An employer claims that full time employees have a mean income of $35,000. A sample of 200 ?employees was considered. Their mean salary was $34,000.

Question

An employer claims that full time employees have a mean income of $35,000. A sample of 200 ?employees was considered. Their mean salary was $34,000. Assume a population standard ?deviation of $4,000. Does her claim seem reasonable at a 0.05 level of significance?

Answer 1

We can use z-test to verify the

hypothesis (H_a) μ>35000

Null Hypothesis(H₀) is mean income = μ = 35000, alpha = 0.05

Sample mean = x = 34000

size of sample = n = 200

population standard deviation = σ = 4000

z = (x - μ)/(σ/√n)

z = (34000-35000)/(4000/√200)

z = -3.535

Hence, Mean value of 34000 is more than three and a half times of standard deviation less than the expected value.

P=(z ≤ -3.53 When H₀ is true) = area under the normal curve to the left of -3.53
P = .0002
Since .0002< .05, this indicates that H₀ is true.
This means that the claim made by the employer is not reasonable.