Question

In a report prepared by the Economic Research Department of a major bank, the Department manager maintains that the average annual family income on Metropoles is $48,432. With α = 0.05, use both the rejection region method and p-value to carry out the following tests:

a) After some research, you find that a random sample of 100 families shows an average income of $49,400 with a standard deviation of $8000. We would like to test if the average annual income is greater than what the report states.

b) A follow researcher, took a sample of 38 families shows an average income of $46,000 with a standard deviation of $6000. We would like to test if the average annual income is less than what the report states.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 48432
Alternative Hypothesis, Ha: μ > 48432

Rejection Region
This is right tailed test, for α = 0.05
Critical value of z is 1.645.
Hence reject H0 if z > 1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (49400 - 48432)/(8000/sqrt(100))
z = 1.21

P-value Approach
P-value = 0.1131
As P-value >= 0.05, fail to reject null hypothesis.

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 48432
Alternative Hypothesis, Ha: μ < 48432

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.645.
Hence reject H0 if z < -1.645

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (46000 - 48432)/(6000/sqrt(38))
z = -2.5

P-value Approach
P-value = 0.0062
As P-value < 0.05, reject the null hypothesis.