A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% c

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A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% c

onfidence 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.

Mathematics

1 answer:

Svet_ta [4.3K] · Answer 1 · 2026-02-23T22:33:49+03:00

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