Question

The mean and standard deviation of a random sample of 100 measurements are equal to 10 and 2, respectively.

a. Find a 90% confidence interval for μ

b. Test whether μ differs from 11 given α= 0:05

c. What is the p-value for the test.

Answer 1

sample mean, xbar = 10
sample standard deviation, s = 2
sample size, n = 100
degrees of freedom, df = n - 1 = 99

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.66

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (10 - 1.66 * 2/sqrt(100) , 10 + 1.66 * 2/sqrt(100))
CI = (9.67 , 10.33)

b)
AS 11 is not included in the CI, reject the null hypothesis
Yes, mu differes from 11

c)
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (10 - 11)/(2/sqrt(100))
t = -5

P-value Approach
P-value = 0.0000