Question

In: Statistics and Probability

A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates...

A university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $25,000. The population standard deviation was $2,500. What is the 90% confidence interval for the mean salary of all graduates from the English Department?

A) [$24,794, $25,206]

B) [$24,988, $25,012]

C) [$22,500, $27,500]

D) [$24,755, $25,245]

Solutions

Expert Solution

Solution

Given that,

= 25000

= 2500

n = 400

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( $\sqrt$ /n)

= 1.645 * (2500 / $\sqrt$ 400 )

= 206

At 90% confidence interval estimate of the population mean is,

- E < < + E

25000- 206 < < 25000 + 206

25794< < 25206

(25794, 25206)

Option a ) is correct.

orchestra answered 1 year ago

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