Question

2. Score on a 20-point assignment is normally distributed with a mean of 15 and standard deviation of 5. You believe that your friends are smarter and that their average is greater. You took a sample of 9 of your friends. The mean for this group is 18.  

• What is the rejection region?
• Would you use the t test?
• The 95% confidence interval for the population mean is

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 15
Alternative Hypothesis, Ha: μ > 15

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

No, t-test cannot be used

sample mean, xbar = 18
sample standard deviation, σ = 5
sample size, n = 9

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

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (18 - 1.96 * 5/sqrt(9) , 18 + 1.96 * 5/sqrt(9))
CI = (14.73 , 21.27)