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 3.75 minutes.a. After observing 150 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 95% confidence interval for the mean assembly time.

Answer 1

(15,595, 16,805)

Explanation:

We have to:

m = 16.2, sd = 3.75, n = 150

m is the mean, sd is the standard deviation and n is the sample size.

the degree of freedom would be:

n - 1 = 150 - 1 = 149

df = 149

at 95% confidence level the t is:

alpha = 1 - 95% = 1 - 0.95 = 0.05

alpha / 2 = 0.05 / 2 = 0.025

now well for t alpha / 2 (0.025) and df (149) = t = 1,976

the margin of error = E = t * sd / (n ^ (1/2))

replacing:

E = 1,976 * 3.75 / (150 ^ (1/2))

E = 0.605

The 95% confidence interval estimate of the popilation mean is:

m - E <u <m + E

16.2 - 0.605 <u <16.2 + 0.605

15,595 <u <16,805

(15,595, 16,805)