Stat 321 Exam
1 October 5, 2004
Please write clearly on your own paper; you will also be asked to hand in this sheet. You may use a calculator, notes, and text. You have 50 minutes to complete the exam. You are to work completely independently on all aspects of the exam.
For questions that call for calculations, present your method of solution in a clear, well-labeled manner and show the details of your calculations. For questions that ask for interpretations and explanations, explain your answers fully unless instructed otherwise.
For problems with multiple parts, be aware that you can usually complete later parts successfully whether or not you do so on earlier parts. Tentative point allocations are provided to help you budget your time.
1. (24 pts) Suppose that a high school senior has applied to two colleges: Cal Poly and UCLA. Her guidance counselor informs her that she has probability .60 of getting into UCLA, .75 of getting into at least one college, and .20 of getting into exactly one college. Let C denote the event that she gets into Cal Poly and U denote the event that she gets into UCLA.
[Note: You may have had different
but similar numbers.]
a) (6 pts) Translate the three given probabilities into set notation involving C and U.
P(U) = .6, P(U È C) = .75, P[(U'ÇC) È (UÇC')]
b) (6 pts) What is the probability that she gets into Cal Poly?
We know that P(U È C) = .75, so P(U'ÇC') = 1- P(UÈC) = .25. This enables
us to fill in the probability table as:
|
|
Get into Cal Poly |
Do not get into Cal Poly |
Total |
|
Get into UCLA |
|
|
.6 |
|
Do not get into UCLA |
.15 |
.25 |
.4 |
|
Total |
|
|
1.0 |
Now, knowing that the probability of
acceptance to exactly one school is .2, we can fill in P(UÇC')=.05 and can complete the table as:
|
|
Get into Cal Poly |
Do not get into Cal Poly |
Total |
|
Get into UCLA |
.55 |
.05 |
.6 |
|
Do not get into UCLA |
.15 |
.25 |
.4 |
|
Total |
.7 |
.3 |
1.0 |
Thus, P(C) = .7.
c) (6 pts) Are the events C and U mutually exclusive? Explain.
No, because P(UÇC) > 0. In other
words, it’s possible for the student to be accepted by both schools.
d) (6 pts) Are the events C and U independent? Justify your answer numerically.
To check for independence, we can check whether P(UÇC) = P(U) ´ P(C). But P(UÇC) = .55 and P(U) ´ P(C) = (.6)(.7) = .42, so these events are not independent.
2. (12 pts) Suppose that an ice cream parlor comes up with a clever promotion: customers roll two fair, six-sided dice, and the price of a small cone (in cents) is the larger number followed by the smaller number, in cents. For example, if you roll a 3 and a 5, in either order, the price would be 53 cents. [Note that there are 6 × 6 = 36 equally likely outcomes in the sample space.]
[Note: You may have been asked about different but similar events.]
a) (6 pts) Determine the probability that the price would exceed 50 cents.
One way to count how many of the 36
possible outcomes produce a price greater than 50 cents is to recognize that
the price exceeds 50 cents whenever at least one die lands on 5 or 6. Thus, the price will be less than 50 cents
whenever both dice land on numbers 1-4.
There are 4 ´ 4 = 16
ways for this to happen. Thus, there are
20 outcomes for which the price exceeds 50 cents, so this probability is 20/36
= 5/9, or about .556.
b) (6 pts) Determine the probability that the price is an odd number.
The price is an odd number for the
following 21 outcomes: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1),
(3,3), (3,4), (3,5), (3,6), (4,1), (4,3), (5,1), (5,3), (5,5), (5,6), (6,1),
(6,3), (6,5). Thus, the probability of
an odd price is 21/36 = 7/12, or about .583.
3. (10 pts) Suppose that in a certain population, 4.23% of the males are color-blind and 0.65% of the females are color-blind. Assume that the population contains 50% males and 50% females. If you select a person at random from this population and find that the person is color-blind, what is the conditional probability that the person is male? Also identify by name the probability rule that you use.
Using self-evident symbols, we want
to find P(M|C), which Bayes’ Theorem reveals can be found as 
, which becomes 
, which is about .867.
4. (26 pts) In a recent study, researchers presented bank managers with identical personnel files and asked them whether the candidate was worthy of promotion. Half of the managers were randomly assigned to receive a file with a male name attached, and the other half were randomly assigned to receive a file with a female name attached. The researchers suspected that the managers would be more likely to recommend promotion for candidates with male names. The resulting experimental data are summarized in the following two-way table:
|
|
Male name |
Female name |
Total |
|
Promotion |
21 |
14 |
35 |
|
No promotion |
3 |
10 |
13 |
|
Total |
24 |
24 |
48 |
a) (8 pts) Consider conducting Fisher’s exact test on these data. Without doing the calculation, write an expression for the p-value that involves combinations and/or permutations.
The p-value can be found as 
.
A simulation analysis (of 10,000 randomizations) to assess the statistical significance of these results produced the following histogram:
b) (6 pts) Based on these simulation results, report the approximate p-value of the test.
A total of 203+33+3 = 239 of the
10,000 randomizations produced a result as extreme as in the actual study (21
or more promotions in the “male” group), so the approximate p-value is .0239.
c) (6 pts) Write a sentence or two interpreting what this p-value represents.
This p-value says that if there were
no effect of the name’s gender, there’s only a 2.39% chance of obtaining such
an extreme result as in this study.
d) (6 pts) Do the experimental data provide reasonably strong evidence that the gender of the name on the file affected how the bank manager assessed the candidate’s qualifications? Explain how your conclusion follows from the simulation analysis.
Because the p-value is quite small
(.0239), it suggests that these results would be unlikely to occur by chance
alone, if there were no effect of the name’s gender. Thus, the data provide fairly string evidence
that the gender of the name did affect the manager’s assessment of the candidate’s
qualifications.
5. (18 pts) When a pair of fair, six-sided dice are rolled, the probability is 8/36 (or 2/9) that the resulting sum is a 7 or 11. Suppose that five independent rolls are made.
a) (6 pts) Determine the probability that a sum of 7 or 11 is obtained on at least one of these rolls.
Let Ai denote the event that roll i results
in a 7 or 11. We want P(A1 È A2 È A3 È A4 È A5), but
the complement is that none of the rolls produce a 7 or 11, so we can find 1-P(A1'
Ç A2' Ç A3' Ç A4' Ç A5'). Because the
rolls are independent, this is 1 – (7/9)^5, which is about .7154.
b) (6 pts) Determine the probability that a sum of 7 or 11 is obtained on every one of these rolls.
We want P(A1 Ç A2 Ç A3 Ç A4 Ç A5), which is (2/9)^5, or about .00054.
c) (6 pts) If your instructor rolls a pair of six-sided dice five times, and every roll results in a sum of 7 or 11, would you have reason to believe that the dice were not fair? Explain your reasoning, based on a probabilistic argument.
Obtaining five consecutive rolls of
7 or 11 would occur only about .054% of the time in the long run with fair
dice. Because such a result is so
unlikely, it would provide very strong evidence that the dice are not fair.
6. (10 pts) Consider the following data on flights on-time and delayed for two airlines at five airports in one month:
|
|
On time |
Delayed |
% on-time |
On time |
Delayed |
% on-time |
|
|
221 |
12 |
94.8 |
4840 |
415 |
92.1 |
|
|
1841 |
305 |
85.8 |
201 |
61 |
76.7 |
|
Total |
2062 |
317 |
86.7 |
5041 |
476 |
91.4 |
Notice that Alaska Airlines has a higher on-time percentage than America West for both airports, yet America West Airlines has a higher on-time percentage than Alaska Airlines overall. Explain how this apparent paradox happens, as if to a peer with no knowledge of statistics, basing your argument on the data provided.
The explanation involves two
components: (1) America West had most of its flights to