16 points Sa QUESTION 2 The human resource (HR) director ok a large corporation

Question

16 points Sa QUESTION 2 The human resource (HR) director ok a large corporation wishes to study absenteeism among its mid-level managers at its central office during the year. A random sample of 50 mid-level managers reveals the mean to be 7.2 days and a standard deviation to be 2.3 days Construct a 95% confidence interval estimate for the mean number of absences for mid-level managers during the year. What is the mwgn of error? Answers should be accurate up to 4 decimal places. Lower limit Upper limit: Margin of error

Explanation / Answer

Soution:

Given in the question

Mean =7.2

SD=2.3

N=50

Confidence interval can be calculated as

7.2-1.96*2.3/sqrt(50) to 7.2+1.96*2.3/sqrt(50)

7.2-1.96*0.3239 to 7.2+1.96*0.3239

7.2-0.6349 to 7.2+0.6349

6.56 to 7.83 is the confidence interval

Margin of error can be calculated as

Zalpha/2*SD/sqrt(n)

1.96*2.3/sqrt(50)

So margin of error is 0.6349 days

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