Question

1. The salary of mayors in Canadas cities is reported to be normally distributed with mean $120,000 and standard deviation $12,121. To test this claim, 5 Canadian cities are selected at random and it is found that the average salary is $132,389. Test at α = 0.05 whether the salary is different from the claim.

Question: Hypothesis test (Four steps) and Construct a 95% confidence interval for the average salary of mayors in Canadas cities.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 120000
Alternative Hypothesis, Ha: μ ≠ 120000

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (132389 - 120000)/(12121/sqrt(5))
z = 2.29

P-value Approach
P-value = 0.022
As P-value < 0.05, reject the null hypothesis.

sample mean, xbar = 132389
sample standard deviation, σ = 12121
sample size, n = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 12121/sqrt(5)
ME = 10624.52

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (132389 - 1.96 * 12121/sqrt(5) , 132389 + 1.96 * 12121/sqrt(5))
CI = (121764.48 , 143013.52)