In an analysis investigating the usefulness of pennies, the cents portions of 102 randomly selected credit card charges from students are recorded, and they have a mean of 47.6 cents and a standard deviation of 33.2 cents. If the amounts from 0 cents to 99 cents are all equally likely, the mean is expected to be 49.5 cents and the population standard deviation is expected to be 28.866 cents. Use a 0.01 significance level to test the claim that the sample is from a population with a standard deviation equal to 28.866 cents. Complete parts (a) through (e) below.
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: σ = 28.866
Alternative Hypothesis, Ha: σ ≠ 28.866
Rejection Region
This is two tailed test, for α = 0.01 and df = 101
Critical value of Χ^2 are 68.146 and 141.351
Hence reject H0 if Χ^2 < 68.146 or Χ^2 > 141.351
Test statistic,
Χ^2 = (n-1)*s^2/σ^2
Χ^2 = (102 - 1)*33.2^2/28.866^2
Χ^2 = 133.605
P-value Approach
P-value = 0.0331
As P-value >= 0.01, fail to reject null hypothesis.
In an analysis investigating the usefulness of pennies, the cents portions of 102 randomly selected credit...