Economics 202
Spring 2008
Homework 9
Your homework is due at the beginning of class next Wednesday, 4/16/08. Homework must be
legible, stapled, and must have your name on it. No late homework will be accepted under any
circumstances.
Chapter 8: 8.30, 8.34, 8.42, 8.44, 8.46, 8.48
NOTE: There are no graphs on this key because it’s hard to get them into the Word file,
but on the exam, you will be required to draw graphs for the power of the test and
hypothesis testing problems.
8.30
H0: p≥0.40
H1: p<0.40
p̅ = 0.38
Decision rule: reject the null hypothesis if z<-1.645
z=(0.38-0.40)/√(.4)(.6)/100 = -.4082
Do not reject the null hypothesis and conclude that the shopping center developer’s claim may be
correct; at least 40% of the adult female population does visit the center one or more times per
week.
8.42
a. Find the critical value. -1.28 = (x – 88)/12/√64), so x = 86.08.
With a mean of 86, the z-value for 86.08 would be .05. P(z>.05) = .5 - .0199 = 0.4801.
In other words, if the true mean is 86, the probability of a Type II error is 48%.
b. Power = 1 - .4801 = .5199. There is a 52% chance that the hypothesis test will be able to
distinguish between a population with a mean of 88 and a population with a mean of 86 (which is
not very good).
c. The power increases and beta decreases as the sample size increases. You could also increase
alpha, the significance level, since alpha and beta are inversely related.
d. The decision rule is, if z < -1.28, reject the null hypothesis, or you could write, if the sample
mean is less than 86.08, reject the null hypothesis.
Thus, since 85.66<86.08, reject the null hypothesis.
8.44
a. First find the critical value with a mean of 256. Since ά = .05 (and it’s a one-tailed test), z =
1.645. So:
1.645 = (x – 256)/(40/√100)
Solve for x. x = 262.58
Now assume that the true mean is 260, and find the z-value that corresponds with the x value
above.
z = (262.58 – 260)/(40/√100) = 0.645
Area = .24055 (I got this by averaging the values for z = 0.64 and z = 0.65).
Thus β = .5 + .24055 = 0.74055
b. The power of the test = 1 – β = .25945
c. Now assume that the true mean is 260, and find the z-value that corresponds with the x value
above.
z = (262.58 – 262)/(40/√100) = 0.145
Area = .05765 (I got this by averaging the values for z = 0.14 and z = 0.15).
Thus β = .5 + .0565 = .5565, and the power of the test is 0.44235. Since the alternative
hypothesis is farther away from the value in the original hypothesis, the probability of Type II
error is smaller, and the power of the test, which is essentially the ability of the test to distinguish
between the two hypotheses, is larger.
d. The critical z value would now be 1.28. Thus, when you solve for the x value, you should get
261.12. Then the corresponding z value for the alternative hypothesis will be 0.28 rather than
0.645. As you can see by calculating this, the β value would decrease, and the power of the test
would increase. In general, ά and β values are inversely related though not proportionate.
Note: In this problem, the x values are all above the mean for the alternate hypothesis. Thus
you need to add .5 to find the area. If (as in the example in class) the x value is below the mean
for the alternate hypothesis, you would subtract the area from .5 to find the β value. (This is why
it is a very good idea to draw pictures of the distribution when you do power of the test
problems; it really helps to keep things straight.)
8.46
a.
H0: μ≤250
H1: μ>250
70
b. 𝑥̅𝛼 = 250 + 1.28 (
) ; 𝑥̅𝛼 = 258.96
√100
Find the z value; z=-0.15.
P(z<-0.15)=.5 - 0.0596=0.4404
Power of the test = 0.5596
c. It would get smaller, since α and β are inversely related. Reducing α will increase β;
increasing β will decrease the power of the test.
d. It would get larger. (I have no idea what the second part of the question is looking for.)
e. Answers will vary, but it depends on what you think an acceptable probability of a Type I and
Type II error is.
8.46
a.
H0: μ≥30,000
H1: μ<30,000
4000
𝑥̅𝛼 = 30,000 + 1.28 (
) ; 𝑥̅𝛼 = 29,744
√400
Find z; z = -.28
P(z>-.28)= .5 + 0.1103 = 0.6103