1. An IBD/TIPP poll conducted to learn about attitudes toward investment and retirement, asked male and female respondents how important they felt level of risk was in choosing a retirement investment. The following joint probability table was constructed from the data provided. “Important” me ans the respondent said level of risk was either important or very important. Male
Total
Important
0.22
0.27
0.49
Not Important
0.28
0.23
0.51
0.5
0.5
Total
a. b. c. d. e.
Female
What is the probability a survey respondent will say level of risk is important? What is the probability a male respondent r espondent will say level of risk is important? What is the probability a female respondent will say level of risk is important? Is the level of risk independent of the gender of the respondent? r espondent? Why or why not? Do male and female attitudes towards risk differ?
important or very important important Solution: Let I = important M = male F = female a. P (I) (I) = .49 (a marginal probability) b. P (I (I | M) = .22/.50 = .44 (a conditional probability) probability) c. P (I (I | F) = .27/.50 = .54 (a conditional probability) probability) d. It is not independent (I) = .49 P (I (I | M) = .44 P (I) and (I) = .49 P (I (I | F) = .54 P (I) e. Since level of importance is dependent on gender, we we conclude that that male and female respondents have different attitudes toward risk. 2. A local bank reviewed its credit card policy with the intension of recalling some of its credit
cards. In the past approximately 5% of cardholders defaulted, leaving the bank unable to collect the outstanding balance. Hence management established a prior probability of .05 that any particular cardholder will default. The bank also found that the probability of missing a monthly payment is .20 for customers who do not not default. Of course, the probability probability of missing a monthly monthly payment for those who default default is 1. a. Given that a customer missed one or more monthly payments, compute the posterior probability that the the customer will default. b. The bank would like to recall its card if the probability that a customer will default is greater than .20. Should the bank bank recall its card if the customer customer misses a monthly payment? Why or why not? not?
Solution:
M = missed payment D1 = customer defaults D2 = customer does not default
P (D1) = .05
a.
P (D1 M)
P (D2) = .95
P (M | D2) = .2
P(D1 ) P (M D1 ) P(D1 ) P(M D1 ) P(D 2 ) P (M D 2 )
P (M | D1) = 1 (.05)(1)
(.05)(1) (.95)(.2)
.05 .24
.21
b. Yes, the probability of default is greater than .20. 3. A survey of magazine subscribers showed that 45.8% rented a car during the past 12 months for
business reasons, 54% rented a car during the past 12 months for personal reasons , and 30% rented a car during the past 12 months for both business and personal reasons. a. What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? b. What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons? Solution: Let:
B = rented a car for business reasons P = rented a car for personal reasons a. P (B P) = P (B) + P (P) - P (B P) = .54 + .458 - .30 = .698
b. P (Neither) = 1 - .698 = .302 4. A study of job satisfaction was conducted for four occupations: Cabinetmaker, lawer, physical
therapist, and systems analyst. Job satisfaction was measured on a scale of 0-100. The data obtained are summarized in the following cross tabulation. Occupation Cabinetmaker Lawyer Physical Therapist Systems Analyst
Under 50 0 6 0 2
50 - 59 2 2 5 1
60 -69 4 1 2 4
70 – 79 3 1 1 3
80 – 89 1 0 2 0
a. Develop a joint probability table. b. What is the probability one of the participants studied received a satisfaction score in the 80s? c. What is the probability of a satisfaction score in the 80s given the study participant was a physical therapist? d. What is the probability one of the participants studied was a lawyer? e. What is the probability one of the participants was a lawyer and received a score under 50? f. What is the probability of receiving a satisfaction score under 50 given a person is a lawyer? g. What is the probability of a satisfaction score 70 or higher? h. What is the probability one of the participants studied received a satisfaction score in 90s?
Solution:
a. Satisfaction Score Occupation Under 50 50-59 60-69 Cabinetmaker .000 .050 .100 Lawyer .150 .050 .025 Physical Therapist .000 .125 .050 Systems Analyst .050 .025 .100 Total .200 .250 .275
70-79 .075 .025 .025 .075 .200
80-89 .025 .000 .050 .000 .075
Total .250 .250 .250 .250 1.000
b. P(80s) = .075 (a marginal probability) c.
P(80s | PT) = .050/.250 = .20 (a conditional probability)
d. P(L) = .250 (a marginal probability) e.
P(L Under 50) = .150 (a joint probability)
f.
P(Under 50 | L) = .150/.250 = .60 (a conditional probability)
g. P(70 or higher) = .275 (Sum of marginal probabilities) h.
P (90s) = 0
5. Business Week conducted a survey of graduates from 30 top MBA programs. On the basis of the
survey, assume that the mean annual salary for male and female graduates 10 years after graduation is $168,000 and $117,000, respectively. Assume the standard deviation for the male graduates is $40,000, and for female graduates it is $25,000. a. What is the probability that a simple random sample of 40 male graduates will provide a sample mean within $10,000 of the population mean, $168,000? b. What is the probability that a simple random sample of 40 female graduates will provide a sample mean within $10,000 of the population mean, $117,000? c. In which of the preceding two cases, part (a) or part (b), do we have a higher probability of obtaining a sample estimate within $10,000 of the population mean? Why? d. What is the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean? Solution
a.
x
/
n
40, 000 / 40
At x = 178,000, z
178,000 168,000
At x = 158,000, z = -1.58 P(z
6324.56
< -1.58) = .0571, thus
6324.56
1.58
P(z
≤ 1.58) = .9429
P(158,000
b.
x
/
≤ x ≤ 178,000) = .9429 - .0571 = .8858
n
25, 000 / 40
At x = 127,000, z
3 952.85
127,000 117,000 3952.85
At x = 107,000, z = -2.53, P(107,000
c.
P(z
2.53
P(z
≤ 2.53) = .9943
< -2.53) = .0057, thus
≤ x ≤ 127,000) = .9943 - .0057 = .9886
In part (b) we have a higher probability of obtaining a sample mean within $10,000 of the population mean because the standard error is smaller.
d. With n = 100,
/
x
At x = 164,000, z P( x
n
40, 000 / 100
164,000 168,000 4000
4000
1
< 164,000) = P(z < -1) = .1587
Therefore, the probability that a simple random sample of 100 male graduates will provide a sample mean more than $4,000 below the population mean is (1-0.1587 = 0.8413) 6. To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain. b. if the population standard deviation is 8.2 years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever n / N .05 ?
Solution
a. n / N = 40 / 4000 = .01 < .05; therefore, the finite population correction factor is not necessary. b. With the finite population correction factor
x
N
n
N 1
n
4000 40 8.2 4000 1
40
. 129
Without the finite population correction factor
x
/
n
. 130
Including the finite population correction factor provides only a slightly different value for than when the correction factor is not used. x
7. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = .17 and a simple random sample of 800 households will be selected from the population. a. Show the sampling distribution of p , the sample proportion of households spending more than $100 per week on groceries. b. What is the probability that the sample proportion will be within .02 of the population proportion? c. Answer part (b) for a sample of 1600 households. Solution
a. E ( p ) = .17
p
p (1 p)
(.17)(1 .17) 800
n
.0133
Distribution is approximately normal because np = 800(.17) = 136 > 5 and n(1 – p) = 800(.83) = 664 > 5 b.
z
.19 .17 .0133
P ( z ≤ 1.51) = .9345
1.51
P ( z < -1.51) = .0655 P (.15
c.
p
z
p
.19) = P (-1.51 z 1.51) = .9345 - .0655 = .8690
p (1 p)
(.17)(1 .17)
n
.19 .17 .0094
2.13
1600
.0094
P ( z ≤ 2.13) = .9834
P ( z < -2.13) = .0166 P (.15
p
.19) = P (-2.13 z 2.13) = .9834 - .0166 = .9668
8. Playbill magazine reported that the mean annual household income of its readers is $119,155. Assume this estimate of the mean annual household income is based on a sample of 80 households, and based on past studies, the population standard deviation is known to be $30,000. a. Develop a 90% confidence interval estimate of the population mean. b. Develop a 95% confidence interval estimate of the population mean. c. Develop a 99% confidence interval estimate of the population mean. d. Discuss what happens to the width of the confidence interval as a confidence level is increased. Does this result seem reasonable? Explain.
Solution
a.
x z /2
n
119,155
1.645 (30,000/ 80)
119,155
5517 or $113,638 to $124,672
1.96 (30,000/ 80)
6574 or $112,581 to $125,729
2.576 (30,000/ 80)
8640 or $110,515 to $127,795
b. 119,155 119,155 c. 119,155 119,155
d. The confidence interval gets wider as we increase our confidence level. We need a wider interval to be more confident that it will contain the population mean.
9. Is your favorite TV program often interrupted by advertising? CNBC presented statistics on the average number of programming minutes in a half hour sitcom. The following data (in minutes) are representative of their findings. 21.06
22.24
20.62
21.66
21.23
23.86
23.82
20.3
21.52
21.52
21.91
23.14
20.02
22.2
21.2
22.37
22.19
22.34
23.36
23.44
Assume the population is approximately normal. Provide a point estimate and a 95% confidence interval for the mean number of programming minutes during a half-hour television sitcom. Solution x
xi
s
x
/ n 22 minutes
( xi x )
n 1
2
1.12
minutes
t .025 ( s / n ) df = 19
22.00
2.093 (1.12/ 20)
22.00
.52 or 21.48 to 22.52 minutes
