Monthly incomes of employees at a particular company have a mean of $5954. The distribution of sample means for samples of size 70 is normal with a mean of $5954 and a standard deviation of $259. Suppose you take a sample of size 70 employees from the company and find that their mean monthly income is $5747. How many standard deviations is the sample mean from the mean of the sampling distribution?

Question

A. 0.8 standard deviations above the mean

B. 0.8 standard deviations below the mean

C. 7.3 standard deviations below the mean

D. 207 standard deviations below the mean

Answer 1

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.