Question

3. Suppose that university facility with the rank of assistant professor earn an average of $74,000 per year with a standard deviation of $6,000. In an attempt to verify this salary level, a random sample of 60 assistant professor was selected. (a) [2] Describe the sampling distribution of the sample mean X. (b) (2) Within what limit would you expect the sample average to lie, within proba bility 0.95? (e) (2) Calculate the probability that the sample mean is greater than $78,000 (d) (2) If your random sample aetually produced a sample mean of $78,000, would you consider this unusual? What conclusion might you draw?

Answer 1

average, µ = $74,000

standard deviation , s = $6,000

sample size, n = 60

standard error, e = s/vn = 6000/v60 = 774.6
a sampling distribution of the sample mean X is a normal distribution .
b at 95% confidence for a two tail distribution, z = 1.96 (from z table)

limit we can expect the sample average to lie= µ±z*e

=74000 ± (1.96 * 774.6)

=(72481.784, 75518.216)

c. the probability that the sample mean X is greater than $78,000

= P(X>78000)

=P()

| 7800 -740) 774.6

=P(z > 5)

= 1-1 = 0          (from z table)

d.If your random sample actually produced a sample mean of $78,000, we would
consider this unusal.