Question

a) A therapist wishes to determine if a new approach to treating phobic clients with a fear of airplane travel will be more effective than the standard therapy she has used for many years. She knows, after many years of experience with phobic patients, that it takes an average of 12.8 therapy sessions before clients are able to board and complete a short trip in an airplane. She decides to try her new approach on her next 15 patients. The number of therapy sessions required to successfully complete treatment for the phobia for these 15 patients is 9.6 therapy sessions, with a standard deviation of 5 sessions. Test the hypothesis that the new therapy decreases the number of sessions required to successfully treat the phobia at the .01 level of significance. Again, be sure to show all work and provide a step by step hypothesis test summary as shown in the text.

b) Find the 95% confidence interval for the data above and explain what the confidence interval tell us about the true mean of the population for this sample.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 12.8
Alternative Hypothesis: μ < 12.8

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (9.6 - 12.8)/(5/sqrt(15))
t = -2.479

P-value Approach
P-value = 0.0133
As P-value >= 0.01, fail to reject null hypothesis.

b)

sample mean, xbar = 9.6
sample standard deviation, s = 5
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.14

ME = tc * s/sqrt(n)
ME = 2.14 * 5/sqrt(15)
ME = 2.763

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (9.6 - 2.14 * 5/sqrt(15) , 9.6 + 2.14 * 5/sqrt(15))
CI = (6.8373 , 12.3627)

we are 95% confident that the true mean of the population for this sample is between 6.8373 and 12.3627