Question

A health insurance company suspects a particular clinic of overcharging patients for a certain procedure. The insurance company has determined that the overall mean cost for the procedure is $1200 with a standard deviation of $220. The insurance company sought the assistance for a statistician to perform a hypothesis test to determine whether the clinic is overcharging patients for the procedure. A random sample of 65 patients who received the procedure was selected and the mean cost of the procedure was found to be $1280. Using a 1% level of significance, test whether the particular clinic is overcharging the patients for the procedure, by answering the following:

a) Construct the 99% confidence level interval for the population mean of the procedure.

b) State the null and alternate hypothesis for this test.

c) Calculate the value of the test statistics for this test

d) Identify the critical region for this test (show it graphically).

e) State the conclusion of the test. Give a reason for your answer

Answer 1

a)
sample mean, xbar = 1280
sample standard deviation, σ = 220
sample size, n = 65

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (1280 - 2.58 * 220/sqrt(65) , 1280 + 2.58 * 220/sqrt(65))
CI = (1209.60 , 1350.40)

b)
Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 1200
Alternative Hypothesis: μ > 1200

c)
Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (1280 - 1200)/(220/sqrt(65))
z = 2.93

d)
Rejection Region
This is right tailed test, for α = 0.01
Critical value of z is 2.33.
Hence reject H0 if z > 2.33

e)
Reject the null hypothesis.
There is significant evidence to conclude that the particular clinic is overcharging the patients for the procedure