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Practice Exam 3
Attempt all questions. You must support all answers with reasons – correct
answers with incorrect or missing reasons will receive NO CREDIT.
1. Here is a Markov model of class mobility. Assume that transitions between social classes
(upper, middle, lower) of the successive generations in a family can be regarded as
transitions of a Markov chain. That is, assume that the occupation of a child will depend
only on the father’s occupation and not on the grandfather’s occupation. Suppose that the
transition matrix is give by:
Child
Father
Upper
Middle
Lower
Upper
.45
.05
.01
Middle
.48
.70
.50
Lower
.07
.25
.49
a. Find the stationary distribution.
Solve
0 = .450 + .051 + .012
1 = .480 + .701 + .502
2 = .070 + .051 + .012
with 0 + 1 + 2 = 1. This turns out to be 0 = .07, 1 = .62, 2 = .31.
b. In the long run, about what percentage of people have upper class jobs?
This is just 0, or 7%.
c. Compute the probability that a family moves from lower class to upper class in
three generations (i.e. grandparent lower class, parent middle class, and child
upper class). .31*.50*.05 = 0.00775.
2. Three cards are dealt from a well-shuffled deck. The deck has 52 cards, 13 of which are
diamonds. Find the chance that:
a. all of the cards are diamonds. P(DDD) = (13/52)*(12/51)*(11/50) = .013.
b. none of the cards is a diamond.
(39/52)*(38/51)*(37/50) = .41.
P(Not
D,
Not
D,
Not D) =
c. the cards are not all diamonds. P( at least 1 Not D) = 1 – P(DDD) = .987.
3. Polygraph (lie-detector) tests may be given to employees in sensitive positions. Suppose
that, if a person is lying, the probability that this is detected by the test is .88, whereas if
the person is telling the truth, the test indicates that the person is telling the truth with
probability .86. Suppose that people tell the truth with probability .99. A person
produces a positive response on the test. What is the probability that the test is incorrect,
and the person is in fact telling the truth?
P(Truth|Test says Lie) =
P(Test says Lie|Truth)P(Truth)/[P(Test says Lie|Truth)P(Truth)+P(Test says Lie|Lie)P(Lie)]
= .14*.99/[.14*.99 + .88*.01] = .94.
4. A random sample of size 20 gives the following values:
-3, 7, 5, 8, 0, 2, 12, -13, -8, 1, 2, 0, 0, 2, 3, -3, -1, -1, 3, 4
-10
-5
0
5
10
a. Make a boxplot of the data.
b. Find the sample mean and SD. mean = 1, SD = 5.44 (if you divide by 19; 5.16 if
you divide by 20).
c. Find an unbiased estimate of the variance of the sample mean.
An unbiased
estimate of the population variance is SD2 (when divided by 19) = 5.44*5.44 =
29.6. An unbiased estimate of the variance of the sample mean is 29.6/20 =1.48.
d. Find an approximate 95% CI for the population mean.
1 +/- 2*sqrt(1.48).
5. A conservative advice columnist ‘Dear Blabby’ receives a letter from an office worker
complaining about people who are ill coming to work. Dear Blabby replies that people
who are ill should not come to work, even if it means that the worker will receive less
pay. Dear Blabby invites readers to write in with their opinions, and receives 10,000
responses.
a. Are the people who write letters a sample of the population? Yes
b. Are the people who write letters a random sample of the population? No
c. What types of bias might be present in this sample?
everyone reading the letter writes in with an opinion.
Nonresponse bias, not
d. Assume that of the 10,000 letter writers, 90% agree with Dear Blabby. If
possible, make an approximate 95% CI for the population percentage who agree
with Dear Blabby. If not possible or appropriate, explain why not.
Not possible, don’t have a random sample.
6. Booklets are packaged in bundles of 100 by weighing them. Suppose that the weight of
each booklet is a random variable with mean 20g and SD 0.4g. Find the probability that
a bundle of exactly 100 booklets would be counted as containing 101 booklets or more –
that is, that the bundle weighs more than 2020g.
Use the normal approximation: Let X = total weight.
P(X > 2020) = P[(X – 2000)/(10*.4) > (2020 – 2000)/(10*.4)]  P(Z > 5)  0.
7. True or false, and explain.
a. If a NULL hypothesis is not rejected at the significance level , the probability that
the NULL is true is . False, the null is either true or not, there is no probability
here.
b. For a simple alternative, the power of the test is determined by the alternative
distribution of the test statistic. True, power is the chance the null is rejected
assuming the alternative is true.
c. The unknown parameter  in a test problem is a random variable. False, it is a
parameter.
d. The test statistic is a random variable.
which is random.
True, it is a function of your sample data,
e. If the p-value for a test statistic is .1, then the NULL hypothesis can be rejected at
level .05. False, reject the null if the p-value is smaller than , not bigger.
8. From long experience with a process for manufacturing an alcoholic beverage, it is
known that the yield is normally distributed with a mean of 500 units and a SD of 100
units. A modification of the process is suggested for which it is claimed that the mean
yield will increase (leaving the SD unchanged). With 49 observations, an average yield
of 535 units is observed. Using a 5% significance level, test the null hypothesis that the
mean yield remains unchanged at 500 units.
H: mean = 500
A: mean > 500
TS = z = (535 – 500)/(100/sqrt(49)) = 2.45.
P(Z > 2.45) = .007; reject the NULL in favor of the alternative.
.