The public relations officer for a particular city claims the average monthly co

Question

The public relations officer for a particular city claims the average monthly cost for childcare outside the home for a single child is $700. A potential resident is interested in whether the claim is correct. She obtains a random sample of 64 records and computes the average monthly cost of childcare to be $689. Assume the population standard deviation to be $40.

Perform the appropriate test of hypothesis for the potential resident using = 0.01. (8 points)

Step 1

Step 2

Step 3

Step 4

b. Find the p-value for the test in a.). (3 points)

c. What effect, if any, would there be on the conclusion in part a.) if you change to 0.05? (3 points)

d. Find the power of the test when is actually $685 and = 0.05. (8 points)

Explanation / Answer

(a)

Step 1:

H0: = 700
Ha: 700

Step 2: Reject null hypothesis if z<-2.58 or z>2.58.

Step 3:

Sample mean = 689
Standard deviation = 40
Standard error of mean = / n
Standard error of mean = 40 / 64
SE = 40/8
Standard error of mean 5
z = (xbar- ) / SE
z = (689-700) / 5
z = -2.2

Step 4:

Since z lies between -2.58 and 2.58, fail to reject the null hypothesis.

b)
P-value = 2 P( z < -2.2) = 2(.0139) =.0278

c)
If we changed alpha to .05, we would reject the null hypothesis because the P-value < .05.
Using alpha = .01, we would not reject the null hypothesis since P-value > .01.

d)
With = 0.05, the critical region for rejection of H0 is:
Reject H0 if
(x-bar - 700) / [40 / 64] > 1.96 or < -1.96

That is: Reject H0 if:
(x-bar-700)/5 > 1.96 or < -1.96
xbar > (1.96)(5)+700 or xbar < (-1.96)(5)+700

Reject H0 if xbar > 709.8 or xbar < 690.2
Power = P(reject H0/ H1 is true)
Power = P( xbar > 709.8 / = 685) + P(xbar < 690.2 / = 685)

Compute P( xbar > 709.8 / = 685):
Mean = 685
Standard deviation = 40
Standard error / n = 40 / 64 = 5
standardize xbar to z = (xbar - ) / ( / n )
P(xbar > 709.8) = P( z > (709.8-685) / 5)
= P(z > 4.96) = 0.0000 --------------(1)
(from normal probability table)

Compute P( xbar < 690.2 / = 685):
Mean = 685
Standard deviation = 40
Standard error / n = 40 / 64 = 5
standardize xbar to z = (xbar - ) / ( / n )
P(xbar < 690.2) = P( z < (690.2-685) / 5)
= P(z < 1.04) = .8508 -------------(2)
(From Normal probability table)

Power = (1)+(2) = .8508

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.