Question

The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes. After observing 125 workers assembling similar devices, the manager noticed their average time was 16.2 minutes. Construct a 95% confidence interval for the mean assembly time.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 16.2

sample standard deviation = s = 3.6

sample size = n = 125

Degrees of freedom = df = n - 1 = 124

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,124 = 1.979

Margin of error = E = t_/2,df * (s /n)

= 1.979 * (3.6 / 125)

= 0.637

The 95% confidence interval estimate of the population mean is,

- E < < + E

16.2 - 0.637 < < 16.2 + 0.637

15.563 < < 16.837

( 15.563 , 16.837)