exercise: The average number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let X = the number of sick days they took for the past year. Should the personnel team believe that the average number is about 10?

Question

exercise: The average number of sick days an employee takes per year is believed to be about 10.  Members of a personnel department do not believe this figure.  They randomly survey 8 employees.  The number of sick days they took for the past year are as follows:  12; 4; 15; 3; 11; 8; 6; 8.  Let X = the number of sick days they took for the past year.  Should the personnel team believe that the average number is about 10?

PsyDAG · Accepted Answer

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.

Anonymous · Answer

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