Question

An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $182.37 with a standard deviation of $72.33 (a) At the 5 percent level of significance, does this sample prove a violation of the guide line that the average patient should pay no more than $150 out-of-pocket?

a) What test should you run?

b) State the null and alternate hypothesiss

c) What the level of significance?

d) Find the critical value

e) Compute the value of test statistic

f) Determine the p-value

e) Interpret the result

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 150

H_a : > 150

t test

= 182.37

= 150

s = 72.33

= level of significance

n = 25

df = n - 1 = 25 - 1 = 24

Test statistic = t

= ( - ) / s / n

= (182.37 - 150) / 72.33 / 25

= 2.24

Test statistic = 2.24

= 0.05

t_0.05,24 = 1.711

Critical value = 1.711

P-value = 0.0173

P-value <

Reject the null hypothesis .