Question

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% confidence interval for the average hourly wage (in $) of all information system managers is
a. 37.54 to 43.96.
b. 38.61 to 42.89.
c. 39.14 to 42.36.
d. 39.40 to 42.10.

Answer 1

Answer: c. 39.14 to 42.36

Step-by-step explanation:

We want to determine a 95% confidence interval for the average hourly wage (in $) of all information system managers

Number of sample, n = 75

Mean, u = $40.75

Standard deviation, s = $7.00

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

40.75 +/- 1.96 × 7/√75

= 40.75 ± 1.96 × 0.82

= 40.75 + 1.6072

The lower end of the confidence interval is 40.75 - 1.6072 =39.14

The upper end of the confidence interval is 40.75 + 1.6072 =42.36