A random sample of 85 group leaders supervisors and similar

A random sample of 85 group leaders, supervisors, and similar personnel revealed that a person spent an average 6.5 years on the job before being promoted. The standard deviation of the population was 1.7 years Determine the 95% confidence interval estimate of the population mean Determine the 92% confidence interval estimate of the population mean.

Solution

Here we are given that average (Xbar) = 6.5

population standard deviation (sigma) = 1.7

sample size (n) = 85

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

Xbar - E < mu < Xbar + E

where Xbar = 6.5

mu is the population mean.

E is the margin of error.

E = (Zc*sigma) / sqrt(n)

where Zc is the critical value for Normal distribution.

Zc we can find by using EXCEL.

confidence level (c) = 95% = 0.95

a = 1 - c = 1 - 0.95 = 0.05

a/2 = 0.05 / 2 = 0.025

1 - a/2 = 1 - 0.025 = 0.975

=NORMSINV(probability)

where probability = 1 - a/2

Zc = 1.96

E = (Zc*sigma) / sqrt(n) = (1.96*1.7)/sqrt(85) = 0.3614

lower limit = Xbar - E = 6.5 - 0.3614 = 6.1386

upper limit = Xbar + E = 6.5 + 0.3614 = 6.8614

The 95% confidence interval estimate of the population mean is (6.1386, 6.8614).

The 92% confidence interval estimate of the population mean is,

Xbar - E < mu < Xbar + E

where Xbar = 6.5

mu is the population mean.

E is the margin of error.

E = (Zc*sigma) / sqrt(n)

where Zc is the critical value for Normal distribution.

Zc we can find by using EXCEL.

confidence level (c) = 92% = 0.92

a = 1 - c = 1 - 0.92 = 0.08

a/2 = 0.08 / 2 = 0.04

1 - a/2 = 1 - 0.04 = 0.96

=NORMSINV(probability)

where probability = 1 - a/2

Zc = 1.75

E = (Zc*sigma) / sqrt(n) = (1.75*1.7) / sqrt(85) = 0.3228

lower limit = Xbar - E = 6.5 - 0.3228 = 6.1772

upper limit = Xbar + E = 6.5 + 0.3228 = 6.8228

The 92% confidence interval estimate of the population mean is (6.1772, 6.8228).