Answer: c. 39.14 to 42.36
Step-by-step explanation:
We aim to find a 95% confidence interval for the average hourly wage (in $) of all information system managers.
Sample size, n = 75
Mean, u = $40.75
Standard deviation, s = $7.00
For a 95% confidence level, the associated z value is 1.96 as sourced from the normal distribution table.
The formula we will implement is:
Confidence interval
= mean +/- z × standard deviation/√n
This results in:
40.75 +/- 1.96 × 7/√75
= 40.75 ± 1.96 × 0.82
= 40.75 + 1.6072
The lower limit of the confidence interval becomes 40.75 - 1.6072 =39.14
The upper limit is therefore 40.75 + 1.6072 =42.36