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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.

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.

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

Total

100.0

0.0

100.0

84.6

15.4

100.0

16.7

83.3

100.0

b. y

c. y 1

27.8

0.0

61.1

16.7

11.1

83.3

Total

100.0

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

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

Total 100.0 100.0 100.0 100.0

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.

3. Descriptive Statistics: Numerical Methods 5.

Σxi 3181 = = $159 20 n

Median 10th $160 11th $162 Median =

c. d.

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.

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=

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

Σxi 1.44 = = .16 n 9

( 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

( 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 =

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.

4. Introduction to Probability 4.

a. 1st Toss

2nd Toss

3rd Toss H

(H,H,H)

(H,H,T)

T H

H T

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

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)

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

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

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.

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.

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.

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) =

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

P(x > 3) = 1 – f (0) – f (1) – f (2)

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.

6. Continuous Probability Distributions 2.

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

b. c. d. e. 9.

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

b. c.

.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

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

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 . Using Standard Normal Probability Table: x −μ 5 At x = 205, z = = = 1 P ( z ≤ 1) = .8413 σx 5 At x = 195, z =

x −μ

σx

−5 = −1 P ( z < −1) = .1587 5

P (195 ≤ x ≤ 205) = .8413 – .1587 = .6826

For ± 10, 190 ≤ x ≤ 210 . Using Standard Normal Probability Table: x − μ 10 At x = 210, z = = = 2 P ( z ≤ 2) = .9772 5 σx At x = 190, z =

x −μ

σx

−10 5

= −2 P ( z < −2) = .0228

P (190 ≤ x ≤ 210) = .9772 – .0228 = .9544

37. a.

Normal distribution: E ( p ) = .12 , σ p = z=

p− p

p (1 − p ) = n

.03 = 2.14 .0140

P(z < -2.14) = .0162

.015 = 1.07 .0140

P(z < -1.07) = .1423

P(z ≤ 2.14) = .9838 σp P(.09 ≤ p ≤ .15) = .9838 – .0162 = .9676

44. a.

53. a. b.

p− p

(.12)(1 − .12) = .0140 540

P(z ≤ 1.07) = .8577 σp P(.105 ≤ p ≤ .135) = .8577 – .1423 = .7154

Normal distribution because of central limit theorem (n > 30) σ 35 = = 5.53 E ( x ) = 115.50 , σ x = 40 n x −μ 10 P(z ≤ 1.81) = .9649, P(z < -1.81) = .0351 z= = = 1.81 σ / n 35 / 40 P(105.50 ≤ x ≤ 125.50) = P(-1.81 ≤ z ≤ 1.81) = .9649 – .0351 = .9298 100 − 115.50 At x = 100, z = P( x ≤ 100) = P(z ≤ -2.80) = .0026 = −2.80 35 / 40 Yes, this is an unusually low spending group of 40 alums. The probability of spending this much or less is only .0026. Normal distribution with E ( p ) = .15 and σ p = P (.12 ≤ p ≤ .18) = ?

p(1 − p) = n

(015 . )(0.85) = 0.0292 150

.18 − .15 P(z ≤ 1.03) = .8485, P(z < -1.03) = .1515 = 1.03 .0292 P(.12 ≤ p ≤ .18) = P(-1.03 ≤ z ≤ 1.03) = .8485 – .1515 =.6970 z=

8. Interval Estimation 7.

Margin of error = z.025 (σ / n ) = 1.96(600/ 50 ) = 166.31 A larger sample size would be needed to reduce the margin of error to $150 or less. Section 8.3 can be used to show that the sample size would need to be increased to n = 62. 1.96(600 / n ) = 150 Solving for n yields n = 62

14.

x ± tα / 2 ( s / n )

df = 53

22.5 ± 1.674 (4.4 / 54) 22.5 ± 1 or 21.5 to 23.5 22.5 ± 2.006 (4.4 / 54) 22.5 ± 1.2 or 21.3 to 23.7 22.5 ± 2.672 (4.4 / 54) 22.5 ± 1.6 or 20.9 to 24.1 As the confidence level increases, there is a larger margin of error and a wider confidence interval.

For the JobSearch data set, x = 22 and s = 11.8862 x = 22 weeks

b. c.

18. b. c. d.

29. a. b. 34.

margin of error = t.025 s / n = 2.023(11.8862) / 40 = 3.8020 The 95% confidence interval is x ± margin of error = 22 ± 3.8020 or 18.20 to 25.80 Skewness = 1.0062, data are skewed to the right. This modest positive skewness in the data set can be expected to exist in the population. Regardless of skewness, this is a pretty small data set. Consider using a larger sample next time. (196 . ) 2 (6.25) 2 n= = 37.52 Use n = 38 22 (196 . ) 2 (6.25) 2 n= = 150.06 Use n = 151 12 Use planning value p* = .50 (196 . ) 2 (0.50)(0.50) n= = 1067.11 Use n = 1068 (0.03) 2

36. a. b.

p = 46/200 = .23

p (1 − p ) .23(1 − .23) = = .0298 , p ± z.025 n 200 = .23 ± .0584 or .1716 to .2884

p (1 − p ) = .23 ± 1.96(.0298) n

39. a.

2 z.025 p∗ (1 − p∗ ) (1.96) 2 (.156)(1 − .156) = = 562 E2 (.03) 2

2 z.005 p∗ (1 − p∗ ) (2.576) 2 (.156)(1 − .156) = = 970.77 Use 971 E2 (.03) 2

9. Hypothesis Testing 1.

a. b. c.

H0: μ ≤ 600 Ha: μ > 600 assuming that you give benefit of doubt to the manager. We are not able to conclude that the manager’s claim is wrong. The manager’s claim can be rejected. We can conclude that μ > 600.

a. b. c.

H0: μ ≤ 14 Ha: μ > 14 Research hypothesis There is no statistical evidence that the new bonus plan increases sales volume. The research hypothesis that μ > 14 is supported. We can conclude that the new bonus plan increases the mean sales volume.

H0: μ ≤ 8000 Ha: μ > 8000 Research hypothesis to see if the plan increases average sales. Claiming μ > 8000 when the plan does not increase sales. A mistake could be implementing the plan when it does not help. Concluding μ ≤ 8000 when the plan really would increase sales. This could lead to not implementing a plan that would increase sales.

b. c.

10. a. b. c. d.

x − μ0 26.4 − 25 = = 1.48 σ/ n 6 / 40

Upper tail p-value is the area to the right of the test statistic Using normal table with z = 1.48: p-value = 1.0000 – .9306 = .0694 p-value > .01, do not reject H0 Reject H0 if z ≥ 2.33 1.48 < 2.33, do not reject H0 x − μ0

= −1.54 s / n 4.5 / 48 b. Degrees of freedom = n – 1 = 47 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between .05 and .10; therefore, p-value is between .10 and .20. Exact p-value corresponding to t = -1.54 is .1303 c. p-value > .05, do not reject H0. d. With df = 47, t.025 = 2.012 Reject H0 if t ≤ -2.012 or t ≥ 2.012 t = -1.54; do not reject H0

30. a.

H0: μ = 600, Ha: μ ≠ 600 x − μ0 612 − 600 df = n – 1 = 39 = = 1.17 t= s/ n 65 / 40 Because t > 0, p-value is two times the upper tail area Using t table: area in upper tail is between .10 and .20; therefore, p-value is between .20 and .40. Exact p-value corresponding to t = 1.17 is .2491 With α = .10 or less, we cannot reject H0. We are unable to conclude there has been a change in the mean CNN viewing audience. The sample mean of 612 thousand viewers is encouraging but not conclusive for the sample of 40 days. Recommend additional viewer audience data. A larger sample should help clarify the situation for CNN.

c. d.

34. a. b.

17 − 18

24. a.

H a: μ ≠ 2 H0: μ = 2 Σxi 22 x= = = 2.2 10 n

c. d.

e. 36. a.

40. a. b.

c. 45. a. b.

Σ ( xi − x ) n −1

= .516

x − μ0

2.2 − 2 = = 1.22 s / n .516 / 10 Degrees of freedom = n – 1 = 9 Because t > 0, p-value is two times the upper tail area Using t table: area in upper tail is between .10 and .20; therefore, p-value is between .20 and .40. Exact p-value corresponding to t = 1.22 is .2535 p-value > .05; do not reject H0. No reason to change from the 2 hours for cost estimating purposes. t=

p − p0

.68 − .75

= −2.80 p0 (1 − p0 ) .75(1 − .75) 300 n Lower tail p-value is the area to the left of the test statistic Using normal table with z = -2.80: p-value =.0026 p-value ≤ .05; Reject H0 .72 − .75 z= = −1.20 .75(1 − .75) 300 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.20: p-value =.1151 p-value > .05; Do not reject H0 .70 − .75 z= = −2.00 .75(1 − .75) 300 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -2.00: p-value =.0228 p-value ≤ .05; Reject H0 .77 − .75 z= = .80 .75(1 − .75) 300 Lower tail p-value is the area to the left of the test statistic Using normal table with z = .80: p-value =.7881 p-value > .05; Do not reject H0 414 = .2702 (27%) 1532 H0: p ≤ .22, Ha: p > .22 p − p0 .2702 − .22 z= = = 4.75 p0 (1 − p0 ) .22(1 − .22) 1532 n Upper tail p-value is the area to the right of the test statistic Using normal table with z = 4.75: p-value ≈ 0 so Reject H0. Conclude that there has been a significant increase in the intent to watch the TV programs. These studies help companies and advertising firms evaluate the impact and benefit of commercials. p=

H0: p = .30 24 p= = .48 50

Ha: p ≠ .30

p − p0

.48 − .30

= 2.78 p0 (1 − p0 ) .30(1 − .30) 50 n Because z > 0, p-value is two times the upper tail area Using normal table with z = 2.78: p-value = 2(.0027) = .0054 p-value ≤ .01; reject H0. We would conclude that the proportion of stocks going up on the NYSE is not 30%. This would suggest not using the proportion of DJIA stocks going up on a daily basis as a predictor of the proportion of NYSE stocks going up on that day.

58.

α = .05. Note however for this two – tailed test, zα / 2 = z.025 = 1.96 At μ0 = 28, At μa = 29, β = .15. z.15 = 1.04 σ =6 ( zα / 2 + z β ) 2 σ 2 (1.96 + 1.04) 2 (6) 2 n= = = 324 (μ0 − μa )2 (28 − 29) 2

59.

α = .02. z.02 = 2.05 At μ0 = 25, At μa = 24, β = .20. z.20 = .84 σ =3 ( zα + z β ) 2 σ 2 (2.05 + .84) 2 (3) 2 n= = = 75.2 Use 76 ( μ0 − μ a ) 2 (25 − 24) 2

65. a.

H0: μ ≥ 6883

x − μ0

Ha: μ < 6883

Degrees of freedom = n – 1 = 39 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between .025 and .01 Exact p-value corresponding to t = -2.268 is 0.0145 (one tail) We should conclude that Medicare spending per enrollee in Indianapolis is less than the national average. Using the critical value approach we would: Reject H0 if t ≤ −t.05 = -1.685 Since t = -2.268 ≤ -1.685, we reject H0.

s/ n

5980 − 6883

H0: μ = 2.357

67.

2518 / 40

= −2.268

Ha: μ ≠ 2.357

Σ ( xi − x ) Σx x = i = 2.3496 s= = .0444 n n −1 x − μ0 2.3496 − 2.3570 t= = = −1.18 s/ n .0444 / 50 Degrees of freedom = 50 – 1 = 49 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between .10 and .20; therefore, p-value is between .20 and .40. Exact p-value corresponding to t = -1.18 is .2437 p-value > .05; do not reject H0. There is not a statistically significant difference between the National mean price per gallon and the mean price per gallon in the Lower Atlantic states. 2

73. a. b.

Ha: p < .24 H0: p ≥ .24 81 p= = .2025 400

p − p0

.2025 − .24

= −1.76 p0 (1 − p0 ) .24(1 − .24) 400 n Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.76: p-value =.0392 p-value ≤ .05; reject H0. The proportion of workers not required to contribute to their company sponsored health care plan has declined. There seems to be a trend toward companies requiring employees to share the cost of health care benefits.

10. Statistical Inference about Means and Proportions with Two populations 7.

μ1 = Population mean 2002 μ 2 = Population mean 2003 H0: μ1 − μ 2 ≤ 0 Ha: μ1 − μ 2 > 0

With time in minutes, x1 − x2 = 172 – 166 = 6 minutes

( x1 − x2 ) − D0 σ

2 1

2 2

(172 − 166) − 0 122 122 + 60 50

= 2.61 p-value = 1.0000 – .9955 = .0045

p-value ≤ .05; reject H0. The population mean duration of games in 2003 is less than the population mean in 2002.

σ 12

122 122 + = 6 ± 4.5 = (1.5 to 10.5) 60 50

x1 − x2 ± z.025

Percentage reduction: 6/172 = 3.5%. Management should be encouraged by the fact that steps taken in 2003 reduced the population mean duration of baseball games. However, the statistical analysis shows that the reduction in the mean duration is only 3.5%. The interval estimate shows the reduction in the population mean is 1.5 minutes (.9%) to 10.5 minutes (6.1%). Additional data collected by the end of the 2003 season would provide a more precise estimate. In any case, most likely the issue will continue in future years. It is expected that major league baseball would prefer that additional steps be taken to further reduce the mean duration of games.

20. a.

σ 22

= (172 − 166) ± 1.96

3, -1, 3, 5, 3, 0, 1

d = ∑ di / n = 14 / 7 = 2

sd =

d =2

With 6 degrees of freedom t.025 = 2.447, 2 ± 2.447 2.082 / 7 = 2 ± 1.93 = (.07 to 3.93)

23. a.

∑( d i − d ) 2 = n −1

(

μ1 = population mean grocery expenditures, H0: μ d = 0

26 = 2.08 7 −1

d − μd

)

μ2 = population mean dining-out expenditures

H a: μ d ≠ 0 =

850 − 0

= 4.91 df = n – 1 = 41 p-value ≈ 0 sd / n 1123 / 42 Conclude that there is a difference between the annual population mean expenditures for groceries and for dining-out.

Groceries has the higher mean annual expenditure by an estimated $850. d ± t.025

25. a.

= 850 ± 2.020

H0: μd = 0

1123 42

= 850 ± 350 = (500 to 1200)

Ha: μd ≠ 0

Use difference data: -3, -2, -4, 3, -1, -2, -1, -2, 0, 0, -1, -4, -3, 1, 1 d =

∑ di −18 = = −1.2 15 n

d − μd sd / n

sd =

−1.2 − 0

= −2.36

1.97 / 15

∑( d i − d ) 2 54.4 = = 1.97 n −1 15 − 1

df = n – 1 = 14

Using t table, the 1-tail area is between .01 and .025, so the Two-tail p-value is between .02 and .05. The exact p-value corresponding to t = -2.36 is .0333 Since the p-value ≤ .05, reject H0. Conclude that there is a difference between the population mean weekly usage for the two media. b.

31. a. b. c.

∑ xi 282 ∑ xi 300 = = 18.8 hours per week for cable television, xR = = = 20 for radio. 15 15 n n Radio has greater usage. xTV =

Professional Golfers: p1 = 688/1075 = .64, Amateur Golfers: p2 = 696/1200 = .58 Professional golfers have the better putting accuracy. p1 − p 2 = .64 − .58 = .06 Professional golfers make 6% more 6-foot putts than the very best amateur golfers. p (1 − p1 ) p2 (1 − p2 ) .64(1 − .64) .58(1 − .58) + = .64 − .58 ± 1.96 = .06 ± .04 (.02 to .10) p1 − p2 ± z.025 1 + n1 n2 1075 1200 The confidence interval shows that professional golfers make from 2% to 10% more 6-foot putts than the best amateur golfers. H0: μ1 – μ2 = 0 ( x − x ) − D0 z= 1 2 =

38.

σ 12 n1

σ 22 n2

Ha: μ1 – μ2 ≠ 0 (4.1 − 3.4) − 0 = 2.79 (2.2) 2 (1.5) 2 + 120 100

p-value = 2(1.0000 – .9974) = .0052 p-value ≤ .05, reject H0. A difference exists with system B having the lower mean checkout time. 41. a.

n1 = 10 n2 = 8 x1 = 21.2 x2 = 22.8 s2 = 3.55 s1 = 2.70 x1 − x2 = 21.2 – 22.8 = -1.6 so Kitchens are less expensive by $1600. 2

⎛ s12 s22 ⎞ ⎛ 2.702 3.552 ⎞ + ⎜ + ⎟ ⎜ ⎟ n1 n2 ⎠ 10 8 ⎠ ⎝ ⎝ df = = = 12.9 . Use df = 12, t.05 = 1.782 2 2 2 2 1 ⎛ 2.702 ⎞ 1 ⎛ 3.552 ⎞ 1 ⎛ s12 ⎞ 1 ⎛ s22 ⎞ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ 9 ⎝ 10 ⎠ 7 ⎝ 8 ⎠ n1 − 1 ⎝ n1 ⎠ n2 − 1 ⎝ n2 ⎠

−1.6 ± 1.782

47. a.

2.702 3.552 = -1.6 ± 2.7 = (-4.3 to 1.1) + 10 8

p1 = .276 Most recent week, p2 = .487 One Week Ago, p3 = .397 One Month Ago Point estimate = p1 − p2 = .276 − .487 = −.211

Margin of error: z.025

p1 (1 − p1 ) p2 (1 − p2 ) .276(1 − .276) .487(1 − .487) + = 1.96 + = .085 n1 n2 240 240

95% confidence interval: -.211 ± .085 (-.296, -.126) b. c.

H0: p1 – p3 ≥ 0 p=

H a: p 1 – p 3 < 0

n1 p1 + n2 p3 (240)(.276) + (240)(.397) = = .3365 n1 + n3 240 + 240

⎛1 1 ⎞ ⎛ 2 ⎞ p (1 − p ) ⎜ + ⎟ = (.3365)(.6635) ⎜ ⎟ = .0431 n n ⎝ 240 ⎠ 2 ⎠ ⎝ 1 .276 − .397 p-value = .0025 z= = −2.81 .0431 With p-value ≤ .01, we reject H0 and conclude that bullish sentiment has declined over the past month. s p1 − p2 =

14. Simple Linear regression 13. a.

30.0 25.0

20.0 15.0 10.0 5.0 0.0 0.0

20.0

40.0

60.0

80.0

100.0

120.0

x b.

The summations needed to compute the slope and the y-intercept are: Σxi = 399 Σyi = 97.1 Σ( xi − x )( yi − y ) = 1233.7 Σ( xi − x ) 2 = 7648 b1 =

Σ( xi − x )( yi − y ) 1233.7 = = 0.16131 Σ( xi − x ) 2 7648

b0 = y − b1 x = 1387143 . − (016131 . )(57) = 4.67675 y$ = 4.68 + 016 . x y$ = 4.68 + 016 . x = 4.68 + 016 . (52.5) = 13.08 or approximately $13,080. The agent’s request for an audit appears to be justified.

14. a.

140.0

b. c.

There appears to be a positive linear relationship between x = features rating and y = PCW World Rating. Σx 784 Σy 777 x= i = = 78.4 y = i = = 77.7 10 10 n n Σ( xi − x )( yi − y ) = 147.20 Σ( xi − x )2 = 284.40 b1 =

d. 18. a.

b. c. 21. a.

b0 = y − b1 x = 77.7 − (.51758)(78.4) = 37.1217 yˆ = 37.1217 + .51758 x yˆ = 37.1217 + .51758(70) = 73.35 or 73

The estimated regression equation and the mean for the dependent variable are: yˆ = 1790.5 + 581.1x y = 3650 The sum of squares due to error and the total sum of squares are SSE = ∑( yi − yˆi ) 2 = 85,135.14 SST = ∑( yi − y ) 2 = 335, 000 Thus, SSR = SST – SSE = 335,000 – 85,135.14 = 249,864.86 r2 = SSR/SST = 249,864.86/335,000 = .746 We see that 74.6% of the variability in y has been explained by the least squares line. r = .746 = +.8637 The summations needed in this problem are: Σxi = 3450 Σyi = 33, 700 Σ( xi − x )( yi − y ) = 712,500 b1 =

b. c.

d. 35. a.

Σ( xi − x )( yi − y ) 147.20 = = .51758 Σ( xi − x ) 2 284.40

Σ( xi − x ) 2 = 93, 750

Σ( xi − x )( yi − y ) 712,500 = = 7.6 Σ( xi − x ) 2 93, 750

b0 = y − b1 x = 5616.67 − (7.6)(575) = 1246.67 y$ = 1246.67 + 7.6 x $7.60 The sum of squares due to error and the total sum of squares are: SSE = ∑( yi − yˆ i ) 2 = 233,333.33 SST = ∑( yi − y ) 2 = 5, 648,333.33 Thus, SSR = SST – SSE = 5,648,333.33 – 233,333.33 = 5,415,000 r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587 We see that 95.87% of the variability in y has been explained by the estimated regression equation. y$ = 1246.67 + 7.6 x = 1246.67 + 7.6(500) = $5046.67

s = 145.89 x = 3.2 Σ( xi − x ) 2 = 0.74 s yˆp = s

2 1 ( xp − x ) 1 (3 − 3.2) 2 145.89 + = + = 68.54 n Σ( xi − x ) 2 6 0.74

ˆy p = 1790.54 + 581.08 x = 1790.54 + 581.08( 3 ) = 3533.78 y$ p ± tα / 2 s y$ p = 3533.78 ± 2.776 (68.54) = 3533.78 ± 190.27 or $3343.51 to $3724.05

sind = s 1 +

2 1 ( xp − x ) 1 (3 − 3.2) 2 145.89 1 + = + + = 161.19 n Σ( xi − x ) 2 6 0.74

y$ p ± tα / 2 sind = 3533.78 ± 2.776 (161.19) = 3533.78 ± 447.46 or $3086.32 to $3981.24

44. a/b. The scatter diagram shows a linear relationship between the two variables. c. The Minitab output is shown below: The regression equation is Rental$ = 37.1 – 0.779 Vacancy% Predictor Constant Vacancy%

Coef 37.066 -0.7791

S = 4.889

SE Coef 3.530 0.2226

R-Sq = 43.4%

T 10.50 -3.50

P 0.000 0.003

R-Sq(adj) = 39.8%

Analysis of Variance Source Regression Residual Error Total

DF 1 16 17

SS 292.89 382.37 675.26

MS 292.89 23.90

F 12.26

P 0.003

Predicted Values for New Observations New Obs 1 2

Fit 17.59 28.26

SE Fit 2.51 1.42

95.0% CI ( 12.27, 22.90) ( 25.26, 31.26)

( (

95.0% PI 5.94, 29.23) 17.47, 39.05)

Values of Predictors for New Observations New Obs 1 2

Vacancy% 25.0 11.3

d. e. f. g.

Since the p-value = 0.003 is less than α = .05, the relationship is significant. r2 = .434. The least squares line does not provide a very good fit. The 95% confidence interval is 12.27 to 22.90 or $12.27 to $22.90. The 95% prediction interval is 17.47 to 39.05 or $17.47 to $39.05.

47. a.

Let x = advertising expenditures and y = revenue yˆ = 29.4 + 1.55 x SST = 1002 SSE = 310.28 SSR = 691.72 MSR = SSR / 1 = 691.72 MSE = SSE / (n – 2) = 310.28/ 5 = 62.0554 F = MSR / MSE = 691.72/ 62.0554= 11.15 F.05 = 6.61 (1 degree of freedom numerator and 5 denominator) Since F = 11.15 > F.05 = 6.61 we conclude that the two variables are related.

Or:

Using F table (1 degree of freedom numerator and 5 denominator), p-value is between .01 and .025 Using Excel or Minitab, the p-value corresponding to F = 11.15 is .0206. Because p-value ≤ α = .05, we conclude that the two variables are related.

c. 10

Residuals

5 0 -5 -10 -15 25

Predicted Values d.

The residual plot leads us to question the assumption of a linear relationship between x and y. Even though the relationship is significant at the .05 level of significance, it would be extremely dangerous to extrapolate beyond the range of the data.

55.

No. Regression or correlation analysis can never prove that two variables are casually related.

57.

The purpose of testing whether β1 = 0 is to determine whether or not there is a significant relationship between x and y. However, rejecting β1 = 0 does not necessarily imply a good fit. For example, if β1 = 0 is rejected and r2 is low, there is a statistically significant relationship between x and y but the fit is not very good.

60. a. 1300 1280

S&P500

1260 1240 1220 1200 1180 1160 10000

10200

10400

10600 DJIA

10800

11000

The Minitab output is shown below: The regression equation is S&P500 = – 182 + 0.133 DJIA Predictor Constant DJIA

Coef -182.11 0.133428

S = 6.89993

SE Coef 71.83 0.006739

R-Sq = 95.6%

T -2.54 19.80

P 0.021 0.000

R-Sq(adj) = 95.4%

Analysis of Variance Source Regression Residual Error Total

DF 1 18 19

SS 18666 857 19523

MS 18666 48

F 392.06

P 0.000

Using the F test, the p-value corresponding to F = 392.06 is .000. Because the p-value ≤ α =.05, we reject H 0 : β1 = 0 ; there is a significant relationship.

With R-Sq = 95.6%, the estimated regression equation provided an excellent fit.

yˆ = −182.11 + .133428DJIA= − 182.11 + .133428(11, 000) = 1285.60 or 1286.

The DJIA is not that far beyond the range of the data. With the excellent fit provided by the estimated regression equation, we should not be too concerned about using the estimated regression equation to predict the S&P500.

15. Multiple Regression 5.

The Minitab output is shown below: The regression equation is Revenue = 88.6 + 1.60 TVAdv Predictor Constant TVAdv

Coef 88.638 1.6039

S = 1.215

SE Coef 1.582 0.4778

R-Sq = 65.3%

T 56.02 3.36

P 0.000 0.015

R-Sq(adj) = 59.5%

Analysis of Variance Source Regression Residual Error Total

DF 1 6 7

SS 16.640 8.860 25.500

MS 16.640 1.477

F 11.27

P 0.015

The Minitab output is shown below: The regression equation is Revenue = 83.2 + 2.29 TVAdv + 1.30 NewsAdv Predictor Constant TVAdv NewsAdv

Coef 83.230 2.2902 1.3010

S = 0.6426

SE Coef 1.574 0.3041 0.3207

R-Sq = 91.9%

T 52.88 7.53 4.06

P 0.000 0.001 0.010

R-Sq(adj) = 88.7%

Analysis of Variance Source Regression Residual Error Total

DF 2 5 7

SS 23.435 2.065 25.500

MS 11.718 0.413

F 28.38

P 0.002

No, it is 1.60 in part (a) and 2.29 above. In part (b) it represents the marginal change in revenue due to an increase in television advertising with newspaper advertising held constant.

Revenue = 83.2 + 2.29(3.5) + 1.30(1.8) = $93.56 or $93,560

The Minitab output is shown below: The regression equation is PCW Rating = 66.1 + 0.170 Performance Predictor Constant Performance S = 2.59221

Coef 66.062 0.16989

SE Coef 3.793 0.05407

R-Sq = 55.2%

T 17.42 3.14

P 0.000 0.014

R-Sq(adj) = 49.6%

Analysis of Variance

Source Regression Residual Error Total

DF 1 8 9

SS 66.343 53.757 120.100

MS 66.343 6.720

F 9.87

P 0.014

The Minitab output is shown below: The regression equation is PCW Rating = 40.0 + 0.113 Performance + 0.382 Features Predictor Constant Performance Features S = 1.67285

Coef 39.982 0.11338 0.3820

SE Coef 7.855 0.03846 0.1093

R-Sq = 83.7%

T 5.09 2.95 3.49

P 0.001 0.021 0.010

R-Sq(adj) = 79.0%

Analysis of Variance Source Regression Residual Error Total

DF 2 7 9

SS 100.511 19.589 120.100

MS 50.255 2.798

F 17.96

P 0.002

Note that the coefficient of Performance changed slightly when Features is included in the model. But there is a huge increase in the Adjusted R-Squared, and both variables have low p-values in part b. Hence we can expect better predictions from the 2-variable model.

yˆ = 40.0 + .113(80) + .382(70) = 75.78 or 76

The Minitab output is shown below: The regression equation is Price = 356 – 0.0987 Capacity + 123 Comfort Predictor Constant Capacity Comfort

Coef 356.1 -0.09874 122.87

S = 51.14

SE Coef 197.2 0.04588 21.80

R-Sq = 83.2%

T 1.81 -2.15 5.64

P 0.114 0.068 0.001

R-Sq(adj) = 78.4%

Analysis of Variance Source Regression Residual Error Total

DF 2 7 9

SS 90548 18304 108852

MS 45274 2615

F 17.31

P 0.002

b1 = -.0987 is an estimate of the change in the price with respect to a 1 cubic inch change in capacity with the comfort rating held constant. b2 = 123 is an estimate of the change in the price with respect to a 1 unit change in the comfort rating with the capacity held constant. yˆ = 356 – .0987(4500) + 123 (4) = 404

23. Note: The Minitab output is shown in Exercise 5 a. F = 28.38 Using F table (2 degrees of freedom numerator and 5 denominator), p-value is less than .01 Actual p-value = .002 Because p-value ≤ α , there is a significant relationship.

t = 7.53 Using t table (5 degrees of freedom), area in tail is less than .005; p-value is less than .01 Actual p-value = .001 Because p-value ≤ α , β1 is significant and x1 should not be dropped from the model. c. t = 4.06 Actual p-value = .010 Because p-value ≤ α , β2 is significant and x2 should not be dropped from the model. NOTE: These answers seem to imply that a variable whose p-value is above alpha should be dropped. THAT IS NOT NECESSARILY TRUE!

29. a.

yˆ = 83.2 + 2.29(3.5) + 1.30(1.8) = 93.555 or $93,555 More accurate answer: In Exercise 5b, the Minitab output shows that b0 = 83.230, b1 = 2.2902, and b2 = 1.3010; hence, yˆ = 83.230 + 2.2902×1 + 1.3010×2. Using this estimated regression equation, we obtain yˆ = 83.230 + 2.2902(3.5) + 1.3010(1.8) = 93.588 or $93,588 The difference between these two estimates ($93,588 – $93,555 = $33) is simply due to the fact that additional significant digits are used in Minitab’s computations.

The Minitab output is shown below: Fit Stdev.Fit 95% C.I. 93.588 0.291 ( 92.840, 94.335)

95% P.I. ( 91.774, 95.401)

b. c.

Note that the value of FIT ( yˆ ) is 93.588. Confidence interval estimate: 92.840 to 94.335 or $92,840 to $94,335 Prediction interval estimate: 91.774 to 95.401 or $91,774 to $95,401

34. a. b. c.

$15,300 Estimate of sales = 10.1 – 4.2(2) + 6.8(8) + 15.3(0) = 56.1 or $56,100 Estimate of sales = 10.1 – 4.2(1) + 6.8(3) + 15.3(1) = 41.6 or $41,600

35. a.

Let Type = 0 if a mechanical repair Type = 1 if an electrical repair The Minitab output is shown below: The regression equation is Time = 3.45 + 0.617 Type Predictor Constant Type S = 1.093

Coef 3.4500 0.6167

SE Coef 0.5467 0.7058

R-Sq = 8.7%

T 6.31 0.87

P 0.000 0.408

R-Sq(adj) = 0.0%

Analysis of Variance Source Regression Residual Error Total

b. c.

DF 1 8 9

SS 0.913 9.563 10.476

MS 0.913 1.195

F 0.76

P 0.408

The estimated regression equation did not provide a good fit. In fact, the p-value of .408 shows that the relationship is not significant for any reasonable value of α. Person = 0 if Bob Jones performed the service and Person = 1 if Dave Newton performed the service. The Minitab output is shown below:

The regression equation is Time = 4.62 – 1.60 Person Predictor Constant Person

Coef 4.6200 -1.6000

S = 0.7138

R-Sq = 61.1%

Analysis of Variance Source DF Regression 1 Residual Error 8 Total 9

d. 36. a.

SE Coef 0.3192 0.4514

T 14.47 -3.54

P 0.000 0.008

R-Sq(adj) = 56.2%

SS 6.4000 4.0760 10.4760

MS 6.4000 0.5095

F 12.56

P 0.008

We see that 61.1% of the variability in repair time has been explained by the repair person that performed the service; an acceptable, but not good, fit. The Minitab output is shown below: The regression equation is Time = 1.86 + 0.291 Months + 1.10 Type – 0.609 Person Predictor Constant Months Type Person

Coef 1.8602 0.29144 1.1024 -0.6091

S = 0.4174

SE Coef 0.7286 0.08360 0.3033 0.3879

R-Sq = 90.0%

T 2.55 3.49 3.63 -1.57

P 0.043 0.013 0.011 0.167

R-Sq(adj) = 85.0%

Analysis of Variance Source Regression Residual Error Total

b. c.

DF 3 6 9

SS 9.4305 1.0455 10.4760

MS 3.1435 0.1743

F 18.04

P 0.002

Since the p-value corresponding to F = 18.04 is .002 < α = .05, the overall model is statistically significant. The p-value corresponding to t = -1.57 is .167 > α = .05; thus, the addition of Person is not statistically significant. Person is highly correlated with Months (the sample correlation coefficient is -.691); thus, once the effect of Months has been accounted for, Person will not add much to the model.

42. a.

The Minitab output is shown below: The regression equation is Speed = 71.3 + 0.107 Price + 0.0845 Horsepwr Predictor Constant Price Horsepwr

Coef 71.328 0.10719 0.084496

S = 2.485

SE Coef 2.248 0.03918 0.009306

R-Sq = 91.9%

T 31.73 2.74 9.08

P 0.000 0.017 0.000

R-Sq(adj) = 90.7%

Analysis of Variance Source Regression Residual Error Total Source Price Horsepwr

DF 1 1

DF 2 13 15

SS 915.66 80.30 995.95

MS 457.83 6.18

F 74.12

P 0.000

SE Fit 2.007

Residual 2.118

Seq SS 406.39 509.27

Unusual Observations Obs Price Speed 2 93.8 108.000

Fit 105.882

St Resid 1.45 X

X denotes an observation whose X value gives it large influence.

The standardized residual plot is shown below. There appears to be a very unusual trend in the standardized residuals. 2

Standardized Residual

-1

-2 90

c. d.

100 105 Fitted Value

110

115

120

The Minitab output shown in part (a) did not identify any observations with a large standardized residual; thus, there does not appear to be any outliers in the data. The Minitab output shown in part (a) identifies observation 2 as an influential observation.

16. Regression Analysis: Model Building 4.

The Minitab output is shown below: The regression equation is Y = 943 + 8.71 X Predictor Constant X

Coef 943.05 8.714

s = 32.29

Stdev 59.38 1.544

R-sq = 88.8%

t-ratio 15.88 5.64

p 0.000 0.005

R-sq(adj) = 86.1%

Analysis of Variance SOURCE Regression Error Total

DF 1 4 5

SS 33223 4172 37395

MS 33223 1043

F 31.86

p 0.005

b. p-value = .005 < α = .01; reject H0 5.

The Minitab output is shown below: The regression equation is Y = 433 + 37.4 X – 0.383 XSQ Predictor Constant X XSQ

Coef 432.6 37.429 -0.3829

s = 15.83

Stdev 141.2 7.807 0.1036

R-sq = 98.0%

t-ratio 3.06 4.79 -3.70

p 0.055 0.017 0.034

R-sq(adj) = 96.7%

Analysis of Variance SOURCE Regression Error Total

b. c.

DF 2 3 5

SS 36643 751 37395

MS 18322 250

F 73.15

p 0.003

Since the linear relationship was significant (Exercise 4), this relationship must be significant. Note also that since the p-value of .003 < α = .05, we can reject H0. The fitted value is 1302.01, with a standard deviation of 9.93. The 95% confidence interval is 1270.41 to 1333.61; the 95% prediction interval is 1242.55 to 1361.47.

12. a.

A portion of the Minitab output follows: The regression equation is Scoring Avg. = 46.3 + 14.1 Putting Avg. Predictor Constant Putting Avg.

Coef 46.277 14.103

S = 0.510596

SE Coef 6.026 3.356

R-Sq = 38.7%

T 7.68 4.20

P 0.000 0.000

R-Sq(adj) = 36.5%

Analysis of Variance Source Regression Residual Error Total

DF 1 28 29

SS 4.6036 7.2998 11.9035

MS 4.6036 0.2607

F 17.66

P 0.000

A portion of the Minitab output follows: The regression equation is Scoring Avg. = 59.0 – 10.3 Greens in Reg. + 11.4 Putting Avg. – 1.81 Sand Saves Predictor Constant Greens in Reg. Putting Avg. Sand Saves S = 0.407808

Coef 59.022 -10.281 11.413 -1.8130

SE Coef 5.774 2.877 2.760 0.9210

R-Sq = 63.7%

T 10.22 -3.57 4.14 -1.97

P 0.000 0.001 0.000 0.060

R-Sq(adj) = 59.5%

Analysis of Variance Source Regression Residual Error Total

DF 3 26 29

SSE(reduced) = 7.2998

SS 7.5795 4.3240 11.9035

MS 2.5265 0.1663

SSE(full) = 4.3240

F 15.19

P 0.000

MSE(full) = .1663

SSE(reduced) – SSE(full) 7.2998 – 4.3240 2 F = number of extra terms = = 8.95 MSE(full) .1663

The p-value associated with F = 8.95 (2 degrees of freedom numerator and 26 denominator) is .001. With a p-value < α =.05, the addition of the two independent variables is statistically significant.

21. Decision Analysis 1.

s1 d1

s2 s3 s1

s2 s3

100 25 100 100 75

EV(d1) = .65(250) + .15(100) + .20(25) = 182.5 EV(d2) = .65(100) + .15(100) + .20(75) = 95 The optimal decision is d1

The decision to be made is to choose the type of service to provide. The chance event is the level of demand for the Myrtle Air service. The consequence is the amount of quarterly profit. There are two decision alternatives (full price and discount service). There are two outcomes for the chance event (strong demand and weak demand). EV(Full) = 0.7(960) + 0.3(-490) = 525 EV(Discount) = 0.7(670) + 0.3(320) = 565 Optimal Decision: Discount service EV(Full) = 0.8(960) + 0.2(-490) = 670 EV(Discount) = 0.8(670) + 0.2(320) = 600 Optimal Decision: Full price service

b. c.

250

EV(Small) = 0.1(400) + 0.6(500) + 0.3(660) = 538 EV(Medium) = 0.1(-250) + 0.6(650) + 0.3(800) = 605 EV(Large) = 0.1(-400) + 0.6(580) + 0.3(990) = 605 Best decision: Build a medium or large-size community center. Note that using the expected value approach, the Town Council would be indifferent between building a medium-size community center and a large-size center.

The Town’s optimal decision strategy based on perfect information is as follows: If the worst-case scenario, build a small-size center If the base-case scenario, build a medium-size center If the best-case scenario, build a large-size center Using the consultant’s original probability assessments for each scenario, 0.10, 0.60 and 0.30, the expected value of a decision strategy that uses perfect information is: EVwPI = 0.1(400) + 0.6(650) + 0.3(990) = 727 In part (a), the expected value approach showed that EV(Medium) = EV(Large) = 605.

Therefore, EVwoPI = 605 and EVPI = 727 – 605 = 122 The town should seriously consider additional information about the likelihood of the three scenarios. Since perfect information would be worth $122,000, a good market research study could possibly make a significant contribution. c.

EV(Small) = 0.2(400) + 0.5(500) + 0.3(660) = 528 EV(Medium) = 0.2(-250) + 0.5(650) + 0.3(800) = 515 EV(Large) = 0.2(-400) + 0.5(580) + 0.3(990) = 507 Best decision: Build a small-size community center.

If the promotional campaign is conducted, the probabilities will change to 0.0, 0.6 and 0.4 for the worst case, base case and best case scenarios respectively. EV(Small) = 0.0(400) + 0.6(500) + 0.4(660) = 564 EV(Medium) = 0.0(-250) + 0.6(650) + 0.4(800) = 710 EV(Large) = 0.0(-400) + 0.6(580) + 0.4(990) = 744 In this case, the recommended decision is to build a large-size community center. Compared to the analysis in Part (a), the promotional campaign has increased the best expected value by $744,000 605,000 = $139,000. Compared to the analysis in part (c), the promotional campaign has increased the best expected value by $744,000 – 528,000 = $216,000. Even though the promotional campaign does not increase the expected value by more than its cost ($150,000) when compared to the analysis in part (c), it appears to be a good investment. That is, it eliminates the risk of a loss, which appears to be a significant factor in the mayor’s decision-making process.

12. a. s1 d1 Normal

s1 d2

W ait

s3 s1

2 d1 Cold

s3 s1

4 d2

s1 d1 Don’t W ait

s2 s3 s1

s2 s3

3500 1000 -1500 7000 2000 -9000 3500 2000 -1500 7000 2000 -9000 3500 2000 -1500 7000 2000 -9000

Using Node 5, EV (node 10) = 0.4(3500) + 0.3(1000) + 0.3(-1500) = 1250 EV (node 11) = 0.4(7000) + 0.3(2000) + 0.3(-9000) = 700 Decision: d1 Blade attachment Expected Value $1250 (at Node 5)

EVwPI = 0.4(7000) + 0.3(2000) + 0.3(-1500) = $2950 EVPI = $2950 – $1250 = $1700

EV (node 6) = 0.35(3500) + 0.30(1000) + 0.35(-1500) = 1000 EV (node 7) = 0.35(7000) + 0.30(2000) + 0.35(-9000) = -100 EV (node 8) = 0.62(3500) + 0.31(1000) + 0.07(-1500) = 2375 EV (node 9) = 0.62(7000) + 0.31(2000) + 0.07(-9000) = 4330 Blade attachment EV (node 3) = Max(1000,-100) = 1000 d1 EV (node 4) = Max(2375,4330) = 4330 d2 New snowplow The expected value of node 2 is EV (node 2) = 0.8 EV(node 3) + 0.2 EV(node 4) = 0.8(1000) + 0.2(4330) = 1666 EV (node 1) = Max(node 2, node 5) = Max(1666,1250) = $1666 Wait The optimal strategy is “Wait until September and then, If normal weather, choose the blade attachment, but if unseasonably cold, choose snowplow”