Question

A confidence interval is desired for the amount of time it takes a component of a product to move from one workstation to the next. The standard deviation of the population is 5 seconds. (Round your answers up to the nearest whole number.) (a) How many measurements should be made to be 99% certain that the margin of error, maximum error of estimation, will not exceed 2 seconds? (b) What sample size is required to be 99% confident for a margin of error of 3 seconds?

Answer 1

Solution :

Given that,

Population standard deviation = = 5

a)

Margin of error = E = 2

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z_/2* / E) ²

n = ( 2.576* 5/ 2)²

n = 41.47

n = 41

Sample size = 41

b)

Margin of error = E = 3

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z_/2* / E) ²

n = ( 2.576* 5/ 3)²

n = 18.43

n = 18

Sample size = 18