Question

An auditor reviewed 20 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $270.35 with a standard deviation of $75.34.

At the 5 percent level of significance, does this sample prove a violation of the guideline that the average patient should pay no more than $250 out-of-pocket?

(a-1)	H0: ? ? $250 versus H1: ? > $250. Choose the right option.

a.	Reject H0 if tcalc < 1.729
b.	Reject H0 if tcalc > 1.729

(a-2)

Calculate the test statistic. (Round your answer to 2 decimal places.)

Test statistic

(a-3)

The null hypothesis should be rejected.

	False
	True

(a-4)

The evidence does not show that the average out-of-pocket expense is greater than $250.

	True
	False

(b)

Is this a close decision?

	No
	Yes

Answer 1

a-1)

H0: <= 250

Ha: > 250

This is right tailed test.

t critical value at 0.05 significance level for 19 df = 1.729

Decision rule = Reject H0 if t > 1.729

a-2)

Test statistics

t = - / S / sqrt(n)

= 270.35 - 250 / 75.34 / sqrt(20)

= 1.21

This is test statistics value

a-3)

Since t < 1.729, we do not have sufficient evidence to reject H0.

So, the statement null hypothesis should be rejected is False.

a-4)

The Statement, evidence does not show that the average out-of-pocket expence is greater

than $250 is TRUE

b)

Since t test statistics value is much less than critical value(1.729), this is not close decision.

No.