solution manual of Statistics for Business and Economics Eleventh Edition PDF

Preview

Summary

modified 2/16/2010 EXCERPTS FROM: Solutions Manual to Accompany Statistics for Business and Economics Eleventh Edition David R. Anderson University of Cincinnati Dennis J. Sweeney University of Cincinnati Thomas A. Williams Rochester Institute of Technology The material from which this was excerpted...

Description

modified 2/16/2010

EXCERPTS FROM: Solutions Manual to Accompany

Statistics for Business and Economics Eleventh Edition

David R. Anderson University of Cincinnati

Dennis J. Sweeney University of Cincinnati

Thomas A. Williams Rochester Institute of Technology

The material from which this was excerpted is copyrighted by SOUTH-WESTERN CENGAGE LearningTM

Contents 1. Data and Statistics ....................................................................................................................... 1 2. Descriptive Statistics: Tabular and Graphical Methods.............................................................. 2 3. Descriptive Statistics: Numerical Methods................................................................................. 5 4. Introduction to Probability .......................................................................................................... 8 5. Discrete Probability Distributions............................................................................................. 11 6. Continuous Probability Distributions ....................................................................................... 13 7. Sampling and Sampling Distributions ...................................................................................... 15 8. Interval Estimation .................................................................................................................... 17 9. Hypothesis Testing.................................................................................................................... 18 10. Statistical Inference about Means and Proportions with Two populations............................. 22 14. Simple Linear regression ........................................................................................................ 25 15. Multiple Regression ................................................................................................................ 30 16. Regression Analysis: Model Building .................................................................................... 35 21. Decision Analysis ................................................................................................................... 37

1. Data and Statistics 12. a. b.

c.

21. a. b. c. d. e.

The population is all visitors coming to the state of Hawaii. Since airline flights carry the vast majority of visitors to the state, the use of questionnaires for passengers during incoming flights is a good way to reach this population. The questionnaire actually appears on the back of a mandatory plants and animals declaration form that passengers must complete during the incoming flight. A large percentage of passengers complete the visitor information questionnaire. Questions 1 and 4 provide quantitative data indicating the number of visits and the number of days in Hawaii. Questions 2 and 3 provide qualitative data indicating the categories of reason for the trip and where the visitor plans to stay. The two populations are the population of women whose mothers took the drug DES during pregnancy and the population of women whose mothers did not take the drug DES during pregnancy. It was a survey. 63 / 3.980 = 15.8 women out of each 1000 developed tissue abnormalities. The article reported “twice” as many abnormalities in the women whose mothers had taken DES during pregnancy. Thus, a rough estimate would be 15.8/2 = 7.9 abnormalities per 1000 women whose mothers had not taken DES during pregnancy. In many situations, disease occurrences are rare and affect only a small portion of the population. Large samples are needed to collect data on a reasonable number of cases where the disease exists.

1

2. Descriptive Statistics: Tabular and Graphical Methods 15. a/b. Waiting Time 0-4 5-9 10 - 14 15 - 19 20 - 24 Totals

Frequency 4 8 5 2 1 20

Relative Frequency 0.20 0.40 0.25 0.10 0.05 1.00

c/d. Waiting Time Less than or equal to 4 Less than or equal to 9 Less than or equal to 14 Less than or equal to 19 Less than or equal to 24 e.

Cumulative Frequency 4 12 17 19 20

Cumulative Relative Frequency 0.20 0.60 0.85 0.95 1.00

12/20 = 0.60

29. a. y

x

1

2

Total

A

5

0

5

B

11

2

13

C

2

10

12

Total

18

12

30

1

2

Total

A

100.0

0.0

100.0

B

84.6

15.4

100.0

C

16.7

83.3

100.0

b. y

x

2

c. y 1

2

A

27.8

0.0

B

61.1

16.7

C

11.1

83.3

Total

100.0

100.0

x

d.

Category A values for x are always associated with category 1 values for y. Category B values for x are usually associated with category 1 values for y. Category C values for x are usually associated with category 2 values for y.

50. a. Fuel Type Year Constructed Elec Nat. Gas Oil Propane Other 1973 or before 40 183 12 5 7 1974-1979 24 26 2 2 0 1980-1986 37 38 1 0 6 1987-1991 48 70 2 0 1 Total 149 317 17 7 14

Total 247 54 82 121 504

b. Year Constructed 1973 or before 1974-1979 1980-1986 1987-1991 Total c.

Frequency 247 54 82 121 504

Fuel Type Electricity Nat. Gas Oil Propane Other Total

Frequency 149 317 17 7 14 504

Crosstabulation of Column Percentages Fuel Type Year Constructed Elec Nat. Gas Oil Propane Other 1973 or before 26.9 57.7 70.5 71.4 50.0 1974-1979 16.1 8.2 11.8 28.6 0.0 1980-1986 24.8 12.0 5.9 0.0 42.9 1987-1991 32.2 22.1 11.8 0.0 7.1 Total 100.0 100.0 100.0 100.0 100.0

d.

Crosstabulation of row percentages. Year Constructed 1973 or before 1974-1979 1980-1986 1987-1991

Fuel Type Elec Nat. Gas Oil Propane Other 16.2 74.1 4.9 2.0 2.8 44.5 48.1 3.7 3.7 0.0 45.1 46.4 1.2 0.0 7.3 39.7 57.8 1.7 0.0 0.8

3

Total 100.0 100.0 100.0 100.0

e.

Observations from the column percentages crosstabulation For those buildings using electricity, the percentage has not changed greatly over the years. For the buildings using natural gas, the majority were constructed in 1973 or before; the second largest percentage was constructed in 1987-1991. Most of the buildings using oil were constructed in 1973 or before. All of the buildings using propane are older. Observations from the row percentages crosstabulation Most of the buildings in the CG&E service area use electricity or natural gas. In the period 1973 or before most used natural gas. From 1974-1986, it is fairly evenly divided between electricity and natural gas. Since 1987 almost all new buildings are using electricity or natural gas with natural gas being the clear leader.

4

3. Descriptive Statistics: Numerical Methods 5.

Σxi 3181 = = $159 20 n

a.

x=

b.

Median 10th $160 11th $162 Median =

c. d.

e.

19. a. b.

Los Angeles Seattle

160 + 162 = $161 2

Mode = $167 San Francisco and New Orleans

⎛ 25 ⎞ i=⎜ ⎟ 20 = 5 ⎝ 100 ⎠ 5th $134 6th $139 134 + 139 Q1 = = $136.50 2 ⎛ 75 ⎞ i=⎜ ⎟ 20 = 15 ⎝ 100 ⎠ 15th $167 16th $173 167 + 173 Q3 = = $170 2

Range = 60 - 28 = 32 IQR = Q3 - Q1 = 55 - 45 = 10 435 x= = 48.33 9 Σ( xi − x ) 2 = 742 Σ( xi − x ) 2 742 = = 92.75 s = 92.75 = 9.63 n −1 8 The average air quality is about the same. But, the variability is greater in Anaheim. s2 =

c. 34. a.

x=

Σxi 765 = = 76.5 10 n

Σ( xi − x ) 2 442.5 = =7 n −1 10 − 1 x − x 84 − 76.5 z= = = 1.07 s 7 Approximately one standard deviation above the mean. Approximately 68% of the scores are within one standard deviation. Thus, half of (100-68), or 16%, of the games should have a winning score of 84 or more points. x − x 90 − 76.5 z= = = 1.93 s 7 s=

b.

5

c.

Approximately two standard deviations above the mean. Approximately 95% of the scores are within two standard deviations. Thus, half of (100-95), or 2.5%, of the games should have a winning score of more than 90 points. Σx 122 x= i = = 12.2 n 10 Σ( xi − x ) 2 559.6 = = 7.89 n −1 10 − 1 x − x 24 − 12.2 = = 1.50 . No outliers. Largest margin 24: z = s 7.89 s=

50. a. 1

S&P 500

0.5

0 -1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

-0.5

-1 DJIA

b.

x=

Σxi 1.44 = = .16 n 9

y=

xi

yi

( xi − x )

0.20 0.82 -0.99 0.04 -0.24 1.01 0.30 0.55 -0.25

0.24 0.19 -0.91 0.08 -0.33 0.87 0.36 0.83 -0.16

0.04 0.66 -1.15 -0.12 -0.40 0.85 0.14 0.39 -0.41

( yi − y )

0.11 0.06 -1.04 -0.05 -0.46 0.74 0.23 0.70 -0.29 Total

Σxi 1.17 = = .13 n 9 ( xi − x ) 2 0.0016 0.4356 1.3225 0.0144 0.1600 0.7225 0.0196 0.1521 0.1681 2.9964

6

( yi − y ) 2 0.0121 0.0036 1.0816 0.0025 0.2166 0.5476 0.0529 0.4900 0.0841 2.4860

( xi − x )( yi − y )

0.0044 0.0396 1.1960 0.0060 0.1840 0.6290 0.0322 0.2730 0.1189 2.4831

sxy =

sx =

Σ( xi − x ) 2 = n −1

2.9964 = .6120 8

sy =

Σ( yi − y ) 2 = n −1

2.4860 = .5574 8

rxy =

c.

Σ( xi − x )( yi − y ) 2.4831 = .3104 = n −1 8

sxy sx s y

=

.3104 = .9098 (.6120)(.5574)

There is a strong positive linear association between DJIA and S&P 500. If you know the change in either, you will have a good idea of the stock market performance for the day.

7

4. Introduction to Probability 4.

a. 1st Toss

2nd Toss

3rd Toss H

(H,H,H)

T

H

(H,H,T)

T

H

H

T H

T

T

H T

H T

(H,T,H) (H,T,T) (T,H,H) (T,H,T) (T,T,H) (T,T,T)

b.

Let: H be head and T be tail (H,H,H) (T,H,H) (H,H,T) (T,H,T) (H,T,H) (T,T,H) (H,T,T) (T,T,T)

c.

The outcomes are equally likely, so the probability of each outcomes is 1/8.

7.

No. Requirement (4.4) is not satisfied; the probabilities do not sum to 1. P(E1) + P(E2) + P(E3) + P(E4) = .10 + .15 + .40 + .20 = .85

21. a.

Use the relative frequency method. Divide by the total adult population of 227.6 million. Age Number Probability 18 to 24 29.8 0.1309 25 to 34 40.0 0.1757 35 to 44 43.4 0.1907 45 to 54 43.9 0.1929 55 to 64 32.7 0.1437 65 and over 37.8 0.1661 Total 227.6 1.0000 P(18 to 24) = .1309 P(18 to 34) = .1309 + .1757 = .3066 P(45 or older) = .1929 + .1437 + .1661 = .5027

b. c. d. 26. a. b. c.

Let D = Domestic Equity Fund P(D) = 16/25 = .64 Let A = 4- or 5-star rating 13 funds were rated 3-star of less; thus, 25 – 13 = 12 funds must be 4-star or 5-star. P(A) = 12/25 = .48 7 Domestic Equity funds were rated 4-star and 2 were rated 5-star. Thus, 9 funds were Domestic Equity funds and were rated 4-star or 5-star P(D ∩ A) = 9/25 = .36

8

d. 28. a. b. 31. a. b. c. d. 34. a.

P(D ∪ A) = P(D) + P(A) - P(D ∩ A) = .64 + .48 - .36 = .76 Let: B = rented a car for business reasons P = rented a car for personal reasons P(B ∪ P) = P(B) + P(P) - P(B ∩ P) = .54 + .458 - .30 = .698 P(Neither) = 1 - .698 = .302 P(A ∩ B) = 0 P (A ∩ B) 0 = =0 P (A B) = P (B) .4 No. P(A | B) ≠ P(A); ∴ the events, although mutually exclusive, are not independent. Mutually exclusive events are dependent. Let O Oc S U J Given:

= flight arrives on time = flight arrives late = Southwest flight = US Airways flight = JetBlue flight P(O | S) = .834 P(O | U) = .751 P(S) = .40 P(U) = .35 P(O ∩ S) P(O | S) = P (S)

P(O | J) = .701 P(J) = .25

∴ P(O ∩ S) = P(O | S)P(S) = (.834)(.4) = .3336

b. c. d.

Similarly P(O ∩ U) = P(O | U)P(U) = (.751)(.35) = .2629 P(O ∩ J) = P(O | J)P(J) = (.701)(.25) = .1753 Joint probability table On time Late Total Southwest .3336 .0664 .40 US Airways .2629 .0871 .35 JetBlue .1753 .0747 .25 Total: .7718 .2282 1.00 Southwest Airlines; P(S) = .40 P(O) = P(S ∩ O) + P(U ∩ O) + P(J ∩ O) = .3336 + .2629 + .1753 = .7718 P(S ∩ Oc ) .0664 P(S Oc ) = = = .2910 .2282 P(Oc ) .0871 = .3817 .2282 .0747 P (J Oc ) = = .3273 .2282 Most likely airline is US Airways; least likely is Southwest

Similarly, P (U Oc ) =

42.

a. b.

M = missed payment D1 = customer defaults D2 = customer does not default P(D1) = .05 P(D2) = .95 P(M | D2) = .2 P(M | D1) = 1 P(D1 )P(M D1 ) (.05)(1) .05 P(D1 M) = = = = .21 P(D1 )P(M D1 ) + P(D 2 )P(M D 2 ) (.05)(1) + (.95)(.2) .24 Yes, the probability of default is greater than .20.

9

43.

Let: S = small car Sc = other type of vehicle F = accident leads to fatality for vehicle occupant We have P(S) = .18, so P(Sc) = .82. Also P(F | S) = .128 and P(F | Sc) = .05. Using the tabular form of Bayes Theorem provides: Prior Conditional Joint Posterior Probabilities Probabilities Probabilities Probabilities Events S .18 .128 .023 .36 Sc .82 .050 .041 .64 1.00 .064 1.00 From the posterior probability column, we have P(S | F) = .36. So, if an accident leads to a fatality, the probability a small car was involved is .36.

56. a. b. c. d. e.

P(A) = 200/800 = .25 P(B) = 100/800 = .125 P(A ∩ B) = 10/800 = .0125 P(A | B) = P(A ∩ B) / P(B) = .0125 / .125 = .10 No, P(A | B) ≠ P(A) = .25

59. a. b.

P(Oil) = .50 + .20 = .70 Let S = Soil test results Events High Quality (A1) Medium Quality (A2) No Oil (A3)

P(Ai) .50 .20 .30 1.00

P(S | Ai) .20 .80 .20

P(Ai ∩ S) .10 .16 .06 P(S) = .32

P(Ai | S) .31 .50 .19 1.00

P(Oil) = .81 which is good; however, probabilities now favor medium quality rather than high quality oil. 60. a. b.

Let Let

F = female. Using past history as a guide, P(F) = .40. D = Dillard's ⎛3⎞ .40 ⎜ ⎟ .30 ⎝4⎠ = .67 = P(F D) = ⎛3⎞ ⎛ 1 ⎞ .30 + .15 .40 ⎜ ⎟ + .60 ⎜ ⎟ ⎝4⎠ ⎝4⎠

The revised (posterior) probability that the visitor is female is .67. We should display the offer that appeals to female visitors.

10

5. Discrete Probability Distributions 2.

a. b. c.

14. a. b. c.

Let x = time (in minutes) to assemble the product. It may assume any positive value: x > 0. Continuous f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150) = 1 - .95 = .05 This is the probability MRA will have a $200,000 profit. P(Profit) = f (50) + f (100) + f (150) + f (200) = .30 + .25 + .10 + .05 = .70 P(at least 100) = f (100) + f (150) + f (200) = .25 + .10 +.05 = .40

19. a. b. c.

E(x) = Σ x f (x) = 0 (.56) + 2 (.44) = .88 E(x) = Σ x f (x) = 0 (.66) + 3 (.34) = 1.02 The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will make more points in the long run with the 3 - point shot.

24. a.

Medium E (x) = Σ x f (x) = 50 (.20) + 150 (.50) + 200 (.30) = 145 Large: E (x) = Σ x f (x) = 0 (.20) + 100 (.50) + 300 (.30) = 140 Medium preferred. Medium x f (x) (x - μ)2 (x - μ)2 f (x) x-μ 50 .20 -95 9025 1805.0 150 .50 5 25 12.5 200 .30 55 3025 907.5 σ2 = 2725.0 Large y f (y) (y - μ)2 (y - μ)2 f (y) y-μ 0 .20 -140 19600 3920 100 .50 -40 1600 800 300 .30 160 25600 7680 σ2 = 12,400 Medium preferred due to less variance.

b.

26. a. b. c. d. e. f. 29. a.

f (0) = .3487 f (2) = .1937 P(x ≤ 2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298 P(x ≥ 1) = 1 - f (0) = 1 - .3487 = .6513 E (x) = n p = 10 (.1) = 1 σ = .9 = .9487 Var (x) = n p (1 - p) = 10 (.1) (.9) = .9, ⎛n⎞ f ( x ) = ⎜ ⎟ ( p) x (1 − p) n − x ⎝ x⎠ 10! f (3) = (.30)3 (1 − .30)10 −3 3!(10 − 3)! f (3) =

b.

10(9)(8) (.30)3 (1 − .30) 7 = .2668 3(2)(1)

P(x > 3) = 1 - f (0) - f (1) - f (2)

11

f (0) =

10! (.30)0 (1 − .30)10 = .0282 0!(10)!

f (1) =

10! (.30)1 (1 − .30)9 = .1211 1!(9)!

f (2) =

10! (.30) 2 (1 − .30)8 = .2335 2!(8)!

P(x > 3) = 1 - .0282 - .1211 - .2335 = .6172 39. a. b. c. d. e. f. 58. a.

b. c. d.

2 x e −2 x! μ = 6 for 3 time periods 6 x e −6 f ( x) = x! 2 −2 2 e 4(.1353) = = .2706 f (2) = 2! 2 66 e −6 = .1606 f (6) = 6! 45 e −4 = .1563 f (5) = 5! f ( x) =

Since the shipment is large we can assume that the probabilities do not change from trial to trial and use the binomial probability distribution. n = 5 ⎛5⎞ f (0) = ⎜ ⎟ (0.01)0 (0.99)5 = 0.9510 ⎝0⎠ ⎛5⎞ f (1) = ⎜ ⎟ (0.01)1 (0.99) 4 = 0.0480 ⎝1⎠ 1 - f (0) = 1 - .9510 = .0490 No, the probability of finding one or more items in the sample defective when only 1% of the items in the population are defective is small (only .0490). I would consider it likely that more than 1% of the items are defective.

12

6. Continuous Probability Distributions 2.

a.

f (x) .15 .10 .05 x 0

b. c. d. e. 9.

10

20

30

40

P(x < 15) = .10(5) = .50 P(12 ≤ x ≤ 18) = .10(6) = .60 10 + 20 E ( x) = = 15 2 (20 − 10) 2 Var( x) = = 8.33 12

a. σ =5

35

b. c.

40

45

50

55

60

65

.683 since 45 and 55 are within plus or minus 1 standard deviation from the mean of 50 (Use the table or see characteristic 7a of the normal distribution). .954 since 40 and 60 are within plus or minus 2 standard deviations from the mean of 50 (Use the table or see characteristic 7b of the normal distribution).

13. a. b. c.

P(-1.98 ≤ z ≤ .49) = P(z ≤ .49) - P(z < -1.98) = .6879 - .0239 = .6640 P(.52 ≤ z ≤ 1.22) = P(z ≤ 1.22) - P(z < .52) = .8888 - .6985 = .1903 P(-1.75 ≤ z ≤ -1.04) = P(z ≤ -1.04) - P(z < -1.75) = .1492 - .0401 = .1091

15. a. b.

The z value corresponding to a cumulative probability of .2119 is z = -.80. Compute .9030/2 = .4515; z corresponds to a cumulative probability of .5000 + .4515 = .9515. So z = 1.66. Compute .2052/2 = .1026; z corresponds to a cumulative probability of .5000 + .1026 = .6026. So z = .26. The z value corresponding to a cumulative probability of .9948 is z = 2.56. The area to the left of z is 1 - .6915 = .3085. So z = -.50.

c. d. e. 41. a. b.

P(defect) = 1 - P(9.85 ≤ x ≤ 10.15) = 1 - P(-1 ≤ z ≤ 1) = 1 - .6826 = .3174 Expected number of defects = 1000(.3174) = 317.4 P(defect) = 1 - P(9.85 ≤ x ≤ 10.15) = 1 - P(-3 ≤ z ≤ 3) = 1 - .9974 = .0026

13

c.

Expected number of defects = 1000(.0026) = 2.6 Reducing the process standard deviation causes a substantial reduction in the number of defects.

14

7. Sampling and Sampling Distributions 3.

459, 147, 385, 113, 340, 401, 215, 2, 33, 348

19. a.

The sampling distribution is normal with E ( x ) = μ = 200 and σ x = σ / n = 50 / 100 = 5 For ± 5, 195 ≤ x ≤ 205 . Us...