Download Homework2 Solution

Stat 503
Solutions to Homework #2 (32 points)
Problem 2.57
The mean and SD of a set of 47 body temperature measurements were as follows:
⎯y = 36.497 °C
s = 0.172 °C
If the 47 measurements were converted to °F,
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(a) What would be the new mean and SD?
The conversion from Celsius to Fahrenheit is F = y’ = 1.8 y + 32.
The new mean would be ⎯y’ = 1.8⎯y + 32 = 1.8(36.497) + 32 = 65.6946 + 32 = 97.6946 ≈ 97.7.
The new SD would be s’ = 1.8 s = 1.8(0.172) = 0.3096 ≈ 0.31
(b) What would be the new coefficient of variation?
The new coefficient of variation would be s’/⎯y’ = 0.3096/97.6946 = 0.003169 ≈ 0.0032.
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Problem 3.1 (Sampling exercise)
Refer to the collection of 100 ellipses shown in the accompanying figure, which can be thought of as
representing a natural population of the mythical organism C. ellipticus. The ellipses have been
given identification numbers 00, 01, …, 99 for convenience in sampling. Certain individuals of C.
ellipticus are mutants and have two tail bristles.
a) Use your judgment to choose a sample of size 10 from the population that you think is
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representatitve of the entire population. Note the number of mutants in the sample.
(Any) ten numbers from the range, e.g.: 03, 10, 30, 35, 42, 70, 64, 66, 84, 95.
Number of mutants (here): 6.
(Grader: do not count mutants.)
(Comment: there are only 39 mutants on the picture. Chances that they are overrepresented in the sample
chosen by judgment!)
b) Use random digits (from Table 1 or you calculator or comuter) to choose a random sample of
size 10 from the population and note the number of mutants in the sample.
One may use Table 1, starting at the first row, reading pairs of digits:
06 04 89 60 63 22 (04-cancelled because repeated) 98 65 32 75.
Number of mutants: 3.
(Grader: any ten numbers in the range, do not count mutants.)
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Problem 3.5
In a certain population of the freshwater sculpin, Cottus rotheus, the distribution of the number of tail
vertebra is as shown in the table.
No. of
Vertebrae
Percent of Fish
20
3
21
51
22
40
23
6
Total
100
Find the probability that the number of tail vertebrae in a fish randomly chosen from the population
(one point each)
(a) equals 21
51/100 = 0.51.
(b) is less than or equal to 22
(3+51+40)/100= 94/100 = 0.94
(c) is greater than 21
40/100 + 6/100 = 46/100 = 0.46
(d) is no more than 21
(same as <= 21) = 3/100 + 51/100 = 54/100 = 0.54.
Problem 3.9
If a woman takes an early pregnancy test, she will either test positive, meaning that the test says she
is pregnant, or test negative, meaning that the test says she is not pregnant. Suppose that if a woman
really is pregnant, there is a 98% chance that she will test positive. Also, suppose that if a woman
really is not pregnant, there is a 99% chance that she will test negative.
a) (not required)
b) Suppose that 1,000 women take early pregnancy tests and that 50 of them really are pregnant.
What is the probability that a randomly chosen woman from this group will test positive?
Draw a probability tree. (see below)
Pr(test positive) = Pr(true positive) + Pr(false positive) = Pr(pregnant and positive) + Pr(not pregnant and
positive) = 0.049 + 0.0095 = 0.0585.
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EVENT
PROBABILITY
positive
true
positive
0.049
negative
false
negative
0.001
positive
false
positive
0.0095
negative
true
negative
0.9405
0.98
pregnant
0.05
0.02
0.01
0.95
not
pregnant
0.99
3.11 (p 88)
Suppose that a medical test has a 92% chance of detecting a disease if a person really has it (i.e.
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92% sensitivity) and a 94% chance of correctly indicating that a disease is absent if the person really
does not have the disease (i.e. 94% specificity). Suppose that 10% of the population has the
disease.
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(c) What is the probability that a random person will test positive?
Using a tree diagram should find Pr(Test Positive) = 0.146.
(d) Suppose a randomly selected person tests positive. What is the probability that this person
really has the disease?
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Pr(Disease | Test Positive) = 0.092/0.146 = 0.63
3.12 (p 92)
In a study of the relationship between health risk and income a large group of people living in
Massachusetts was asked a series of questions. Some of the results are shown in the following table.
Smoke
Don’t Smoke
Total
Low
634
1846
2480
Income
Medium
332
1622
1954
High
247
1868
2115
Total
1213
5336
6549
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(a) What is the probability that someone in this study smokes?
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Pr(smoke) = 1213/6549 = 0.1852
(b) What is the conditional probability that someone in this study smokes given that this person has
a high income?
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Pr(smoke | high income) = Pr(smoke and high income) / Pr(high income) = 247/2115=0.1168.
(c) Is being a smoker independent of having a high income? Why or why not?
No, because Pr(smoke | high income) ≠ Pr(smoke).
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