Recently, a university surveyed recent graduates of the English Department for t
ID: 3269250 • Letter: R
Question
Recently, 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 with a standard deviation of $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English Department? Select one: a. [$22,500, $27,500] b.middot[$24,755, $25,245] c. [$24,988, $25,012] d. [$24,600, $25,600] The correct answer is: [$24,755, $25,245] To determine the size of a sample, the standard deviation of the population must be estimated by either taking a pilot survey or by approximating it based on knowledge of the population. Select one: True FalseExplanation / Answer
Average salary = $25,000
Standard deviation of the salary = $2,500
Standard error of the mean salary = Standard deviation / Sqrt(N)
= 2500 / sqrt(400) = 2500/ 20 = 125
z value for 95% confidence interval is 1.96
Margin of error = z value * standard error = 1.96 * 125 = 245
95% confidence interval is (Mean - Margin of error, Mean + Margin of error)
= (25000 - 245, 25000 + 245)
= (24755, 25245)
Hence (b) is the correct answer.
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