17. A writer for a career magazine is working on an article about the projected
career earnings for college graduates in various fields. He selects a random
sample of 25 educators and surveys them about their current salary, the number
of years they have worked in the field, and their pay raise structure.
Based upon this information he computes the average projected career earnings for
graduates with degrees in education to be $2.5 million dollars with a standard deviation
of 0.4 million dollars.
Assuming all conditions for inference are met, which of the following is a 90 percent
confidence interval for the mean projected career earnings for graduates with degrees
in education?
(A) 2.5 = 1.7112
(B) 2.5 +1.645(0.04)
(C) 2.5 = 1.960
V25
(D) 2.5 +1.645 0.4
1 25
(E) 2.5 1.960(0.04)​

Respuesta :

Answer:

A

Step-by-step explanation:

The 90 percent confidence interval for the mean projected career earnings for graduates with degrees is; CI = 2.5 ± 1.645(0.4/√25)

How to calculate confidence Intervals?

Formula for confidence Interval is;

CI = x' ± z(s/√n)

z at 90% CL is; z = 1.645

We are given;

Sample size; n = 25

sample mean; x' = 2.5

standard deviation; s = 0.4

Thus;

Confidence interval; CI = 2.5 ± 1.645(0.4/√25)

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