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? c) [2] Calculate the probability that the sample mean X is greater than $78,000. (d) [2] If your random sample actually produced a sample mean of $78,000, would you consider this unusual? What conclusion might you draw ?

Answer 1

Solution:

We are given

µ = 74000

σ = 6000

n = 60

Part a

Mean of sampling distribution of sample mean Xbar = µ = 74000

Standard deviation of sampling distribution of sample mean Xbar = σ/sqrt(n) = 6000/sqrt(60) = 774.5967

Part b

For probability 0.95, Z values by using z-table are given as -1.96 and 1.96.

Xbar lower = µ - Z*[ σ/sqrt(n)] = 74000 - 1.96*774.5967 = 72481.79

Xbar upper = µ + Z*[ σ/sqrt(n)] = 74000 + 1.96*774.5967 = 75518.21

Answer: 72481.79 to 75518.21

Part c

P(Xbar>78000) = 1 – P(Xbar<78000)

Z = (Xbar - µ)/[ σ/sqrt(n)]

Z = (78000 – 74000)/[ 6000/sqrt(60)]

Z = 4000/774.5967

Z = 5.163978

P(Z<5.163978) = P(Xbar<78000) =1.000

(by using z-table)

P(Xbar>78000) = 1 – P(Xbar<78000)

P(Xbar>78000) = 1 – 1

P(Xbar>78000) = 0

Required probability = 0.0000

Part d

Yes, we consider this as unusual because we get the required probability less than 0.05.