A textile fiber manufacturer is investigating a new drapery

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of µ = 12 kilograms with standard deviation of = 0.5 kilograms.

(a) What should be the sample size, so that with probability 0.95 we will estimate the mean thread elongation with error at most 0.15 kg?

(b) What should be the sample size, so that with probability 0.95 we will estimate the mean thread elongation with error at most 0.05 kg?

Please Show step by step, Thank You

Solution

a)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.5
ME =0.15
n = ( 1.96*0.5/0.15) ^2
= (0.98/0.15 ) ^2
= 42.684 ~ 43      

b)
Standard Deviation ( S.D) = 0.5
ME =0.05
n = ( 1.96*0.5/0.05) ^2
= (0.98/0.05 ) ^2
= 384.16 ~ 385