A PPE N D IX D: A N SWE RS TO S E LF - RE V I E W

CHAPTER 1 c. Class frequencies.

1.1 a. Inferential statistics, because a sample was used to draw a d. The largest concentration of commissions is $1,500 up to
conclusion about how all consumers in the population $1,600. The smallest commission is about $1,400 and the
would react if the chicken dinner were marketed. largest is about $1,800. The typical amount earned is
b. On the basis of the sample of 1,960 consumers, we esti- $1,550.
mate that, if it is marketed, 60% of all consumers will pur- 2.3 a. 2 6 = 64 < 73 < 128 = 2 7 , so seven classes are
chase the chicken dinner: (1,176/1,960) × 100 = 60%. recommended.
1.2 a. Age is a ratio-scale variable. A 40-year-old is twice as old b. The interval width should be at least (488 − 320)/7 = 24.
as someone 20 years old. Class intervals of either 25 or 30 are reasonable.
b. The two variables are: (1) if a person owns a luxury car, and c. Assuming a class interval of 25 and beginning with a lower
(2) the state of residence. Both are measured on a nominal scale. limit of 300, eight classes are required. If we use an interval
of 30 and begin with a lower limit of 300, only 7 classes are
CHAPTER 2 required. Seven classes is the better alternative.
2.1 a. Qualitative data, because the customers’ response to the
taste test is the name of a beverage.
Distance Classes Frequency Percent
b. Frequency table. It shows the number of people who prefer
each beverage. 300 up to 330 2 2.7%
c. 330 up to 360 2 2.7
360 up to 390 17 23.3
40 390 up to 420 27 37.0
420 up to 450 22 30.1
450 up to 480 1 1.4
30
480 up to 510 2 2.7
Frequency

Grand Total 73 100.00

20
d. 17
e. 23.3%, found by 17/73
10 f. 71.2%, found by (27 + 22 + 1 + 2)/73
2.4 a.
0
20
Cola-Plus Coca-Cola Pepsi Lemon-Lime 20
Beverage 15 13
Number of
suppliers

10
d. 10
6
5
1
0
Lemon-Lime 2 5 8 11 14 17
Pepsi 15%
20% Imports ($ millions)
b.

40
Coca-Cola Cola-Plus
Percent of total

25% 40% 30

20

10

2.2 a. The raw data or ungrouped data. 0

2 5 8 11 14 17
b.
Number of Imports ($ millions)
Commission Salespeople The plots are: (3.5, 12), (6.5, 26), (9.5, 40), (12.5, 20), and
(15.5, 2).
$1,400 up to $1,500 2
c. The smallest annual volume of imports by a supplier is
1,500 up to 1,600 5
about $2 million, the largest about $17 million. The highest
1,600 up to 1,700 3 frequency is between $8 million and $11 million.
1,700 up to 1,800 1 2.5 a. A frequency distribution.
Total 11

818
 Confirming Pages

4
b. 3–5 1. a. A  bout 9.9%, found by √ 1.458602236, then 1.099 −
Hourly Wages Cumulative Number
1.00 = .099
Less than $8 0 b. About 10.095%
Less than $10 3 c. Greater than, because 10.095 > 9.9
Less than $12 10 20 120,520
2. 8.63%, found by √ − 1 = 1.0863 − 1
Less than $14 14 23,000
Less than $16 15 3–6 a. 22 thousands of pounds, found by 112 − 90
824
b. x = = 103 thousands of pounds
15 8
3732
x = 14 c. Variance = 5
= 46.625
8
y = 14
Cumulative frequencies

12 $16,900
3–7 a. μ = = $3,380
5
x = 12 (3,536 − 3,380) 2 + . . . + (3,622 − 3,380) 2
8 y = 10 b. σ2 =
5
(156) 2 + (−207) 2 + (68) 2
+ (−259) 2 + (242) 2
4 =
5
197,454
= = 39,490.8
5
0 c. σ = √39,490.8 = 198.72
8 10 12 14 16
d. There is more variation in the Pittsburgh office because the
Hourly wages (in dollars) standard deviation is larger. The mean is also larger in the
c. About seven employees earn $11.00 or less. Pittsburgh office.
3–8 2.33, found by:
Σx 28
CHAPTER 3 x= = =4
n 7
Σx 2
Σ(x − x )
3–1 1. a. x = s2 =
n n−1
$267,100 14
b. x= = $66,775 =
4 7−1
c. Statistic, because it is a sample value. = 2.33
d. $66,775. The sample mean is our best estimate of the s = √2.33 = 1.53
­population mean. 14.15 − 14.00
Σx 3–9 a. k = = 1.5
2. a. μ= .10
N 13.85 − 14.0
498 k= = −1.5
b. μ= = 83 .10
6 1
c. Parameter, because it was computed using all the popu- 1− = 1 − .44 = .56
(1.5) 2
lation values.
b. 13.8 and 14.2
3–2 1. a. $878
3–10 a. Frequency distribution.
b. 3, 3
ΣfM $244
2. a. 17, found by (15 + 19)/2 = 17 b. x = = = $12.20
b. 5, 5 M 20
303.20
c. s = √
c. There are 3 values that occur twice: 11, 15, and 19.
= $3.99
There are three modes. 20 − 1
3–3 a.
CHAPTER 4
4–1 1. a. 79, 105
Frequency

b. 15
c. From 88 to 97; 75% of the stores are in this range.
4–2 a. 7.9
b. Q1 = 7.76, Q3 = 8.015
4–3 The smallest value is 10 and the largest 85; the first quartile is
Mode
Median
Mean

25 and the third 60. About 50% of the values are between 25
and 60. The median value is 40. The distribution is positively
skewed. There are no outliers.
Weekly sales 407
4–4 a. x = = 81.4,
b. Positively skewed, because the mean is the largest aver- 5
age and the mode is the smallest. 923.2
3–4 a. $237, found by: s = √ = 15.19, Median = 84
5−1
(95 × $400) + (126 × $200) + (79 × $100) 3(81.4 − 84.0)
= $237.00 b. sk = = −0.51
95 + 126 + 79 15.19
b. The profit per suit is $12, found by $237 − $200 cost − 5
$25 commission. The total profit for the 300 suits is c. sk = [−1.3154] = −0.5481
(4)(3)
$3,600, found by 300 × $12.
d. The distribution is somewhat negatively skewed.

819

lin39470_appD_818-831.indd 819 03/15/21 02:50 PM

 4–5 a. 5–4 a. Need for corrective shoes is event A. Need for major dental
Revenue vs. seating capacity work is event B.
8 P(A or B) = P(A) + P(B) − P(A and B)
= .08 + .15 − .03
7
= .20
6 b. One possibility is:
5
Revenue

4
3
B Both A
2 .15 .08
.03
1
0
5500 6000 6500 7000 7500
Capacity
5–5 (.95)(.95)(.95)(.95) = .8145
b. The correlation coefficient is 0.90. 5–6 a. .002, found by:

( 12 )( 11 )( 10 )( 9 ) = 11,880 = .002
c. $7,500 4 3 2 1 24
d. Strong and positive. Revenue is positively related to s­ eating
capacity.
b. .14, found by:

( 12 )( 11 )( 10 )( 9 ) = 11,880 = .1414
8 7 6 5 1,680
CHAPTER 5
5–1 a. Count the number who think the new game is playable. c. No, because there are other possibilities, such as three
b. Seventy-three players found the game playable. Many women and one man.
other answers are possible. 225
c. No. Probability cannot be greater than 1. The probability 5–7 a. P(B2 ) = = .45
500
that the game, if put on the market, will be successful is
b. The two events are mutually exclusive, so apply the special
65/80, or .8125.
rule of addition.
d. Cannot be less than 0. Perhaps a mistake in arithmetic.
100 225
e. More than half of the players testing the game liked it. P(B1 or B2 ) = P(B1 ) + P(B2 ) = + = .65
(Of course, other answers are possible.) 500 500
4 queens in deck c. The two events are not mutually exclusive, so apply the
4
5–2 1. = = .0769 general rule of addition.
52 cards total 52
P(B1 or A1 ) = P(B1 ) + P(A1 ) − P(B1 and A1 )
Classical.
182 100 75 15
2. = .338 Empirical. = + − = .32
539 500 500 500
3. The probability of the outcome is estimated by applying the d. As shown in the example/solution, movies attended per
subjective approach to estimating a probability. If you think month and age are not independent, so apply the general
that it is likely that you will save $1 million, then your prob- rule of multiplication.
ability should be between .5 and 1.0. P(B1 and A1 ) = P(B1 )P(A1 ∣ B1 )

=(
500 )( 100 )
(50 + 68) 100 15
5–3 a. i. = .059 = .03
2,000
302 80
ii. 1 − = .849 5–8 a. P(visited often) = = .41
2,000 195
b. 90
b. P(visited a store in an enclosed mall) = = .46
195
c. The two events are not mutually exclusive, so apply the
general rule of addition.

B E P(visited often or visited a Sears in an enclosed mall)

= P(often) + P(enclosed mall) − P(often and enclosed mall)
80 90 60
= + − = .56
195 195 195
d. P(visited often | went to a Sears in an enclosed mall)
60
= = .67
90
e. Independence requires that P(A | B) = P(A). One possibility
is: P(visit often | visited an enclosed mall) = P(visit often).
Does 60/90 = 80/195? No, the two variables are not inde-
D pendent. Therefore, any joint probability in the table must
be computed by using the general rule of multiplication.
f. As shown in part (e), visits often and enclosed mall are not
independent, so apply the general rule of multiplication.
∼D
P(often and enclosed mall) = P(often)P(enclosed ∣ often)
=(
195 )( 80 )
80 60
c. They are not complementary, but are mutually exclusive. = .31

820
 g. 60 / b.
90 .31
Often 1
6
25
/90

Probability
90
/195 .13
Visits Occasional
yes 5/
90
.03 0
Never 1 2 3 4 5 6

Enclosed 20/ .10 Number of spots

105
no Often
105/ 6
195 35
/105 c. or 1.
6
.18 6–2 a. It is discrete because the values $1.99, $2.49, and $2.89 are
Visits Occasional clearly separated from each other. Also the sum of the proba-
50/ bilities is 1.00, and the outcomes are mutually exclusive.
105 b.
.25 x P(x) xP(x)
Never 1.99 .30 0.597
5–9 a. P(A3 )P(B2 ∣ A3 ) 2.49 .50 1.245
P(A3 ∣ B2 ) =
P(A1 )P(B2 ∣ A1 ) + P(A2 )P(B2 ∣ A2 ) + P(A3 )P(B2 ∣ A3 ) 2.89 .20 0.578
(.50)(.96) Sum is 2.42
b. =
(.30)(.97) + (.20)(.95) + (.50)(.96)
.480 Mean is 2.42
= = .499 c.
.961 x P(x) (x − µ) (x − µ)2P(x )
5–10 1. (5)(4) = 20
2. (3)(2)(4)(3) = 72 1.99 .30 −0.43 0.05547
5–11 1. a. 60, found by (5)(4)(3). 2.49 .50 0.07 0.00245
b. 60, found by: 2.89 .20 0.47 0.04418
5! 5·4·3·2·1
= 0.10210
(5 − 3)! 2·1
2. 5,040, found by: The variance is 0.10208, and the standard deviation is 31.95
10! 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 cents.
=
(10 − 4)! 6·5·4·3·2·1 6–3 a. It is reasonable because each employee either uses direct
3. a. 35 is correct, found by: deposit or does not; employees are independent; the proba-
n! 7! bility of using direct deposit is 0.95 for all; and we count the
7C3 = = = 35
r!(n − r)! 3!(7 − 3)! number using the service out of 7.
b. Yes. There are 21 combinations, found by: b. P(7) = 7C7 (.95)7 (.05)0 = .6983
n! 7! c. P(4) = 7C4 (.95)4 (.05)3 = .0036
7C5 = = = 21
r!(n − r)! 5!(7 − 5)! d. Answers are in agreement.
50! 6–4 a. n = 8, π = .40
4. a. 50P3 = = 117,600 b. P(x = 3) = .2787
(50 − 3)!
50! c. P(x > 0) = 1 − P(x = 0) = 1 − .0168 = .9832

( 3!5! ) ( 2!2! )
b. 50C3 = = 19,600 8! 4!
3! (50 − 3)!
8C3 4C2
6–5 P(3) = =
CHAPTER 6 12C5 12!
6–1 a. 5!7!
Number of (56)(6)
Spots Probability = = .424
792
1 6–6 µ = 4,000(.0002) = 0.8
1
6 0.81e−0.8
1 P(1) = = .3595
2 1!
6
1 CHAPTER 7
3 7–1 a.
6 P(x )
1
4
6
.167
1
5
6
1
6
6
6 8 14
Total = 1.00
6

821
 b. P(x) = (height)(base) b. .9686, found by .4686 + .5000. First calculate z:
=(
14 − 8 )
1 149.5 − 160
(14 − 8) z= = −1.86
5.66
= ( ) (6) = 1.00
1 Area from Appendix B.3 is .4686.

6 7–8 a. .7769, found by:
a + b 14 + 8 22 1
c. µ = = = = 11 − (15)
2 2 2 P( Arrival < 15) = 1 − e 10
2
(b − a) (14 − 8) 2 36
σ=√ =√ =√ = √3 = 1 − .2231 = .7769
12 12 12 b. .0821, found by:
= 1.73 1
− (25)
d. P(10 < x < 14) = (height)(base) P( Arrival > 25) = e 10 = .0821
=(
14 − 8 )
1
(14 − 10) c. .1410, found by
1 P (15 < x < 25) = P (Arrival < 25) − P (Arrival < 15)
= (4) = .9179 − .7769 = .1410
6
= .667 d. 16.09 minutes, found by:
e. P(x < 9) = (height)(base) 1
(x)

=(
14 − 8 )
−
1 .80 = 1 − e 10
(9 − 8)
1
= 0.167 −ln 0.20 = x
10
7–2 a. 
z = (64 − 48)∕12.8 = 1.25. This person’s difference of x = −(−1.609)(10) = 1.609(10) = 16.09
16 ounces more than average is 1.25 standard deviations
above the average. CHAPTER 8
b. z = (32 − 48)∕12.8 = −1.25. This person’s difference of 8–1 a. Students selected are Lehman, Edinger, Nickens, Chontos,
16 ounces less than average is 1.25 standard deviations be- St. John, and Kemp.
low the average. b. Answers will vary.
7–3 a. $46,400 and $48,000, found by $47,200 ± 1($800). c. Skip it and move to the next random number.
b. $45,600 and $48,800, found by $47,200 ± 2($800). 8–2 The students selected are Berry, Francis, Kopp, Poteau, and
c. $44,800 and $49,600, found by $47,200 ± 3($800). Swetye.
d. $47,200. The mean, median, and mode are equal for a normal 8–3 a. 10, found by:
distribution. 5!
5C2 =
e. Yes, a normal distribution is symmetrical. 2!(5 − 2)!
7–4 a. Computing z: b.
154 − 150 Sample
z= = 0.80
5 Service Mean
Referring to Appendix B.3, the area is .2881. So Snow, Tolson 20, 22 21
P(150 < temp < 154) = .2881.
Snow, Kraft 20, 26 23
b. Computing z:
164 − 150 Snow, Irwin 20, 24 22
z= = 2.80 Snow, Jones 20, 28 24
5
Referring to Appendix B.3, the area is .4974. So Tolson, Kraft 22, 26 24
P(164 > temp) = .5000 − .4974 = .0026 Tolson, Irwin 22, 24 23
7–5 a. Computing the z-values: Tolson, Jones 22, 28 25
146 − 150 156 − 150 Kraft, Irwin 26, 24 25
z= = −0.80 and z = = 1.20
5 5 Kraft, Jones 26, 28 27
P (146 < temp < 156) =  P (−0.80 < z < 1.20) Irwin, Jones 24, 28 26
= .2881 + .3849 = .6730
b. Computing the z-values: c.
162 − 150 156 − 150 Mean Number Probability
z= = 2.40 and z = = 1.20
5 5 21 1 .10
P(156 < temp < 162) = P(1.20 < z < 2.40) 22 1 .10
= .4918 − .3849 = .1069 23 2 .20
7–6 85.24 (instructor would no doubt make it 85). The closest area 24 2 .20
to .4000 is .3997; z is 1.28. Then: 25 2 .20
x − 75 26 1 .10
1.28 =
8 27 1 .10
10.24 = x − 75 10 1.00
x = 85.24
7–7 a. .
0 465, found by µ = nπ = 200(.80) = 160, and d. Identical: population mean, μ, is 24, and mean of sample
σ2 = nπ (1 − π) = 200(.80)(1 − .80) = 32. Then, means, is also 24.
√32 = 5.66
σ= e. Sample means range from 21 to 27. Population values go
169.5 − 160 from 20 to 28.
z= = 1.68 f. No, the population is uniformly distributed.
5.66
Area from Appendix B.3 is .4535. Subtracting from .5000 g. Yes.
gives .0465.

822
 8–4 The answers will vary. Here is one solution. x−μ

c. z =
σ∕√n
Sample Number
d. Reject H0 if z < −1.96 or z > 1.96.
1 2 3 4 5 6 7 8 9 10 16.017 − 16.0 0.0170
e. z = = = 0.80
8 2 2 19 3 4 0 4 1 2 0.15∕√50 0.0212
19 1 14 9 2 5 8 2 14 4 f. Do not reject H0.
8 3 4 2 4 4 1 14 4 1 g. We cannot conclude the mean amount dispensed is differ-
0 3 2 3 1 2 16 1 2 3 ent from 16.0 ounces.
10–2 a. H0: µ ≤ 16.0; H1: µ > 16.0
2 1 7 2 19 18 18 16 3 7
b. Reject H0 if z > 1.645.
Total 37 10 29 35 29 33 43 37 24 17 c. The sampling error is 16.04 − 16.00 = 0.04 ounce.
x 7.4 2 5.8 7.0 5.8 6.6 8.6 7.4 4.8 3.4 16.040 − 16.0 .0400
d. z = = = 1.89
0.15∕√50 .0212
Mean of the 10 sample means is 5.88.
e. Reject H0.
3 f. The mean amount dispensed is more than 16.0 ounces.
Frequency

2 g. p-value = .5000 − .4706 = .0294. The p-value is less than

α (.05), so H0 is rejected. It is the same conclusion as in part (d).
1 10–3 a. H0: µ ≤ 305; H1: µ > 305
b. df = n − 1 = 20 − 1 = 19
The decision rule is to reject
2 3 4 5 6 7 8 H0 if t > 1.729.
Mean years
31.08 − 31.20
8–5 z = = −1.20 Region of
0.4∕ √16
Do not rejection
The probability that z is greater than −1.20 is .5000 + .3849 =
.8849. There is more than an 88% chance the filling operation reject H0 α = .05
will produce bottles with at least 31.08 ounces.
0 1.729 t
CHAPTER 9 Critical
9–1 a. Unknown. This is the value we wish to estimate. value
b. The sample mean of $20,000 is the point estimate of the
population mean daily franchise sales. X−μ 311 − 305
$3,000 c. t = = = 2.236
c. $20,000 ± 1.960 = $20,000 ± $930 s∕√n 12∕√20
√40
Reject H0 because 2.236 > 1.729. The modification
d. The estimate of the population mean daily sales for the
­increased the mean battery life to more than 305 days.
­Bun-and-Run franchises is between $19,070 and $20,930.
10–4 a. H0: µ ≥ 9.0; H1: µ < 9.0
About 95% all possible samples of 40 Bun-and-Run fran-
b. 7, found by n − 1 = 8 − 1 = 7
chises would include the population mean.
c. Reject H0 if t < −2.998.
18 11.6
9–2 a. x = = 1.8 s=√ = 1.1353
10 10 − 1
b. The population mean is not known. The best estimate is the Region of
sample mean, 1.8 days. rejection
1.1353
Do not
c. 1.80 ± 2.262 = 1.80 ± 0.81 reject H0
√10
The endpoints are 0.99 and 2.61.
d. t is used because the population standard deviation is –2.998 0 Scale of t
unknown. Critical
e. The value of 0 is not in the interval. It is unreasonable to value
conclude that the mean number of days of work missed is 0

per employee.
420 d. t = −2.494, found by:
9–3 a. p = = .30
0.36
s=√
1,400
= 0.2268
b. .30 ± 2.576 (.0122) = .30 ± .03 8−1
c. The interval is between .27 and .33. About 99% of the similarly 70.4
constructed intervals would include the population mean. x= = 8.8
8
9–4 n = (
) = 206.6. The sample should be
2.576(.279) 2 Then
.05 8.8 − 9.0
t= = −2.494
rounded to 207. 0.2268∕√8
.375(1 − .375) 250 − 40
9–5 .375 ± 1.96 √ √ 250 − 1
= Since −2.494 lies to the right of −2.998, H0 is not rejected.
40 We have not shown that the mean is less than 9.0.
.375 ± 1.96(.0765)(.9184) = .375 ± .138 e. The p-value is between .025 and .010.
10–5 .0054, found by determining the area under the curve b­ etween
CHAPTER 10 10,078 and 10,180.
10–1 a. H0: µ = 16.0; H1: µ ≠ 16.0 xc − μ 1
z=
b. .05 σ∕√n

823
 10,078 − 10,180 538.875
= = −2.55 sd = √ = 8.774
400∕√100 8−1
The area under the curve for a z of −2.55 is .4946 (­Appendix B.3), 8.875
t= = 2.861
and .5000 − .4946 = .0054. 8.774∕ √8
d. p-value = .0122
CHAPTER 11 e. Do not reject H0. We cannot conclude that the students lost
11–1 a. H0: µW ≤ µM H1: µW > µM weight.
The subscript W refers to the women and M to the men. f. The distribution of the differences must be approximately
b. Reject H0 if z > 1.645. normal.
$1,500 − $1,400
c. z = = 2.11 CHAPTER 12
($250) 2 ($200) 2
√ 50
+ 12–1 Let Mark’s assemblies be population 1, then H0: σ21 ≤ σ22;
40 H1: σ21 > σ22; df1 = 10 − 1 = 9; and df2 also equals 9. H0 is
d. Reject the null hypothesis. ­rejected if F > 3.18.
e. p-value = .5000 − .4826 = .0174 (2.0) 2
f. The mean amount sold per day is larger for women. F= = 1.78
(1.5) 2
11–2 a. H0: µd = µa H1: µd ≠ µa
b. df = 6 + 8 − 2 = 12 H0 is not rejected. The variation is the same for both employees.
Reject H0 if t < −2.179 or t > 2.179. 12–2 a. H0: µ1 = µ2 = µ3
H1: At least one treatment mean is different.
42 10
c. x1 = = 7.00 s1 = √ = 1.4142 b. Reject H0 if F > 4.26.
6 6−1 240
c. x = = 20
80 36
= 10.00 s2 = √
12
x2 = = 2.2678
8 8−1 SS total = (18 − 20) 2 + . . . + (32 − 20) 2
(6 − 1)(1.4142) 2 + (8 − 1)(2.2678) 2 = 578
s2p = SSE = (18 − 17) 2 + (14 − 17) 2 + . . . + (32 − 29) 2
6+8−2
= 74
= 3.8333
SST = 578 − 74 = 504
7.00 − 10.00
t= = −2.837 d.
Sum of Degrees of Mean
√ (6 + 8)
1 1
3.8333 Source Squares Freedom Square F

d. Reject H0 because −2.837 is less than the critical value. Treatment 504 2 252 30.65
e. The p-value is less than .02. Error 74 9 8.22
f. The mean number of defects is not the same on the two shifts. Total 578 11
g. Independent populations, populations follow the normal
distribution, populations have equal standard deviations. The F-test statistic, 30.65.
11–3 a. H0: µc ≥ µa H1: µc < µa e. 
H0 is rejected. There is a difference in the mean number of
bottles sold at the various locations.
[ (3562∕10) + (8572∕8)] 2 12–3 a. H0: µ1 = µ2 =µ3

b. df = = 8.93
(3562∕10) 2 (8572∕8) 2 H1: Not all means are equal.
+ b. H0 is rejected if F > 3.98.
10 − 1 8−1
so df = 8 c.
c. Reject H0 if t < −1.860. ANOVA: Single Factor
$1,568 − $1,967 −399.00
d. t = = = −1.234 Groups Count Sum Average Variance
3562 8572 323.23
√ 10 + 8
Northeast 5 205 41 1
Southeast 4 155 38.75 0.916667
e. Do not reject H0. West 5 184 36.8 0.7
f. There is no difference in the mean account balance of those
who applied for their card or were contacted by a ANOVA
telemarketer. Source of
11–4 a. H0: µd ≥ 0, H1: µd > 0 Variation SS df MS F p-Value
b. Reject H0 if t > 2.998. Between Groups 44.16429 2 22.08214 25.43493 7.49E-05
c. Within Groups 9.55 11 0.868182
Name Before After d (d − d ) (d − d )2
Total 53.71429 13
Hunter 155 154 1 −7.875 62.0156
d. H0 is rejected. The treatment means differ.
Cashman 228 207 21 12.125 147.0156
e. (41 − 36.8) ± 2.201 √0.8682( 51 + 51 ) = 4.2 ± 1.3 = 2.9
Mervine 141 147 −6 −14.875 221.2656  and 5.50. The means are significantly different. Zero is not
Massa 162 157 5 −3.875 15.0156 in the interval.
Creola 211 196 15 6.125 37.5156 These treatment means differ because both endpoints of the con-
Peterson 164 150 14 5.125 26.2656 fidence interval are of the same sign.
Redding 184 170 14 5.125 26.2656 12–4 a. For types:
Poust 172 165 7 −1.875 3.5156 H0: µ1 = µ2 = µ3
71 538.8750 H1: The treatment means are not equal.
Reject H0 if F > 4.46.
For months:
71
d= = 8.875 H0: µ1 = µ2 = µ3= µ4 = µ5
8 H1: The block means are not equal.

824

b. Reject H0 if F > 3.84. b. The slope is 2.2. This indicates that an increase of $1 million in

c. The analysis of variance table is as follows: advertising will result in an increase of $2.2 million in sales.
The intercept is 1.5. If there was no expenditure for advertis-
Source df SS MS F p-value ing, sales would be $1.5 million.
Types 2 3.60 1.80 0.39 0.2397 c. Ŷ = 1.5 + 2.2(3) = 8.1
13–4 H0: β1 ≤ 0; H1: β > 0. Reject H0 if t > 3.182.
Months 4 31.73 7.93 1.71 0.6902
Error 8 37.07 4.63 2.2 − 0
t= = 5.1850
Total 14 72.40 0.4243
Reject H0. The slope of the line is greater than 0.
d. Fail to reject both hypotheses. The p-values are more than .05. 13–5 a.
e. There is no difference in the mean sales among types or
months. y ŷ ( y − ŷ ) ( y − ŷ )2
Σ( y − ŷ) 2
12–5 a. There are four levels of Factor A. The p-value is less than 7 5.9 1.1 1.21 sy · x = √
n−2
.05, so Factor A means differ. 3 3.7 −0.7 .49
b. There are three levels of Factor B. The p-value is less than 1.80
.05, so the Factor B means differ.
8 8.1 −0.1 .01 =√ = .9487
10 10.3 −0.3 .09 4−2
c.  There are three observations in each cell. There is an inter-
action between Factor A and Factor B means because the 1.80
p-value is less than .05.
b. r 2 = (.9648)2 = .9308
CHAPTER 13 c. Ninety-three percent of the variation in sales is accounted for
13–1 a. Advertising expense is the independent variable, and sales by advertising expense.
revenue is the dependent variable. 13–6 6.58 and 9.62, since for an x of 3 is 8.1, found by ŷ =
b. 1.5 + 2.2(3) = 8.1, then x = 2.5 and Σ(x − x) 2 = 5. t from
­Appendix B.5 for 4 − 2 = 2 degrees of freedom at the .10 level is
y 2.920.
12 1 (x − x ) 2
ŷ ± t(sy · x ) √ n +
Sales revenue

9 Σ(x − x ) 2
1 (3 − 2.5) 2
6 = 8.1 ± 2.920(0.9487) √ +

4 5
3 = 8.1 ± 2.920(0.9487)(0.5477)

0 x = 6.58 and 9.62 (in $ millions)

1 2 3 4
Advertising expense
CHAPTER 14

c. 14–1 a. $389,500 or 389.5 (in $000); found by
x y (x − x ) (x − x ) 2
(y − y ) (y − y ) 2
(x − x )(y − y ) 2.5 + 3(40) + 4(72) − 3(10) + .2(20) + 1(5) = 3,895
b. The b2 of 4 shows profit will go up $4,000 for each extra
2 7 −0.5 .25 0 0 0 hour the restaurant is open (if none of the other variables
1 3 −1.5 2.25 −4 16 6 change). The b3 of −3 implies profit will fall $3,000 for each
3 8 0.5 .25 1 1 0.5 added mile away from the central area (if none of the other
4 10 1.5 2.25 3 9 4.5 variables change).
10 28 5.00 26 11.0 14–2 a. The total degrees of freedom (n − 1) is 25. So the sample
size is 26.
10 28 b. There are 5 independent variables.
x = = 2.5 y= =7
4 4 c. There is only 1 dependent variable (profit).
5 d. SY.12345 = 1.414, found by √2. Ninety-five percent of the re-
sx = √ = 1.2910 siduals will be between −2.828 and 2.828, found by
3
26 ±2(1.414).
sy = √ = 2.9439 e. R2 = .714, found by 100∕140. 71.4% of the deviation in profit
3
Σ(X − X )(y − y ) 11 is accounted for by these five variables.
r= = f. R2adj = .643, found by
(n − 1)sx sy (4 −1)(1.2910)(2.9439)
1−[
(26 − (5 + 1)) ]/[ (26 − 1) ]
= 0.9648 40 140

d. There is a strong correlation between the advertising ­expense
and sales. 14–3 a. H0: β1 = β2 = β3 = β4 = β5 = 0
13–2 H0: ρ ≤ 0, H1: ρ > 0. H0 is rejected if t > 1.714. H1: Not all of the βs are 0.
.43 √25 − 2 The decision rule is to reject H0 if F > 2.71. The computed
t= = 2.284 value of F is 10, found by 20∕2. So, you reject H0, which
√1 − (.43) 2
­indicates at least one of the regression coefficients is differ-
H0 is rejected. There is a positive correlation between the percent ent from zero.
of the vote received and the amount spent on the campaign.  Based on p-values, the decision rule is to reject the null
13–3 a. See the calculations in Self-Review 13–1, part (c). hypothesis if the p-value is less than .05. The computed
rsy (0.9648)(2.9439) value of F is 10, found by 20∕2, and has a p-value of .000.
b= = = 2.2
sx 1.2910 Thus, we reject the null hypothesis, which indicates that
at least one of the regression coefficients is different
− 2.2 ( ) = 7 − 5.5 = 1.5
28 10
a= from zero.
4 4

825
 b. For variable 1: H0: β1 = 0 and H1: β1 ≠ 0 f. Do not reject H0.
The decision rule is to reject H0 if t < −2.086 or t > 2.086. g. p-value = 2(.5000 − .2454) = .5092
Since 2.000 does not go beyond either of those limits, we There is no difference in the proportion of adults and chil-
fail to reject the null hypothesis. This regression coeffi- dren that liked the proposed flavor.
cient could be zero. We can consider dropping this vari- 15–3 a. Observed frequencies
able. By parallel logic, the null hypothesis is rejected for b. Six (six days of the week)
variables 3 and 4. c. 10. Total observed frequencies ÷ 6 = 60/6 = 10.
 For variable 1, the decision rule is to reject H0: β1 = 0 if d. 5; k − 1 = 6 − 1 = 5
the p-value is less than .05. Because the p-value is .056, we e. 15.086 (from the chi-square table in Appendix B.7).

f. χ2 = Σ [ ]=
cannot reject the null hypothesis. This regression coefficient (fo − fe ) 2 (12 − 10) 2 (9 − 10) 2
could be zero. Therefore, we can consider dropping this +· · ·+ = 0.8
fe 10 10
variable. By parallel logic, we reject the null hypothesis for g. Do not reject H0.
variables 3 and 4. h. Evidence fails to show a difference in the proportion of
c. We should consider dropping variables 1, 2, and 5. ­Variable 5 ­absences by day of the week.
has the smallest absolute value of t or largest p-value. So 15–4 H0: PC = .60, PL = .30, and PU = .10.
­delete it first and compute the regression equation again. H1: Distribution is not as above.
14–4 a. ŷ = 15.7625 + 0.4415x1 + 3.8598x2 Reject H0 if χ2 > 5.991.
ŷ = 15.7625 + 0.4415(30) + 3.8598(1)
= 32.87 (fo − fe ) 2
b. Female agents make $3,860 more than male agents.
c. H0: β3 = 0 Category fo fe fe
H1: β3 ≠ 0 Current 320 300 1.33
df = 17; reject H0 if t < −2.110 or t > 2.110 Late 120 150 6.00
3.8598 − 0
t= = 2.621 Uncollectible 60 50 2.00
1.4724
The t-statistic exceeds the critical value of 2.110. Also, the 500 500 9.33
p-value = .0179 and is less than .05. Reject H0. Gender
Reject H0. The accounts receivable data do not reflect the na-
should be included in the regression equation.
tional average.
15–5 a. Contingency table
CHAPTER 15 b.  H0: There is no relationship between income and whether
15–1 a. Yes, because both nπ and n(1 − π) exceed 5: nπ = 200(.40) the person played the lottery. H1: There is a relationship
= 80, and n(1 − π) = 200(.60) = 120. between income and whether the person played the
b. H0: π ≥ .40 lottery.
H1: π < .40 c. Reject H0 if χ2 > 5.991.
c. Reject H0 if z < −2.326.
(46 − 40.71) 2 (28 − 27.14) 2 (21 − 27.14) 2
d. χ2 = + +
40.71 27.14 27.14
(14 − 19.29) 2 (12 − 12.86) 2 (19 − 12.86) 2
Region of + + +
19.29 12.86 12.86
rejection = 6.544
α = .01 e. Reject H0. There is a relationship between income level
and playing the lottery.

CHAPTER 16
–2.326 0 16–1 a. Two-tailed because H1 does not state a direction.
Critical value b.

d. z = −0.87, found by:
y
.37 − .40 −.03
z= = = −0.87 .25
.40(1 − .40) √.0012
√
Region of Region of
200 rejection rejection
.20
Probability of success

Do not reject H0.

e. The p-value is .1922, found by .5000 − .3078.
15–2 a. H0: πa = πch .15
1
H : πa ≠ πch
b. .10
c. Two-tailed .10
d. Reject H0 if z < −1.645 or z > 1.645.
87 + 123 210 .05
e. pc = = = .60
150 + 200 350
87 123 x
pa = = .58 pch = = .615 0 1 2 3 4 5 6 7 8 9 10 11 12
150 200
.58 − .615 Number of successes
z = = −0.66
.60(.40) .60(.40)
√ 150
+ Adding down, .000 + .003 + .016 = .019. This is the largest
200 cumulative probability up to but not exceeding .050, which is

826
 half the level of significance. The decision rule is to reject H0 Do not reject H0 if the computed z is between 1.96 and −1.96
if the number of plus signs is 2 or less or 10 or more. (from Appendix B.3); otherwise, reject H0 and accept H1. n1 = 8,
c. Reject H0; accept H1. There is a preference. the number of observations in the first sample.
16–2 H0: median = $3,000, H1: median ≠ $3,000
Binomial distribution with n = 20, and π = 0.5 XL-5000 D2
Distance Rank Distance Rank
Cumulative
probabilities 252 4 262 9
x P(x) in the tails 263 10 242 2
279 15 256 5
0 0.000
273 14 260 8
1 0.000 0.000 271 13 258 7
2 0.000 0.000 265 11.5 243 3
3 0.001 0.001 257 6 239 1
4 0.005 0.006 280 16 265 11.5
5 0.015 0.019 Total 89.5 46.5
6 0.037 0.052
W = 89.5
7 0.074 8(8 + 8 + 1)
8 0.120 89.5 −
2
9 0.160 z=
(8)(8)(8 + 8 + 1)
√
10 0.176
11 0.160 12
12 0.120 21.5
= = 2.26
13 0.074 9.52
14 0.037 0.052 Reject H0; accept H1. There is evidence of a difference in the
15 0.015 0.019 distances traveled by the two golf balls.
16 0.005 0.006 16–5
17 0.001 0.001 Ranks
18 0.000 0.000 Englewood West Side Great Northern Sylvania
19 0.000 0.000 17 5 19 7
20 0.000 20 1 9.5 11

16 3 21 15
Reject H0: median = $3,000 if number of successes is 5 or
13 5 22 9.5
less, or the number of success is 15 or more. In this example,
5 2 14 8
the number of successes is 13. Therefore, fail to reject H0.
18 12
16–3 a. n = 10 (because there was no change for A. A.)
b. ΣR1 = 89 ΣR2 = 16 ΣR3 = 85.5 ΣR4 = 62.5
n1 = 6 n2 = 5 n3 = 5 n4 = 6
Absolute
Before After Difference Difference Rank R− R+ H0: The population distributions are identical.
H1: The population distributions are not identical.
17 18 −1 1 1.5 1.5

[ 6 ]
21 23 −2 2 3.0 3.0 12 (89) 2 (16) 2 (85.5) 2 (62.5) 2
H= + + + − 3(22 + 1)
25 22 3 3 5.0 5.0 22(22 + 1) 6 5 5
15 25 −10 10 8.0 8.0 = 13.635
10 28 −18 18 10.0 10.0 The critical value of chi-square for k − 1 = 4 − 1 = 3 degrees of
16 16 — — — — — freedom is 11.345. Since the computed value of 13.635 is
10 22 −12 12 9.0 9.0 greater than 11.345, the null hypothesis is rejected. We con-
clude that the number of transactions is not the same.
20 19 1 1 1.5 1.5
16–6 a.
17 20 −3 3 5.0 5.0
Rank
24 30 −6 6 7.0 7.0
23 26 −3 3 5.0 5.0 x y x y d d2
48.5 6.5 805 23 5.5 1 4.5 20.25
777 62 3.0 9 −6.0 36.00
H0: Production is the same. 820 60 8.5 8 0.5 0.25
H1: Production has increased. 682 40 1.0 4 −3.0 9.00
The sum of the positive signed ranks is 6.5; the negative 777 70 3.0 10 −7.0 49.00
sum is 48.5. From Appendix B.8, one-tailed test, n = 10, 810 28 7.0 2 5.0 25.00
the critical value is 10. Since 6.5 is less than 10, reject the 805 30 5.5 3 2.5 6.25
null hypothesis and accept the alternate. New procedures 840 42 10.0 5 5.0 25.00
did increase production. 777 55 3.0 7 −4.0 16.00
c.  No assumption regarding the shape of the distribution is 820 51 8.5 6 2.5 6.25
necessary. 0 193.00
16–4 H0: There is no difference in the distances traveled by the
XL-5000 and by the D2.
H1: There is a difference in the distances traveled by the
XL-5000 and by the D2.

827
 6(193) 17–4 a.
rs = 1 − = −.170 For 2015
10(99)
b. H0: ρ = 0; H1: ρ ≠ 0. Reject H0 if t < −2.306 or t > 2.306. Item Weight

10 − 2 Cotton ($0.25/$0.20)(100)(.10) = 12.50

t = −.170√ = −0.488 Autos (1,200/1,000)(100)(.30) = 36.00
1 − (−0.170) 2 Money turnover (90/80)(100)(.60) = 67.50

H0 is not rejected. We have not shown a relationship Total 116.00
­between the two tests.

For 2018
CHAPTER 17 Item Weight
17–1 1.
Country Amount Index (Based=US) Cotton ($0.50/$0.20)(100)(.10) = 25.00
China 831.7 1026.8 Autos (900/1,000)(100)(.30) = 27.00
Japan 104.7 129.3 Money turnover (75/80)(100)(.60) = 56.25
United States 81 100.0 Total 108.25

India 101.5 125.3 b. Business activity decreased 7.75% from 2015 to 2018.
Russia 71.5 88.3
17–5 In terms of the base period, Jon’s salary was $14,637 in 2000
China produced 926.8% more steel than the U.S. and $23,894 in 2018. This indicates that take-home pay in-
2. a. creased at a faster rate than the rate of prices paid for food,
transportation, etc.
Year Average Hourly Earnings Index (1995 = Base) 17–6 $0.37, round by ($1.00/272.776)(100). The purchasing power
has declined by $0.63.
2010 22.76 100.0
17–7
2012 23.73 104.3 Year IPI PPI
2014 24.65 108.3 2007 109.667 93.319
2016 25.93 113.9 2008 97.077 92.442
2018 27.53 121.0 2009 94.330 96.386
The 2018 average increased 21.0% from 2010. 2010 100.000 100.000
2011 102.840 104.710
b. 2012 105.095 106.134
2013 107.381 107.612
Year Average Hourly Earnings Index (1995 – 2000 = Base) 2014 110.877 107.010
2010 22.76 97.9 2015 106.289 104.107
2012 23.73 102.1 2016 107.150 106.079
2014 24.65 106.0 2017 110.906 109.474
2018 115.027 111.008
2016 25.93 111.6
2018 27.53 118.4 The Industrial Production Index (IPI) increased
15.027% from 2010 to 2018. The Producer
The 2018 average increased 18.4% from the average of 2010 and 2012.
Price Index (PPI) increased 11.008%.
17–2 1. a. P1 = ($85/$75)(100) = 113.3
P2 = ($45/$40)(100) = 112.5
CHAPTER 18
P = (113.3 + 112.5)/2 = 112.9 18–1 a.
b. P = ($130/$115)(100) = 113.0 $8,000
$85(500) + $45(1,200)
c. P = (100) 7,500
$75(500) + $40(1,200)
7,000
$96,500
= (100) = 112.9 6,500
85,500
6,000
$85(520) + $45(1,300)
d. P = (100) 5,500
$75(520) + $40(1,300) 5,000
$102,700 4,500
= (100) = 112.9
$91,000 4,000
e. P = √ (112.9)(112.9) = 112.9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
$4(9,000) + $5(200) + $8(5,000) Time period
17–3 a. P = (100)
$3(10,000) + $1(600) + $10(3,000)
b. Over the 18 months, the graph of the time series does not
$77,000 show any trend or seasonal patterns.
= (100) = 127.1
c. Because the graph does not show any trend or seasonal
60,600
b. The value of sales went up 27.1% from 2010 to 2018. patterns, the pattern is random and stationary. Therefore,
the best time series forecasting method is an averaging
method, such as a simple moving average.

828

d. and e. The MAD, or estimate of forecasting error is 751.7885. the best time series forecasting method is an averaging
method. Simple exponential smoothing is a good choice.
Period Revenue 4-Month ABS (error) Bias
d. and e. The forecast for August is $6,849.7643. The error of
the forecast is the MAD, 823.3141.
March $5,874
April 7,651 Period Revenue Forecast (0.2) ABS (error) Bias
May 5,546
March $5,874
June 7,594
April 7,651 $5874.0000 1777.0000 1777.0000
July 6,450 $6,666.25 216.25 −216.25 May 5,546 6229.4000 683.4000 −683.4000
August 5,580 6,810.25 1230.25 −1230.25 June 7,594 6092.7200 1501.2800 1501.2800
September 6,560 6,292.50 267.50 267.50 July 6,450 6392.9760 57.0240 57.0240
October 7,209 6,546.00 663.00 663.00 August 5,580 6404.3808 824.3808 −824.3808
November 7,679 6,449.75 1229.25 1229.25 September 6,560 6239.5046 320.4954 320.4954
December 5,192 6,757.00 1565.00 −1565.00 October 7,209 6303.6037 905.3963 905.3963
January 7,177 6,660.00 517.00 517.00 November 7,679 6484.6830 1194.3170 1194.3170
February 7,693 6,814.25 878.75 878.75 December 5,192 6723.5464 1531.5464 −1531.5464
January 7,177 6417.2371 759.7629 759.7629
March 7,232 6,935.25 296.75 296.75
February 7,693 6569.1897 1123.8103 1123.8103
April 7,742 6,823.50 918.50 918.50 March 7,232 6793.9517 438.0483 438.0483
May 7,142 7,461.00 319.00 −319.00 April 7,742 6881.5614 860.4386 860.4386
June 6,227 7,452.25 1225.25 −1225.25 May 7,142 7053.6491 88.3509 88.3509
July 6,639 7,085.75 446.75 −446.75 June 6,227 7071.3193 844.3193 −844.3193
August 6,937.50 MAD Bias July 6,639 6902.4554 263.4554 −263.4554
751.7885 −231.75 August 6,849.7643 MAD Bias
823.3141 4878.8217

f. The 8-month moving average MAD is 808.5278.

f. 
Using an alpha = 0.7, the forecast for August is $6,610.2779.
Period Revenue 8-Month ABS (error) Bias The error of the forecast is the MAD, 977.1302.
March $5,874
Period Revenue Forecast (0.7) ABS (error) Bias
April 7,651
May 5,546 March $5,874
June 7,594 April 7,651 $5874.0000 1777.0000 1777.0000
July 6,450 May 5,546 7117.9000 1571.9000 −1571.9000
August 5,580 June 7,594 6017.5700 1576.4300 1576.4300
September 6,560 July 6,450 7121.0710 671.0710 −671.0710
October 7,209 August 5,580 6651.3213 1071.3213 −1071.3213
November 7,679 $6,558.000 1121.000 1121.000 September 6,560 5901.3964 658.6036 658.6036
December 5,192 6,783.625 1591.625 −1591.625 October 7,209 6362.4189 846.5811 846.5811
January 7,177 6,476.250 700.750 700.750 November 7,679 6955.0257 723.9743 723.9743
February 7,693 6,680.125 1012.875 1012.875 December 5,192 7461.8077 2269.8077 −2269.8077
March 7,232 6,692.500 539.500 539.500 January 7,177 5872.9423 1304.0577 1304.0577
April 7,742 6,790.250 951.750 951.750 February 7,693 6785.7827 907.2173 907.2173
May 7,142 7,060.500 81.500 81.500 March 7,232 7420.8348 188.8348 −188.8348
June 6,227 7,133.250 906.250 −906.250 April 7,742 7288.6504 453.3496 453.3496
July 6,639 7,010.500 371.500 −371.500 May 7,142 7605.9951 463.9951 −463.9951
August 6,880.500 MAD Bias June 6,227 7281.1985 1054.1985 −1054.1985
808.5278 1538.000 July 6,639 6543.2596 95.7404 95.7404
August 6,610.2779 MAD Bias
g. Based on the comparison of the MADs, the 4-month 977.1302 1051.8255
­moving average has the lower MAD and would be pre-
ferred over the 8-month average. g. 
Based on the comparison of the MADs, the exponential
18–2 a. smoothing model with alpha of 0.2 is preferred because it as
has a lower MAD than the exponential smoothing model with
$8,000 alpha of 0.7.
7,500 18–3 a.
7,000 Annual sales (billions $)
6,500 7,00,000
6,000 6,00,000
5,500 5,00,000
5,000 4,00,000
4,500 3,00,000
4,000
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 2,00,000
Time period 1,00,000
0

b. Over the 18 months, the graph of the time series does not
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

show any trend or seasonal patterns.

Time period, years 1992–2017

c. Because the graph does not show any trend or seasonal
patterns, the pattern is random and stationary. Therefore,
b. The time series graph shows a gradual increase in U.S. total
grocery store annual sales between 1992 and 2017.
829

c. A trend model is appropriate because we would like to es- 18–4 a.
timate the average annual increase shown by the trend Visitors (1,000s)
pattern in the time series graph.

d. Sales = 298,829.3723 + 12,426.7986 (time period). The
quarterly over five years
MAD is 10,932.39. Notice that the error as a percent of the 140.0
forecast is very small. 120.0
100.0
SUMMARY OUTPUT 80.0
Regression Statistics 60.0
Multiple R 0.9906 40.0
R Square 0.9812 20.0
Adjusted R Square 0.9804 0.0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Standard Error 13424.0404
Observations 26 Time period

b.  The pattern in the time series is clearly seasonality. During a
ANOVA
4-quarter time span, winter and summer are always the high-
df SS MS F p-value est number of visitors; spring and fall are always the lowest
number of visitors.
Regression 1 2.25847E+11 2.26E+11 1253.279 0.0000
c. In this time series, there is virtually no trend. So using the
Residual 24 4324916631 1.8E+08 overall average as the base of the seasonal indexes would
Total 25 2.30172E+11 be a logical choice.

d. Computing the indexes by dividing each period’s visitors by
100 shows the following results.
Coefficients Standard Error t Stat P-value
Intercept 298829.3723 5421.0012 55.1244 0.0000 Indexes
Time Period 12426.7986 351.0228 35.4017 0.0000 Season Time Period Visitors (Base = 100)
Winter 1 117.0 1.17
Period Total Sales Forecast ABS Error Spring 2 80.7 0.807
Summer 3 129.6 1.296
1 $337,370 311,256.17 26,113.83 Fall 4 76.1 0.761
2 341,318 323,682.97 17,635.03
Winter 5 118.6 1.186
3 350,523 336,109.77 14,413.23
Spring 6 82.5 0.825
4 356,409 348,536.57 7,872.43
Summer 7 121.4 1.214
5 365,547 360,963.37 4,583.63
Fall 8 77.0 0.77
6 372,570 373,390.16 820.16
7 378,188 385,816.96 7,628.96 Winter 9 114.0 1.14
8 394,250 398,243.76 3,993.76 Spring 10 84.3 0.843
9 402,515 410,670.56 8,155.56 Summer 11 119.1 1.191
10 418,127 423,097.36 4,970.36 Fall 12 75.0 0.75
11 419,813 435,524.16 15,711.16 Winter 13 120.7 1.207
12 427,987 447,950.96 19,963.96 Spring 14 79.6 0.796
13 441,136 460,377.75 19,241.75 Summer 15 129.9 1.299
14 457,667 472,804.55 15,137.55 Fall 16 69.5 0.695
15 471,699 485,231.35 13,532.35 Winter 17 125.2 1.252
16 491,360 497,658.15 6,298.15 Spring 18 80.2 0.802
17 511,222 510,084.95 1,137.05 Summer 19 127.6 1.276
18 510,033 522,511.75 12,478.75 Fall 20 72.0 0.72
19 520,750 534,938.55 14,188.55
20 547,476 547,365.34 110.66
Quarter Seasonal Index
21 563,645 559,792.14 3,852.86
22 574,547 572,218.94 2,328.06 Winter 1.191
23 599,603 584,645.74 14,957.26 Spri ng 0.8146
Summer 1.2552
24 613,159 597,072.54 16,086.46
Fall 0.7392
25 625,295 609,499.34 15,795.66
26 639,161 621,926.14 17,234.86 e. 
The winter index is 1.191. It means that on average, the
MAD number of visitors is 19.1% above the quarterly average of
10,932.39 100,000 visitors, or 191,100 (100,000 × 1.191) visitors.

In the spring, the number of visitors is 18.54% below the
e. 
The predicted annual change in total U.S. grocery sales dollars
quarterly average of 100,000 visitors, or 81,460 (100,000 ×
is $12,426.7986 million.
0.8146) visitors. The summer index is 1.2552. It means
f. Sales = 298,829.3723 + 12,426.7986 (time period). The
that on average, the number of visitors is 25.52% above
next three years–2018, 2019, and 2020–are periods 27,
the quarterly average of 100,000 visitors, or 125,520
28, and 29.
(100,000 × 1.2552) visitors. In the spring, the number of
2018 sales = 298,829.3723 + 12,426.7986 (27) = 634,352.94 ­visitors is 26.08% below the quarterly average of 100,000
2019 sales = 298,829.3723 + 12,426.7986 (28) = 646.779.73 visi­tors, or 73,920 (100,000 × 0.7392) visitors.
2020 sales = 298,829.3723 + 12,426.7986 (29) = 659.206.53
830
CHAPTER 19 Because LCL is a negative value, we set LCL = 0. The shift with
19–1 seven defects is out of control.
300 100 19–4 P(x ≤ 2 | π = .30 and n = 20) = .036
250
200 75
CHAPTER 20

Percent
Count

150 50 20–1
Probability Expected
100 Event Payoff of Event Value
25
50
Market rise $2,200 .60 $1,320
Market decline 1,100 .40 440

No respect

Med. error
Poor food

Poor care

$1,760

Nothing
to do
Dirty

20–2 a. Suppose the investor purchased Rim Homes stock, and the
Count 84 71 63 45 35 2 value of the stock in a bear market dropped to $1,100 as
Percent 28 24 21 15 12 0 anticipated (Table 20–1). Instead, had the investor pur-
chased Texas Electronics and the market declined, the
Cum. % 28 52 73 88 100 100 value of the Texas Electronics stock would be $1,150. The
Seventy-three percent of the complaints involve poor food, difference of $50, found by $1,150 − $1,100, represents
poor care, or dirty conditions. These are the factors the admin- the investor’s regret for buying Rim Homes stock.
istrator should address. b. Suppose the investor purchased Texas Electronics stock,
19–2 a. and then a bull market developed. The stock rose to
$1,900, as anticipated (Table 20–1). However, had the
Sample Times ­investor bought Kayser Chemicals stock and the market
1 2 3 4 Total Average Range value increased to $2,400 as anticipated, the difference of
$500 represents the extra profit the investor could have
1 4 5 2 12 3 4 made by purchasing Kayser Chemicals stock.
2 3 2 1 8 2 2 20–3
1 7 3 5 16 4 6 Expected
9 12 Probability Opportunity
Event Payoff of Event Value
9 12 Market rise $500 .60 $300
x= =3 R= =4
3 3 Market decline 0 .40 0
UCL and LCL = x ± A2 R $300
= 3 ± 0.729(4)
UCL = 5.916 LCL = 0.084 20–4 a.
Probability Expected
Mean chart Event Payoff of Event Value
5.916 UCL
Market rise $1,900 .40 $ 760
Market decline 1,150 .60 690
3 X $1,450

b.
Probability Expected
.084 LCL Event Payoff of Event Value
9 10 11
Time (a.m.) Market rise $2,400 .50 $1,200
Market decline 1,000 .50 500
LCL = D3 R = 0(4) = 0
$1,700
UCL = D4R = 2.282(4) = 9.128
Range chart 20–5 For probabilities of a market rise (or decline) down to .333,
9.128 UCL Kayser Chemicals stock would provide the largest expected
profit. For probabilities .333 to .143, Rim Homes would be the
best buy. For .143 and below, Texas Electronics would give the
largest expected profit. Algebraic solutions:
4 R
Kayser: 2,400p + (1 − p)1,000
Rim: 2,200p + (1 − p)1,100
0 LCL 1,400p + 1,000 = 1,100p + 1,100
9 10 11 p = .333
Time (a.m.)
Rim: 2,200p + (1 − p)1,100
b. Yes. Both the mean chart and the range chart indicate that Texas: 1,900p + (1 − p)1,150
the process is in control.
25 1,100p + 1,100 = 750p + 1,150
19–3 c = = 2.083 p = .143
12
UCL = 2.083 + 3 √2.083 = 6.413
LCL = 2.083 − 3 √2.083 = −2.247

831