Question

An agency is studying the income of store managers in the retail industry. A random sample of 25 managers reveals a sample mean of $45,420. The sample standard deviation is $2,050. Use 90% confidence level to determine the confidence interval.

Answer 1

The Confidence Interval = ($44,745.55 , $46,094.45)

Explanation:

The formula for Confidence Interval =

Confidence Interval = Mean ± z × Standard deviation/√n

Where n = number of samples = 25 managers

Standard deviation = $2,050

Mean = $45,420

z = z score of the given confidence interval

= z score of 90% confidence interval

= 1.645

Confidence Interval = $45,420 ± 1.645 × $2,050/√25

= $45,420 ± 1.645 × $2,050/5

= $45,420 ± 674.45

Confidence Interval =

$45,420 - 674.45 = $44,745.55

$45,420 + 674.45 = $46,094.45

Therefore, the Confidence Interval = ($44,745.55 , $46,094.45)