Question

Suppose that university faculty at the rank of assistant professor earn an average of $74000 per year with a standard deviation of $6000. In an attempt to verify this salary level, a random sample of 60 assistant professors was selected from a personnel data base for all universities in Canada.

1. Find the mean and the standard deviation of the sampling mean x ̄ .

2. Within what limits do you expect sample average to lie, with the probability 0.95?

3. Calculate the probability that the sample mean x ̄ is greater than $78000.

4. If your random sample actually produced a sample mean of $78000, would you consider this unusual?

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)

=(72481.82, 75518.18)

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

= P(X>78000)

=P()

=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.