Question

The American Management Association wishes to have information on the mean income of middle managers in the retail industry. A random sample of 256 managers reveals a sample mean of $45,420. The standard deviation of this population is $2,050. The association would like answers to the following questions: a) Can we determine what the actual population mean is? b) What is a reasonable range (with 95% confidence) of values for the population mean? c)Explain in words to the Association what these results tell them.

Answer 1

Solution:

Given:

Sample Size = n = 256

Sample mean =

Population standard deviation =

Part a) Can we determine what the actual population mean is?

From the sample we can not obtain exact population mean value. But sample mean is an unbiased estimator of population mean, thus best point estimate of population mean is sample mean.

Thus best point estimate of population mean =

Part b) What is a reasonable range (with 95% confidence) of values for the population mean?

where

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Thus

Part c) Explain in words to the Association what these results tell them.

We are 95% confident that the true population mean income of middle managers in the retail industry will be within the limits ($45,168.875 , $45,671.125).

That is: if we construct 100 confidence intervals of sample size 256, then 95 out of 100 confidence intervals would contain the true population mean income of middle managers in the retail industry within the limits ($45,168.875 , $45,671.125).