.2304) + 2.576 10.2640(1-0.2640), 0.2304(1- 0.2304)、 + 985 842

(-0.0184631, 0.0855743)

Therfore 99% confidence interval for the decrease in the defective rate after the reconfiguration is (-0.018, 0.086)

(rounded 3 decimal places )

Ans :

A 99% confidence interval for the decrease in the defective rate after the reconfiguration is (-0.018< p1 - p2 > 0.086)