Question

A sample of 100 workers in a large plant gave a mean assembly time of 294 seconds with a standard deviation of 12 seconds in a time and motion study. Find a 95%percent confidence interval for the mean assembly time for all the workers in the plant

Answer 1

The confidence interval for population mean can be estimated the next way:

$x-Cr_a*{\frac s {\sqrt{n}}}<u<x+Cr*{\frac s {\sqrt{n}}}$ , where x - sample mean, $Cr_a$ - critical value at a confidence level, s - sample standard deviation, n - sample size, u - population mean

Since the sample size is big, we can use z-score(standard normal distribution) as critical value, so

$P(N(0,1)<Cr)=1-{\frac a 2}=0.975\implies Cr=1.96$

So, we have

$294-1.96*{\frac {12} {\sqrt{1000}}}<u<294+1.96*{\frac {12} {\sqrt{1000}}}\implies293.26<u<294.74$