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Economics 202
Spring 2007
Homework 9
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. (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.)
EXTRA ANSWER (by request)
8.47
a. This is a two-tailed test, so you have to solve for BOTH critical values, in the same
way that you usually do. This should give you two x-bar values, 24.1385 and 23.6815.
Then find the z values that go with these two values, IF THE MEAN IS 24.5. The z
value that corresponds with 23.6815 is -5.04. The z value that corresponds with 24.1385
is -1.12. Thus, the probability of making a Type II error is the probability that z is less
than -5.04 (which is zero) plus the probability that z is greater than -1.12, which is .3686
+ .5 = 0.8686.
b. There is an 87% chance that you will reject the null hypothesis if the machine is
overfilling, so that’s probably acceptable.