Solution Manual for Statistics Alive 3rd Edition by Wendy J. Steinberg, Matthew Price

Solution Manual For All Chapters

Module 1. Math Review, Vocabulary, and Symbols 2.

Fraction

Percentage

Decimal

1/6

12.67%

.167

1/10

10%

.10

1/7

14.3%

.143

Percentage

Decimal

2%

.02

.99%

.0099

1%

.01

4.

Fraction 1/50 99/10,000 1/100

6. a. 77.94 b. 1086.27 c. 39.63

8. a. 110.00 b. 12.64 c. 276.77

Solution Manual For All Chapters 10. a.

4 52 (3  2)  (4)(5)(6)  120

b. (4)(5 + 3 × 2) = (4)(5 + 6) = (4)(11) = 44 c.

4  5  (3  2)=(2)(5)+6=10+6=16

d. 42(5 + 3 × 2) = 16(5 + 6) = (16)(11) = 176 e. (4)(8)(22) = (32)(4) = 128

12. a. √5*6+6-32-2 = 5 b. (4+72 -17)/6 = 6 c. 6-2*2+9/3 = 5 d. 100-10*5-5 = 45 e. √(2-8)*16 -2 = 6 14. a. 98 − (75)(1/3) = 98 − 25 = 73 b. .672 + (2)(1/6) = .672 + .333 = 1.005 = 1.00 c. (1/3) + (5)(1/4) = 2.667 + 1.25 = 3.917 = 3.92

16. a. .155 + 5/2 = .155 + (5) (1/2) = .155 + 2.5 = 2.655 b. 3.28 -3/2 = 3.28 – (3) (1/2) = 3.28 – 1.5 = 4.78 c. 720/9-45 = (720) (1/9) – 45 = 80 – 45 = 35

18. a. b + 20 = .5  b = .5 − 20  b = −19.5

Solution Manual For All Chapters b. F = 1.8C + 32  F − 32 = 1.8C  (F − 32)/1.8 = C . . . OR . . . C = (F − 32)/1.8 c. 50 = 60/a  50a = 60  a = 60/50  a = 1.2

20. a. 7 =.5b +2.5  14 = b + 5  b = 9 b. 64 = 4c  c = 16 c. 3a = .25V  a = (.25V)/3

22. a. ∑(1 + N) = N1 + (N)(N) = N + N2 b. ∑(2X + Y) = 2∑X + ∑Y c. ∑(aX)2 = ∑(a2X2) = Na2∑X2

Module 2. Measurement Scales 2. a. Eye color = N b. Time to solve a puzzle = R c. Genre of favorite TV show = N d. Level of depression = I e. Position in starting lineup = O

4. a. Favorite radio station = N

Solution Manual For All Chapters b. How much sleep you had last night = R c. How many calories you ate today = R d. Athletic ability = I e. Position in graduation processional = O

6. a. Type of disease = N b. Business profit = R c. Pass or fail status = N d. Conscientiousness = I e. Car brands = N

8. a. Hair length = C b. Yearly income = C c. How much sleep you had last night = C d. How many calories you ate today = C e. Athletic ability = C

10. a. Amount of irritation you felt today = C b. Number of times you felt irritated today = D c. How many different ice cream flavors you’ve tasted = D d. How much weight you have gained or lost in the past year = C e. How badly you need a vacation right now = C

Solution Manual For All Chapters

12. a. GPA = 3.15−3.25 b. Test score = 82.5−83.5 c. Miles to work = 15−25 d. Feet of snow = 3.125−3.375 e. Shoe size = 101/2–111/2

14. a. IQ = 115.75 − 116.25 b. Height = 66.45− 66.55 c. Age = 19.5 − 20.5 d. Driving speed = 55 − 65 e. Olympic performance = 9.695 − 9.705

Module 3. Frequency and Percentile Tables

2.

Classes Missed

Frequency

Cumulative Frequency

8

1

32

5

1

31

4

3

30

3

4

27

2

9

23

Solution Manual For All Chapters 1

9

14

0

5

5

Number TV Sets

Frequency

Cumulative Frequency

5

1

25

4

2

24

3

5

22

2

9

17

1

6

8

0

2

2

Classes Missed

Frequency

Rel. Frequency

Cum. Rel. Frequency

8

1

.03125

1.00

5

1

.03125

.96875

4

3

.09375

.93750

3

4

.12500

.84375

2

9

.28125

.71875

1

9

.28125

.43750

0

5

.15625

.15625

4.

6.

8.

Number TV Sets

Frequency

Cumulative Frequency

Percent

Cum. Percent

Solution Manual For All Chapters 5

1

25

4.0

100.0

4

2

24

8.0

96.0

3

5

22

20.0

88.0

2

9

17

36.0

68.0

1

6

8

24.0

32.0

0

2

2

8.0

8.0

10. a.

Second

Frequency

110–119

1

100–109

2

90–99

1

80–89

0

70–79

2

60–69

5

50–59

1

40–49

4

30–39

3

20–29

1

b.

Second

Frequency

111–125

1

96–110

3

Solution Manual For All Chapters 81–95

0

66–80

5

51–65

3

36–50

6

20–35

2

c.

Second

Frequency

100–119

3

80–99

1

60–79

7

40–59

5

20–39

4

d. The 20-s interval gives the clearest picture of the data.

12. a.

2-Spam Interval No. Spams Freq 26-27 4 24-25 7 22-23 19 20-21 18-19 16-17 14-15

Freq

b.

5-Spam Interval No. Spams Freq

c.

10-Spam Interval No. Spams

1

25-29

1

20-29

1

20-24

3

10-19

1

15-19

5

0-9

1 1 2 2

10-14 5-9 0-4

2 7 12

Solution Manual For All Chapters 12-13 10-11 8-9 6-7 4-5 2-3 0-1

1 1 2 2 6 7 2

d. The 5-spam interval seems most informative. The 10-spam interval might also be informative. The 2-spam interval is not informative.

14. The 75th percentile rank has 75% of the scores at or below that score. In a set of 50 scores, that would be (.75)(50) = the 37.5th score. Counting up from the bottom of the table, the 37.5th score is approximately 88. Using the formula, the score is P(75) = 87.5 + (1/3)(37 – 34)(.5) P(75) = 87.5 + (1/3)(3)(.5) P(75) = 87.5 + 1(.5) P(75) = 87.5 + .5 P(75) = 88

16. Answers will vary.

Even Solutions to Textbook Exercises

Solution Manual For All Chapters

Module 4. Graphs and Plots 2. a.

Stem

Leaf

5

8

6

00023367

7

122346688889

8

4456788

9

01

b.

Solution Manual For All Chapters

c.

4. a.

Stem

Leaf

2

2,4

1

0,1,1,2,2,2,3,4,5,8

0

3,4,4,5,5,6,6,7,7,8,8,8,9

b.

Solution Manual For All Chapters

Solution Manual For All Chapters c.

6. a. Number of children in American families = PS b. Shoe size for a random sample of men = N c. Age of residents of a nursing home = NS d. Number of crimes committed by a random sample of citizens = PS e. Percentage of applicants admitted to a highly selective college = PS 8. Positively skewed.

10. Positively skewed.

12. a.

Solution Manual For All Chapters

b. Blue = 30.59 Red = 21.18 Green = 17.65 Purple = 14.12 Pink = 8.24 Orange = 4.74 Yellow = 3.53

Solution Manual For All Chapters 14. a.

Solution Manual For All Chapters

b.

16. a. Eye color in your statistics class = B, P b. Types of tools owned by 50 H.O.H. = B, P c. Amount of water drunk by dogs = SL, H, F d. Miles run by racers = SL, H, F e. Relative positions in a contestant lineup = SL, H, F OR: No graph is appropriate for ordinal scores

18. a. A,B,C, or D answers on a multiple choice test answer key = B, P b. Diagnoses for 100 visitors to a medical clinic = B, P c. Number of months people dated their future spouse prior to marriage = SL,

Solution Manual For All Chapters H, F d. Number of pictures a digital camera will take before the battery is exhausted = SL, H, F e. Preferred brand of laundry detergent = B, P

Module 5. Mode, Median, and Mean 2. a. 230 b.

225  10

(40  38) (2)  225  10  225  10(.143)  225  1.43  226.43 14 14

4.

a. 2 b. 1.5 + (1) [(10-4) /7] = 1.5 + .857 = 2.357

6. a. 1 and 2 b. 1.5  1

(16  14) 2  1.5   1.5  .222  1.72 9 9

Solution Manual For All Chapters

c.

64  2.00 32

8. a. 2 b. 2 c. 2.08

10. a. positively skewed b. The mean should be higher than the median.

12. a. Positively skewed b. Yes 14. Modes—report both the major and the lesser modes. The bimodality will cause the mean to overestimate one group and underestimate the other group.

16. Modes—report one for each of the three types of visits because the three types probably result in very different recovery rates.

Module 6. Range, Variance, and Standard Deviation 2. a. 97 − 67 = 30 b.

Solution Manual For All Chapters +192

6.5

42.25

84

−1.5

2.25

97

11.5

132.25

76

−9.5

90.25

70

−15.5

240.25

67

−18.5

342.25

88

2.5

6.25

95

9.5

90.25

92

6.5

42.25

94

8.5

72.25

∑ = 855

∑ = 1060.50

M = 855/10 = 85.50 s2 = 1060.50/10 = 106.05 (using /N) OR s = 1060.50/9 = 117.831 (using /n − 1) 2

c. s  106.05  10.298 (using / N ) OR

s  117.833  10.855 (using / n  1)

4. a. 8 − 0 = 8 b.

X

X−M

(X − M)2

Solution Manual For All Chapters 0

−2

4

1

−1

1

4

2

4

2

0

0

8

6

36

1

−1

1

2

0

0

4

2

4

5

3

9

0

−2

4

2

0

0

2

0

0

1

−1

1

2

0

0

1

−1

1

2

0

0

1

−1

1

0

−2

4

3

1

1

1

−1

1

1

−1

1

2

0

0

1

−1

1

2

0

0

2

0

0

4

2

4

0

−2

4

Solution Manual For All Chapters 1

−1

1

3

1

1

0

−2

4

3

1

1

3

1

1

∑ = 64

∑ = 90

M = 64/32 = 1.75 s2 = 90/32 = 2.812 (using /N) OR s2 = 90/31 = 2.903 (using /n − 1) c. s  2.812  1.677  using/N  OR

s  2.903  1.704  using/n - 1

6. a. range = 5 b. variance = 1.434 (1.493, using /n-1) c. standard deviation = 1.197 (1.222, using /n-1)

8. a. range = 7

Solution Manual For All Chapters b. variance = 3.128 (3.292, using /n-1) c. standard deviation = 1.768 (1.814, using /n-1)

Module 7. Percent Area and the Normal Curve

2. 2.15 + 0.13 = 2.28% and 2.15 + 0.13 = 2.28%. They are the same.

4. 50% below the median + 34.13% = 84.13% 6. 13.59 + 2.15 + 0.13 = 15.87% and 34.13 + 34.13 + 13.59 + 2.15 + 0.13 = 84.13%. They are not the same.

8. 13.59 + 2.15 = 15.74% 10. 50%

Module 8. z Scores 2. z = (123 – 100)/15 = 23/15 = +1.53. a. 93.70% b. 6.30% c. 43.70%

4. z = (570 - 500)/100 = +0.70. a. 75.80% b. 24.20%

Solution Manual For All Chapters c. 25.80%

6. z = (67.00 – 69.50)/2.00 = -2.50/2.00 = -1.25.

8.

a.

10.56%

b.

89.44%

c.

49.44%

z

107  0 107   .656  .66 163 163

a. 25.46% b. 74.54% c. 24.54%

10. z 

245  266 21   1.75 12 12

a. 4.01% b. 95.99% c. 45.99% 12. z = (234-266)/12 = -2.67 a. at the same point in pregnancy as or earlier than Bo = 0.38% b. at a later point in pregnancy than Bo = 99.62% c. between their due date and the point at which Bo gave birth = 49.62%

Solution Manual For All Chapters 14. Hillsboro: z 

27  22 5   .56 9 9

Lowville: z 

27  18 9   .90 10 10

Media: z 

27  19 8   1.33 6 6

Decia’s fine will be lowest in Hillsboro.

Even Solutions to Textbook Exercises Module 9. Score Transformations and Their Effects 2. a. The mean went up by 5 points, to 59.10. When every score is increased by a constant additive, the mean goes up by that same additive. b. The standard deviation remained the same, at 13.41. When every score is increased by a constant additive, the standard deviation is not affected.

4. a.

The mean is 3 times the previous mean and then 5 points less, at 268.3.

Solution Manual For All Chapters

b.

When every score is increased by a constant multiplier, the mean goes up by that same multiplier. When every score is decreased by a constant subtractive, the mean goes down by that same subtractive.

The standard deviation is multiplied by 3, to 25.59. When every score is increased by a constant multiplier, the standard deviation goes up by that same multiplier. When every score is decreased by a constant subtractive, the standard deviation is not affected.

6. a.

b.

The mean is 10 points higher and then divided by 3, to 26.00. When every score is increased by a constant additive, the mean goes up by that same additive. When every score is decreased by a constant divisor, the mean goes down by that same divisor.

The standard deviation is divided by 3, to 3.86. When every score is increased by a constant additive, the standard deviation is not affected.

When every score is decreased by a constant divisor, the standard deviation goes down by that same divisor.

8. Macey’s SAT score is 450.

10. Burt’s and Curt’s SAT scores differ by 20 points.

12. Catherine’s extraversion T-score on the CPI is 63.33. (NOTE: The score will be reported as 63, as Tscores are rounded to whole numbers.)

14. Svetlana’s and Dymitri’s responsibility scores on the CPI differ by 4 T-score points.

Even Solutions to Textbook

Solution Manual For All Chapters

Exercises

Module 10. Probability Definitions and Theorems 2. a. No b. Yes c. No d. Yes e. Yes 4. a.

Yes

b.

No

c.

Yes

d.

Yes

e.

No

6. a. Yes b. No c. Yes d. Yes

Solution Manual For All Chapters e. No 8. a. Yes b. Yes c. No d. Yes e. No 10. a. MT b. AT c. MT d. AT 12. a. MT b. AT c. AT d. MT

Module 11. The Binomial Distribution

2. The probability is .5.

4. RRR

Solution Manual For All Chapters RRB RBR RBB BRR BRB BBR BBB The probability of each 3-draw outcome is (.5)(.5)(.5) = .125. a. 3 ways to get 1 red. Therefore, the probability is 3 times .125 = .375. b. 1 way to get 0 reds. Therefore, the probability is .125. c. 1 way to get 3 reds. Therefore, the probability is .125.

6. a. b. c.

THHH, HTHH, HHTH, HHHT = 4/16 = .25 TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, HTTT, HTTH, HTHT, HHTT = 11/16 = .6875 THHH, HTHH, HHTH, HHHT, HHHH = 5/16 = .3125

a. b. c.

RRR = 1/8 = .125 BBB = 1/8 = .125 BBR, BRB, RBB = 3/8 = .375

a. b.

NNNN = 1/16 = .0625 TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, HTTT, HTTH, HTHT, HHTT = 11/16 = .6875 TTHH, THTH, THHT, NTTH, HTHT, HHTT = 6/16 = .375

8.

10.

c.

12.

Solution Manual For All Chapters

( p  q)6  p6  6 p5q1 

65 4 2 65 4 3 3 65 43 2 4 65 43 pq  pq  pq  1 2 1 2  3 1 2  3  4 1 2  3  4

= 1p6 + 6p5q1 + 15p4q2 + 20p3q3 + 15p2q4 + 6p1q5 + 1q6 a. 3 reds = 20(.5)3(.5)3 = 20(.5)(.5)(.5)(.5)(.5)(.5) = 20(.015625) = .3125 b. 0 reds = 1(.5)6 = (.5)(.5)(.5)(.5)(.5)(.5) = .015625 c. 3 reds = 20(.5)3(.5)3 = 20(.5)(.5)(.5)(.5)(.5)(.5) = 20(.015625) = .3125 2 reds = 15(.5)2(.5)4 = 15(.5)(.5)(.5)(.5)(.5)(.5) = 15(.015625) = .234375 1 red = 6(.5)1(.5)5 = 6(.5)(.5)(.5)(.5)(.5)(.5) = 6(.015625) = .09375 0 reds = 1(.5)6 = (.5)(.5)(.5)(.5)(.5)(.5) = .015625 3 reds + 2 reds + 1 red + 0 reds = .3125 + .234375 + .09375 + .015625 = .65625

14. (p + q)N = pN + NpN-1q1 + [N(N-1)/1x2]pN-2q2 + q3 = 1(p)3 + 3(p)2(q)1 + [3(2)/1x2](p)1q2 + 1(q)3 a. b. c.

1(.5)3 = .125 3(.5)2(.5)1 = 3(.25)(.5) = .375 3(.5)2(.5)1 + 1(.5)3 = 3(.25)(.5) + 1(.5)3 = .375 + .125 = .50

16. (p + q)N = pN + NpN-1q1 + [N(N-1)/1x2]pN-2q2 + [N(N-1)(N-2)/1x2x3]pN-3q3 + [N(N-1)(N-2)(N-3)/1x2x3x4]pN-4q4 + q5 = 1(p)5 + 5(p)4(q)1 + [5(4)/1x2](p)3q2 + [5(4)(3)(2)/1x2x3x4p1q4 + 1(q)5 a. b.

[5(4)/1x2](.5)3(.5)2 = 10(.125)(.25) = .3125 1(.5)5 + 5(.5)4(.5)1 + [5(4)/1x2](p)3q2 10(.125)(.25)

[5(4)(3)/1x2x3]p2q3

=

+

.03125 + 5(.0625)(.5) +

Solution Manual For All Chapters

c.

= .03125 + .15625 + .3125 = .482 1(.5)5 = .03125

18. a. .246 + .246 + .164 + .070 + .018 + .002 = .746 b. .070 + .164 = .234 c. .070 + .018 + .002 = .090 d. .246 20. a.

.0156 + .0938 + .2344 = .3438

b.

.2344

c.

.2344 + .0938 + ..0156 = .3438

a.

.0312 + .0039 = .0351

b.

.0039

c.

.2188

a.

.1318

b.

0002

c.

0044 + .0002 = .0046

d.

.3560

22.

24.

26. .200 28. .531 30. .N = 5, p = .40. All five cum laude = .0102 32 Can’t use table because p = .333, and there is no column for that probability.

Solution Manual For All Chapters 34 N = 7, hit or miss p = .50 by mere guessing. At least five correct = .1614 + .0547 + .0078 = .2239

Module 12. Sampling, Variables, and Hypotheses 2. Stratified random and cluster 4. Simple random 6. cluster 8. systematic random 10. IV = rap or classical music DV = amount of aggression displayed 12. IV = bullied during childhood or not DV = amount of depression in their 20s 14. IV = socioeconomic status DV = number of books signed out

16. IV = time of appointment DV = rate of showing up for appointments 18. a. Men prefer a lower house temperature than women do. b. There is a difference in the preferred house temperature between men and women. c. There is no significant difference in the preferred house temperature between

Solution Manual For All Chapters men and women.

20. a. Teens who listen to rap music display more aggression in a simulated conflict situation than those who listen to classical music. b. Teens who listen to rap music display a different amount of aggression in a simulated conflict situation than those who listen to classical music. c. There is no significant difference in the amount of aggression displayed in a simulated conflict situation between teens who listen to rap music or classical music.

22. a. Boys who were bullied as children are more depressed in their 20s than boys who were not bullied as children. b. The amount of depression for boys in their 20s differs for boys who were bullied during childhood and those who were not. c. There is no significant difference in the depression level between boys in their 20s who were bullied during childhood and those who were not.

24. a. Children from lower socioeconomic neighborhoods check out fewer library books that do children form higher socioeconomic neighborhoods.

Solution Manual For All Chapters b. Children from different socioeconomic neighborhoods differ in the number of library books they check out. c. There is no significant difference in the number of library books checked out by children from lower and higher socioeconomic neighborhoods.

26. a. Clients keep more later appointments than earlier appointments. b. Time of appointment affects clients’ rate of keeping the appointments. c. Time of the appointment has no effect on clients’ rate of keeping the appointments.

Module 13. Errors and Significance 2. Type 2—there is really a height difference between men and women in favor of men, but you did not find it in your study.

4. Type 1—Stan really is not competent, but the company thought that he was competent.

6. Type 2. There really is a difference (in the opposite direction), but the researcher did not find the difference in this study.

8. Type 1. There really is no significant difference between males and female infants in the age at which they first walk, but the pediatrician found a difference.

Solution Manual For All Chapters

Even Solutions to Textbook Exercises Module 14. The z Score as a Hypothesis Test 2. a. There is no significant difference between Girard’s IQ and the average IQ. b. Girard’s IQ is significantly above average. c.

Z

142  100 42   2.80 15 15

Far less than 1% of people have IQs of 142 or higher. d. This IQ is way above the average IQ, so I would reject the null hypothesis.

4, a. b. c. d.

Null hypothesis = Professor Laffing is not funnier than the nationwide average funniness.

Professor Laffing is significantly funnier than the nationwide average. State the research (alternative) z = (54-40)/6 = 14/6 = +2.33. The probability of this funniness level (or higher) from a population with the given funniness rating is .0099.

It does seem that someone as funny as Dr. Laffing is funny enough to reject the null hypothesis, because the probability that he would be this funny by chance alone is .0099, or less than 1%.

Solution Manual For All Chapters

Module 15. Standard Error of the Mean 2. It will be symmetric, with a normally distributed shape.

4. a.

6 6   1.20 25 5

b.

6 6   1.00 36 6

c.

6 6   0.14 49 7

a.

5/SQR 49 = 5/7 = 0.714

b.

5/SQR 64 = 5/8 = 0.625

c.

5/SQR 81 = 5/9 = 0.555

6.

8. Yes, the central limit theorem tells us that the sampling distribution will be

Solution Manual For All Chapters normally distributed.

Module 16. Normal Deviate Z Test 2. a. There is no significant difference between the arithmetic achievement of Ms. Teachwell’s class and the statewide achievement. b. M = 79.80

M 

7.0 7.0   1.28 30 5.477

Z ND 

79.80  77.0 2.80   2.19 1.28 1.28

c. 1.43% d. The students made lucky guesses on this test. This class is particularly able in arithmetic regardless of the teaching.

4. a. There is no significant difference between the mean SAT score of this group of selective colleges and the announced mean SAT score for all selective colleges. b. M = 562

M 

50 50   9.13 30 5.477

Z ND 

562  600 38   4.16 9.13 9.13

c. Much less than 1/10 of 1%.

Solution Manual For All Chapters

6. M = 6% M = /SQR n = 2/SQR 49 = 2/7 = 0.286 ZND = 6-7/0.286 = -1/0.286 = -3.50 Probability is less than .0013.

8. M = 69.75 inches M = /SQR n = 2.0/SQR 100 = 2/10 = 0.20 ZND = 69.75-69.5/0.20 = 0.25/0.20 = +1.25 Probability is .0956.

Even Solutions to Textbook Exercises

Module 17. One-Sample t Test 2. a. There is a difference between the height of 3-year-olds at this day care center and 3-year-olds nationwide. b. Nondirectional critical t at .05 α = 2.262

Solution Manual For All Chapters c. M = 35 s=3 𝑆𝑀 =

3 √10

=

3 = 0.95 3.16

𝑡=

d.

37 − 35 3 = = 3.16 0.95 0.95

Yes

4. a. The job placement rate of biology graduates in these colleges differs from the nationwide job placement rate of biology graduates. b. Nondirectional critical t at .05 α = 2.145 c. M = 32 s = 10 𝑆𝑀 =

10

=

10 = 2.58 3.87

√15 32 − 26 6 𝑡= = = 2.33 2.58 2.58

d. Yes 6. a. Faculty at Almost Ivy college publish significantly more refereed journal articles than faculty at 4-year colleges nationwide. b. Directional critical t at .01 α = 2.390 c. 𝑆𝑀 =

0.80 √60

=

0.80 = 0.103 7.75

Solution Manual For All Chapters

t

3.7  3.2 .50   4.85 .103 .103

d. Yes

8. a.

Directional hypothesis = The reading achievement of Merriweather’s 2 nd graders is significantly lower than the nationwide achievement.

b.

critical t for  .05 = 1.6794.

c.

𝑆𝑀 =

d.

𝑡=

e.

Yes, Merriweather’s 2nd graders are reading at a significantly lower level

3.1 √46

=

74.6−76.2 0.46

3.1 6.78

=

= 0.46 1.6

0.46

= 3.48

than 2nd graders nationwide.

10. a.

Nondirectional hypothesis = The sales mount of the local franchises are different from that of the company nationwide.

b.

critical t for  .01 = 3.2498.

c.

𝑆𝑀 =

d.

𝑡=

e.

No, this franchise’s sales volume was not significantly different than the

1.2 √10

=

33.6−32.4 0.38

1.2 3.16

=

= 0.38 1.2

0.38

= 3.16

national average. NOTE: This result barely missed statistical significance, and would in fact have been statistically significant had this sample size been larger, or had the study been conducted as a directional test, or had the chosen alpha level been higher. This is a good example of the silliness of dichotomous decision-making, as mentioned briefly at the end of Module 17 and discussed in detail in Module 18.

12.

Solution Manual For All Chapters

a. No (-2.37 needed) b. No (-1.99 needed)

Even Solutions to Textbook Exercises Module 18. Interpreting and Reporting OneSample t: Error, Confidence, and Parameter Estimates 2. a. 24 df b. 3.75% c. 96.25% d. t(24) = 1.89, p = .0375

4.

Solution Manual For All Chapters a.

df = 23

b.

actual Type 1 error = 7.50%

c.

confidence = 100-7.50 = 92.50%

d.

t(23) = 1.89, p = .075

6. Confidence = 100-2 = 98% confident. Sample size is already taken account of by the t-table when the results were reported.

8. a. 6.2 min b. 6.2 min c. CI = 6.2 ± (1.73)(.33) = 6.2 ± .57 = 5.63 and 6.77 d. Larger, because we are more confident that the interval includes the mingle time for the new transfers. e. Smaller, because as sample size increases, the standard error and the tabled critical t both decrease, and hence the confidence interval decreases.

10. a.

point estimate = 12.2 years

b.

12.2 years will fall in the center of the interval

c.

99% CI = M  (tcrit at directional  (M) = 12.2 (2.7045) (0.365) = 12.2  0.987 = 11.213 and 13.187

d.

With 25 employees rather than 41, the confidence interval will be wider because the M will increase, which in turn happens due to the smaller divisor in the M formula.

12. 95% CI = M  (tcrit at directional  sM) = 74.8  (2.1448) (1.263) = 74.8  2.709 = 72.091 and 77.509

Solution Manual For All Chapters

Even Solutions to Textbook Exercises Module 19. Standard Error of the Difference Between the Means 2.

As the two sample sizes increases, the standard error of the difference between the means will decrease. This is because sample size is in the denominator of the two M values that make up the M1 – M2.

4.

The difference might be due to different sample sizes in the two studies. If sample size is small, the M1 – M2 is large, and if the M1 – M2 is large, the t will be small. This is because the M1 – M2 is the divisor in the t-test. For this reason, small sample sizes can mask the difference that exists in the numerator. The opposite happens with a large sample size.

6.

 M  M  1.322  1.942  1.74  3.76  5.50  2.35 1

2

Because sex is not assigned, extraneous variables may coexist with sex to explain the results.

8. M1 – M2 = SQR [(1.45)2 + (1.22)2] = SQR (2.1025 + 1.4884) = SQR (3.5909) = 1.895

10. M1 – M2 = SQR [(0.85)2 + (0.96)2] = SQR (0.7225 + 0.9216) = SQR (1.6441) = 1.2822

Solution Manual For All Chapters

Even Solutions to Textbook Exercises Module 20. t Test With Independent Samples and Equal Sample Sizes 2. a. IV = assistance or no assistance DV = amount of purchase b. H0: There is no significant difference in the amount of furniture purchased by customers given shopping assistance and those not given shopping assistance.

HA: Customers given shopping assistance will buy less furniture than those not given shopping assistance. c. 𝑠𝑀1 =

𝑠𝑀2 =

750 √13 514 √13

=

750 = 208.01.

3.61

=

514 = 235.75 3.61

𝑠𝑀1−𝑀2 = √208.012 + 235.752 = √43269.23 + 55576.92 = 314.40 𝑡=

529 − 514 = 0.05 314.40

Do not reject H0 at either .05 α or .01 α.

4. a. Nondirectional

Solution Manual For All Chapters b. Homeowner’s willingness to display the campaign signs will differ by the size of the sign. c. 𝑠𝑀1 =

𝑠𝑀2 =

6 √10

=

6 √10

6 3.16

=

= 1.90

6 = 1.90 3.16

𝑠𝑀1−𝑀2 = √1.902 + 1.902 = √3.60 + 3.60 = 2.68 𝑡=

32 − 40 = −2.98 2.68

Critical t = 2.88. Reject H0.

6. 𝑠𝑀1 = 𝑠𝑀2 =

. 02 √15 . 04 √15

=

. 02 = 0.01 3.87

=

. 04 = 0.01 3.87

𝑠𝑀1−𝑀2 = √0.012 + 0.012 = .01 𝑡=

. 24 − .27 = −2.60 0.01

No, there are no differences between men and women.

8. 𝑠𝑀1 = 𝑠𝑀2 =

1.51 √12 1.76 √12

=

1.51 = 0.44 3.46

=

1.76 = 0.51 3.46

𝑠𝑀1−𝑀2 = √0.442 + 0.512 = .67

Solution Manual For All Chapters 6.92 − 7.25 = −0.49 0.67

𝑡=

No, customers are not significantly more satisfied with either internet package

10. 𝑠𝑀1 = 𝑠𝑀2 =

3.29 √8 4.46 √8

=

3.29 = 1.16 2.83

=

4.46 = 1.58 2.83

𝑠𝑀1−𝑀2 = √1.162 + 1.582 = 1.96 𝑡=

8.63 − 13.75 = −2.61 1.96

Yes, the police car results in significantly lower speed than the warning sign does.

Even Solutions to Textbook Exercises Module 21. t Test With Unequal Sample Sizes

2. Phonics: M = 86.73, SS = 986.90 Whole language: M = 83.60, SS = 784.40

Solution Manual For All Chapters t = 0.87 Critical t = 2.069 Curriculum makes no significant difference in reading enrichment.

4. Rainy nights: M = 13.75, SS = 186.244 Clear nights: M = 8.55, SS = 296.44 t = 3.36 Critical t = 2.467 a. The dogs do howl significantly more on rainy nights than on clear nights. b. Mrs. Campbell may have slept through some howling incidents. There may be disagreement about what is classifiable as a separate howling incident.

𝑡= 6.

0.22−0.25 . √( 1

.

1

1 )( 2 1

1 ) 1

= −.01

No, there is no difference between the groups

𝑡= 8

10−11 √(

12

1

1 1 )( ) 2 12

= −.04

No, customers are not significantly more satisfied with either internet package

Even Solutions to Textbook Exercises

Solution Manual For All Chapters

Module 22. t Test With Related Samples 2. a. Independent samples b. Matched samples 4.

6

a.

repeated measures – pre to post

b.

independent

Mental rotation: Cannot be conducted as a repeated measures study because the IV is gender, which cannot change. Subjects are able to experience only one of the conditions.

8. Mean of D = 0.25, ∑D = 4, ∑D2 = 419 t = 0.199 Critical t = 1.75 a. There is no significant difference in the grades assigned by the two professors. b. No, Dr. .Hard’s students do not receive significantly lower grades than Dr. Easy’s students do.

10. Mean of D = -0.333; D = -4; ∑D2 = 68.67 t = Mean of D / SQR { [D2 – (D)2/n] / (n) (n-1)} = -.333 / SQR { [70 – (-4)2/12] / (12)(11) } = -.462 No, customers are not significantly more satisfied with either internet package 12. Mean of D = -5.125; D = -41; D2 = 225 t = Mean of D / SQR { [D2 – (D)2/n] / (n) (n-1)} = -5.125 / SQR { [225 – (-41)2/8] / (8)(7) }

Solution Manual For All Chapters = -9.944 Yes, police car results in significantly lower speed.

Even Solutions to Textbook Exercises Module 23. Interpreting and Reporting TwoSample t: Error, Confidence, and Parameter Estimates 2. a. 30 df b. 2% c. 98% d. t(30) = 2.182, p = .02

4. a. 14 df b. 3.75% c. 96.25% d. t(14) = 1.95, p = .0375

6. a. 18 df b. 1.75%

Solution Manual For All Chapters c. 98.25% d. t(18) = 2.33, p = .0175 8.

96%

10. a. 6 points b. 6 c. CI = 6 ± (1.96)(2.15) = 6 ± 4.21 = 1.79 and 10.21 d. Smaller, because we are less confident that the interval contains the actual difference between the means for the 350 incoming women and 350 incoming men. e. Larger, because as sample size decreases, the standard error and the tabled critical t both increase, and hence the confidence interval increases.

12. a.

point estimate = 0.0313 seconds

b.

-0.0313 seconds will fall in the middle of the interval

c.

99% CI = 0.0313  (2.7633) (0.01018) = 0.0313  .0281 = 0.0032 and 0.0594

14. a.

point estimate = -0.3333 points

b.

-0.3333 points will fall in the center of the interval

c.

95% confidence = -0.3333  (2.0739) (0.6605) = -0.3333  1.3698 = 1.4031 (with free email less satisfied) and +1.3365 (note the sign flip; this means that those with free basic TV are less satisfied)

16. a.

point estimate = -5.125 mph

Solution Manual For All Chapters b.

-5.125 will fall in the center of the interval

c.

95% CI = -5.125  (2.1448) (1.96112) = -5.125  4.2962 = -9.4212 and 0.863

18. a. 8.2% b. 8.2 c. CI = 8.2 ± (2.88)(2.62) = 8.2 ± 7.55 = .65 and 15.75

Even Solutions to Textbook Exercises Module 24. ANOVA Logic: Sums of Squares, Partitioning, and Mean Squares 2. 683.78 – 457.52 = 226.26 4. 245.22 + 106.33 = 351.55 6. Yes, it is possible. No, it will not be statistically significant.

8. 100/25 = 4.00 10. 3.00 = 300/MSwith. Thus, MSwith = 100 12. 2.00 = MSbet/400. Thus, MSbet = 800 14. a. 200/25 = 8.00

Solution Manual For All Chapters b. 100/50 = 2.00 c.

16. The numerator stays the same while the denominator increases. This will decrease the value of F

Even Solutions to Textbook Exercises Module 25. One-Way ANOVA: Independent Samples and Equal Sample Sizes 2. a. IV = route taken to work DV = time (in minutes) b. H∅: There will be no significant difference in the time it takes to get to work between those who take different routes.

HA: There will be a significant difference in the time it takes to get to work due to the different routes. c.

Source

SS

df

MS

F

Solution Manual For All Chapters Between

36

2

18

Within

125

21

5.95

Total

161

23

3.02

e. F(2, 21) = 3.02, p > .01 4. a. IV = GPA bracket DV = hours studying during week before finals b. H0: GPA bracket will make no significant difference in the time spent studying.

HA: GPA bracket will make a significant difference in the time spent studying. c.

Source

SS

df

MS

F

Between

8

3

2.67

1.37

Within

70

36

1.94

Total

78

39

e. F(3, 36) = 1.37, p > .01

6. a. There is no significant difference in number of barking events due to size of dog. b.

Source

SS

df

MS

F

Between

65

2

32.5

8.36

Solution Manual For All Chapters Within

35

9

Total

100

11

3.89

c. F(2,9) = 8.36, p < .01

8. a. There is no significant difference in salary due to college major. b.

Sum of Squares

Mean df

Square

Between Majors

5,097.844

3

1699.281

Within Majors

6,175.875

28

220.567

Total

11,273.720

31

c. F(3,28) = 7.704, p <.01

10. a. 5 groups b. 10 subjects per group c. 3.45 d. >99% confident, p < .01

12.

F

Sig.

7.704

<.01

Solution Manual For All Chapters

Sum of

Mean

Squares

df

Square

Between Drugs

304.54

3

101.513

Within Drugs

1024.89

76

13.485

Total

1329.43

79

Even Solutions to Textbook Exercises Module 26. Tukey HSD Test

2. Public: M = 32.50 Highway: M = 48.75 Back roads: M = 64.875

HSD  4.61 P

27.45  4.61 3.43  (4.61)(1.85)  8.53 8 H

P

–

H

16.25

–

BR

32.375

16.125

BR

–

F

Sig.

7.528

<.01

Solution Manual For All Chapters The statistical significance lies between all pairs: (1) public transportation and highway, (2) public transportation and back roads, and (3) highway and back roads.

4. 2.0–2.5 GPA: M = 27.3 2.5–3.0 GPA: M = 29.8 3.0–3.5 GPA: M = 32.8 3.5–4.0 GPA: M = 33.3

HSD  4.75

7.76  4.75 .776  (4.75)(.88)  4.18 10

2.0–2.5

2.5–3.0

3.0–3.5

2.0–2.5

–

2.5–3.0

2.5

–

3.0–3.5

5.5

3.0

–

3.5–4.0

6.0

3.5

0.5

3.5–4.0

–

The statistical significance lies (1) between 2.0–2.5 and 3.0–3.5 and (2) between 2.0–2.5 and 3.5–4.0.

6.

Small Dogs: M = 7.25 Medium Dogs: M = 2.75 Large Dogs: M = 5.0 HSD = 4.68 [SQR (0.795/12)] = 4.68 [SQR (0.066)] = (4.68) (0.257) = 1.204 The significant F is due to the difference between all dogs.

8.

Solution Manual For All Chapters English: M = 42.875, Accounting: M = 73.750, Psychology: M = 46.250, Chemistry: M = 63.500.

HSD = 4.84 [SQR (220.557/8)] = 4.84 [SQR (27.570)] = (4.84) (5.251) = 25.414 (All values were multiplied by 1000) The significant F is due to the difference between (1) accounting and English, and between (2) accounting and psychology.

10.

HSD  4.56 G1

G2

41.83  4.56 1.99  (4.56)(1.41)  6.43 21 G3

G1

–

G2

6.53

–

G3

0.06

6.59

–

G4

4.32

2.21

4.38

G4

–

The statistical significance lies (1) between G1 and G2 and (2) between G2 and G3.

Even Solutions to Textbook Exercises Module 27. Scheffé Test 2. Pub versus HW:

Solution Manual For All Chapters SS = 1056.25 MS = 528.125 F = 19.24 Pub versus BR: SS = 4192.562 MS = 2096.281 F = 76.23 HW versus BR: SS = 1041.312 MS = 520.656 F = 18.94 Critical F(2, 21) = 4.87 The statistical significance is (1) between public transportation and highway, (2) between public transportation and back roads, and (3) between highway and back roads.

4. 2.0–2.5 versus 2.5–3.0 GPA: SS = 31.25 MS = 10.42 F = 1.34 2.0–2.5 versus 3.0–3.5 GPA: SS = 151.25 MS = 50.42 F = 6.50 2.0–2.5 versus 3.5–4.0 GPA: SS = 180 MS = 60 F = 7.73

Solution Manual For All Chapters 2.5–3.0 versus 3.0–3.5 GPA: SS = 45 MS = 15 F = 1.93 2.5–3.0 versus 3.5–4.0 GPA: SS = 1.25 MS = 0.42 F = 0.05 Critical F(3, 36) = 3.89 The statistical significance is (1) between 2.0–2.5 and 3.0–3.5 GPA and (2) between 2.0–2.5 and 3.5–4.0 GPA.

Even Solutions to Textbook Exercises Module 28. Main Effects and Interaction Effects 2. Answer will vary.

4. Two 6. Three 8. One 10. Four

Solution Manual For All Chapters 12. a. IV1 = commute miles IV2 = gender DV = number of absences in first M-W-F class b. There can be two possible main effects and one possible interaction effect. c. Two-way ANOVA d. 3 × 2 e.

f.

Male

Female

1.9

1.3

1.6

2.2

1.6

1.9

2.8

2.1

2.45

2.3

1.67

g. There appears to be a main effect for gender. The line height for males is higher than for females, and the column mean for males is greater than for

Solution Manual For All Chapters females. There also appears to be a main effect for commute miles. For each line, the greater the commute miles, the greater are the absences. The row means also increase consistently as commute miles increase. h. There does not appear to be an interaction effect. The lines are quite parallel, and the row and column means increase consistently rather than inconsistently.

14. a.

IVs = parents’ SES and mother’s religiosity

b.

two possible main effects and one possible interaction effect

c.

two way ANOVA or a 3x3 ANOVA.

d.

tree diagram of design

e. Graph of group means

Solution Manual For All Chapters

f. table of group means

Low Religiosity

Medium Religiosity

High Religiosity

16

19

19

Row Total 18

20

20

23

21

24

24

27

25

20

21

23

Below Average SES Average SES Above Average SES Column Total g. Row totals suggest a main effect for parental SES. Column totals suggest a main effect for mother’s religiosity. Graph shows the same average mean differences for each IV. h. Means within the body of the chart follow a fairly consistent pattern across levels of the two IVS, so there does not appear to be an interaction effect. Graph also shows reasonably parallel lines, so no indication of an interaction. Nevertheless, in both the table and the graph, we see a slight inconsistency for the lowest SES and lowest religiosity, when compared to the other means within those same IVs. Thus, there might be a slight interaction. A statistical test (Module 29) would determine the interaction’s existence.

16.1 a.

no main effect for IV1

Solution Manual For All Chapters b.

no main effect for IV2

c.

no interaction effect

16.2 a. no main effect for IV1 b. no main effect for IV2 c. yes, an interaction effect

16.3 a.

yes, main effect for IV1

b.

yes, main effect for IV2

c.

no interaction effect

16.4 a. yes, main effect for IV1 b. yes, main effect for IV2 c. yes, interaction effect 18.1. a. No b. Yes c. No 18.2. a. Yes b. Yes c. Yes 18.3. a. Yes

Solution Manual For All Chapters b. Yes c. No 18.4. a. No b. No c. Yes

20.1. a. No b. Yes c. Yes 20.2. a. No b. No c. Yes 20.3. a. Yes b. Yes c. No 20.4. a. No b. Yes c. No

Even Solutions to Textbook

Solution Manual For All Chapters

Exercises .

Module 29. Factorial ANOVA 2. a. DV = number of cans of beverage IV1 = location of party (on campus, off campus) IV2 = fraternity membership (member, new pledge) b. 2 × 2 ANOVA c. There are two possible main effects and one possible interaction effect. d. XF, On = 2.50 XF, Off = 4.25 XNP, On = 2.625 XNP, Off = 4.125

Solution Manual For All Chapters e. I expect a main effect for party location but not for fraternity membership status. I do not expect an interaction effect. f., g.

Source

SS

df

MS

F

Sig.

Frat membership

0.000

1

0.000

0.000

>.05

Party location

21.25

1

21.25

18.341

<.01

Frat mem × party loc

0.125

1

0.125

0.109

>.05

Within

32.250

28

1.152

Total

53.500

31

h. There is a main effect for party location. i. There is no interaction effect. j. Ffrat mem(1, 28) = 0.000, p > .05 n.s.

Fparty loc(1, 28) = 21.25, p < .01 Ffrat mem × party loc(1, 28) = 0.125, p > .05 n.s.

4. a. DV = milliseconds IV1 = number of stimuli IV2 = age IV3 = amount of sleep b. 3 × 3 × 2 ANOVA c. There are three possible main effects and four possible interaction effects. d., e.

Source

SS

df

MS

F

Sig.

Solution Manual For All Chapters Stimuli

2,003,710.49

2

1,001,855.24

374.05

<.01

Age

1,082,840.16

2

541,420.08

202.14

<.01

Sleep

799,192.90

1

799,192.90

298.39

<.01

Stimuli × age

82,673.44

4

20,668.36

7.72

<.01

Stimuli × sleep

41,187.47

2

20,593.73

7.69

<.01

Age × sleep

6,986.07

2

3,493.03

1.30

>.05

Stimuli × age × sleep

18,424.47

4

4,606.12

1.72

>.05

Within

192,844.40

72

2,678.39

Total

4,227,859.40

f. There is a main effect for number of stimuli, a main effect for age, and a main effect for amount of sleep. g. There is an interaction effect for number of stimuli by age, and an interaction effect for number of stimuli by amount of sleep. h. Fstimuli(2, 72) = 374.05, p < .01 Fage(2, 72) = 202.14, p < .01 Fsleep(1, 72) = 298.39, p < .01.

Fstimuli × age(4, 72) = 7.72, p < .01 Fstimuli × sleep(2, 72) = 7.69, p < .01 Fage × sleep(2, 72) = 1.30, p > .05 n.s.

Fstimuli × age × sleep(4, 72) = 1.72, p > .05 n.s.

6. a.

ANOVA summary table

Solution Manual For All Chapters Source

Spay/Neut er Size Spay/Neut er x Size

Sums of Squares 5.01 4 192.

361 7.86 1

Df

Mean Square

F

Sig.

1

5.014

.533

2

96.18 1 3.931

10.22 6 .418

.46 8 .00 0 .66 0

66

9.405

2 Within Total

620.

750 825.

986

71

b. FSpay/Neuter(1,66) = .533, p = .468 FSize(2,66) = 10.226, p = .000 FSpay/Neuter x Size(2,66) = .418, p = .660

8. a. ANOVA summary table

Source Degree Level Major Degree Level x Major Within

Sums of Squares 306.250

D f 1

7,996.31 3 190.625

3

12,901.7 50

5 6

3

Mean Square 306.250

F

Sig.

1.329

2,665.4 38 63.542

11.56 9 .276

.25 4 .00 0 .84 3

230.388

Solution Manual For All Chapters Total

21,394.9 37

6 3

b. FDegree Level(1,56) = 1.329, p = .254 FMajor(3,56) = 11.569, p = .000 FDegree Level x Major(3,56) =.276, p = .843

Module 30. One-Variable Chi-Square: Goodness of Fit 2. a. There is a significant difference between the sex ratio of this dorm and the college. b. 1 df c.

Men

Women

Obs

Exp

Obs

Exp

86

90

132

128

d.

(86  90) 2 (132  128) 2   0.31 90 128

e. Critical χ2 = 3.84. The research hypothesis is not supported. Do not reject the null hypothesis.

4.

Solution Manual For All Chapters a. There is a significant difference in the source of the highest degree between faculty at this college and faculty nationwide. b. 2 df c.

Alumni

Within State

Out of State

Obs

Exp

Obs

Exp

Obs

Exp

63

42

113

92

206

248

d.

(63  42) 2 (113  92) 2 (206  248)2    22.40 42 92 248

e. Critical χ2 = 5.99. The research hypothesis is supported. Reject the null hypothesis.

6. a.

Male

Female

Obs

Exp

Obs

Exp

194

228

168

134

b.

(194  228) 2 (168  134)2   13.70 228 134

c. Critical χ2 = 3.84 at .05 α and 6.63 at .01 α. The research hypothesis is supported at the .01 level. Confidence >99%.

8.

Expected:

Solution Manual For All Chapters Text Messaging = (65) (.82) = 53.3 Phone Calls = (65) (.10) = 6.5 Internet Surfing = (65) (.08) = 5.2

a.

Text Messaging

Phone Calls

Internet Surfing

Obs

Exp

Obs

Exp

Obs

Exp

43

53.3

14

6.5

8

5.2

b. 2 = [(O-E)2/E] = (43-53.3)2/53.3 + (14-6.5)2/6.5 + (8-5.2)2/5.2 = (10.3)2/53.3 + (7.52/6.5 + (2.8/5.2 = 106.09/53.3 + 56.25/6.5 + 7.84/5.2 = 1.990 + 8.654 + 1.508 = 12.152

c. Confidence >.005

10.

Expected: Football = (450) (.30) = 135 Basketball = (450) (.25) = 112.5 Baseball = (450) (.11) = 49.5 Soccer = (450 (.05) = 22.5 Track = (450) (.03) = 13.5 No Interest = (450) (.26) = 117

a. Table

Solution Manual For All Chapters Football

Basketball

Baseball

Soccer

Track

Exp

Obs

Exp

Obs

Exp

Obs

112.5

60

49.5

36

22.5

25

No Interest Obs

Exp

Obs

Exp

Obs

Exp

105

135

117

13.5

107

117

b. 2 = [(O-E)2/E] = (105-135)2/135 + (117-112.5)2/112.5 + (60-49.5)2/49.5 + (36-22.5)2/22.5 + (25-13.5)2/13.5 + (107-117)2/117 = (-30)2/135 + (4.5)2/112.5 + (10.5)2/49.5 + (13.5)2/22.5 + (11.5)2/13.5 + (10)2/117 = 900/135 + 20.25/112.5 + 110.25/49.5 + 182.25/22.5 + 132.25/13.5 + 100/117 = 6.667 + 0.18 + 2.227 + 8.1 + 9.796 + 0.855 = 26.970

c. Confidence >.005

Module 31. Two-Variable Chi-Square: Test of Independence 2. Educational level and venue in which antiques are bought 4. When go to college and dropout status 6. a. College major and having a job related to that major

Solution Manual For All Chapters b. There is no significant relationship between one’s major in college and having a job related to that major c. (4 − 1)(2 − 1) = (3)(1) = 3 df d.

Business

Humanities

Social Science

Science

48

49

96

47

(48.0)

(60.0)

(96.0)

(36.0)

32

51

64

13

(80)

(100)

(160)

(60)

80

100

160

60

240

Related 160

Not Related

e. 2 = 13.44 f. Critical 2 = 11.34. Reject the null hypothesis. There is a significant relationship between one’s major in college and having a job related to that major

8. a. Gender and choice of escalator versus stairs b. Men and women differ in choosing stairs or escalator. c. (2 − 1)(2 − 1) = (1)(1) = 1 df d.

Stairs

Escalator

21

13

(17)

(17)

31

38

(35)

(34)

34

Men

Women

69

400

Solution Manual For All Chapters 52

51

103

e. 2 = 2.81 f. Critical 2 = 3.84 at .05 a and 6.63 at .01 a . Do not reject the null hypothesis. There is no relationship between gender and choice of escalator versus stairs.

10. a. Gender and grade received b. There is no significant relationship between gender and grade received. c. (2 − 1)(2 − 1) = (1)(1) = 1 df d.

“A” Grade

“Non-A” Grade

7

8

(6)

(9)

3

7

(4)

(6)

10

15

15

Boys 10

Girls 25

e. 2 = 0.70 f. Critical 2 = 3.84 at .05 a and 6.63 at .01 a. Do not reject the null hypothesis. There is no relationship between gender and grade received. g. Gender is a post hoc independent variable, so confounding variables might coexist.

Solution Manual For All Chapters

12.

Male Female

Text Messaging 256 ( 258.3 ) 367 ( 364.7 )

Column Total

623

Phone Calls 31 ( 30.7 ) 43 ( 43.3 )

Internet Surfing 19 ( 17.0 ) 22 ( 24.0 )

74

41

Row Total

306 432 738

df = (r-1) (c-1) = 1x2 = 2 2 = 0.443. Do not reject null hypothesis. There is not a significant relationship between cell phone usage and gender.

14.

Football

East High Central High West High Column Total

Basketba Baseball ll 105 117 60 ( 113.1 ) ( 120.3 ) ( 52.3 ) 97 85 40 ( 89.5 ) ( 95.2 ) ( 41.3 ) 114 134 46 ( 113.4 ) (120.6 ) ( 52.4 ) 316

336

146

Soccer

Track

Row Total

25 ( 27.6 ) 19 ( 21.8 ) 33 ( 27.6 )

No Interest 107 ( 90.2 ) 59 ( 71.4 ) 86 ( 90.4 )

36 ( 46.5 ) 56 ( 46.8 ) 38 ( 46.6 ) 130

77

252

1257

df = (r-1) (c-1) = 2x5 = 10 2  = 26.967. Reject null hypothesis. There is a significant relationship between sports preference and school.

450 356 451

Solution Manual For All Chapters

Module 32. Measures of Effect Size

2. a.

d

7.42  6.00 1.42   1.04 1.369 1.369

(2.84) 2 8.066 8.066 r    .268  .52 2 (2.84)  22 8.066  22 30.066 b. Yes, there is a practical difference. Effect size is large.

4.

a.



4192.585  .879  .94 4768.96

b. Yes, there is a practical difference. Effect size is very large.

6.

a.



235.0  .457  .68 514.4

b. There is a practical difference. Effect size is large. .

8. a.

22.40  (382)(3  1)

22.40  .029  .17 764

Solution Manual For All Chapters b. There is no practical difference. Effect size is small.

10. a.



2.81  .027  .165 103

b. There is no practical difference. Effect size is small.

12. a.



155.16  .018  .134 8,644

V

(.134)2  (.134)2  .018  .134 2 1

b. There is no practical difference. Effect size is small.

Even Solutions to Textbook Exercises Module 33. Power and the Factors Affecting It 2. a. 16% b. 75%

Solution Manual For All Chapters 4. a. Increase power b. The numerator of the t test will be larger 6. a. Increase power b. – 8. 132 subjects 10. r = .371 12. 90% 14. 36 subjects 16. 80% 18. V = 0.50

Module 34. Relationship Strength and Direction 2. a. Yes b. No c. No d. Yes 4. a. no b. yes c. yes

Solution Manual For All Chapters d. no 6.

8.

10. a. − b. +

Solution Manual For All Chapters c. + d. + e. 0 12. a. + b. + c. d. 0 e. +

14. a.

Solution Manual For All Chapters

b. Because correlation is measuring direction and strength of the score pattern between two variables, not similarity in score strength between the variables.

16. Answers will vary.

Even Solutions to Textbook Exercises

Solution Manual For All Chapters

Module 35. Pearson r 2. a.

because the data are in ranks rather than in scores

b.

because the data are in frequency counts rather than in scores

c.

because there are two groups who received different treatments, rather than one group who was scored on two variables

4. a.

b. I expect a moderate or strong negative relationship. c. There is no significant relationship between intelligence and prejudice. d. ∑X = 1,299 ∑X2 = 142,363

∑Y = 284

∑XY = 30,322

∑Y2 = 6,868

r = −.83

e. Critical r for a two-tailed hypothesis = .576 at .05 α and .708 at .01 α. Reject the null hypothesis with >99% confidence. f. r(10) = −.83, p < .01

6.

Solution Manual For All Chapters a. ∑X = 2,929 ∑X2 = 858,811

∑Y = 252

∑XY = 73,884

∑Y2 = 6,378

r = +.46

b. Critical r for a one-tailed hypothesis = .549 at .05 α and .716 at .01 α. Do not reject the null hypothesis.

8.

X = Miles Per Gallon; Y = Planned Driving Distance X = 331; Y = 3,985; XY = 115,175; X2 = 10,255; Y2 = 1,751,721 rXY = +.24

Even Solutions to Textbook Exercises Module 36. Correlation Pitfalls 2 The sample size is too small  Type 2 The study uses only a restricted portion of the real score range  Type 2 The sample is too heterogeneous  Type 1 4 a.

(+1.00)2 = 1.00

b. (−.60)2 = .36

Solution Manual For All Chapters

c. (+.40)2 = .16

6. The small sample size increased the tabled value needed for statistical significance. .

8. The unreliability of the measuring instruments (visual estimation) will decrease the calculated correlation coefficient.

10. “Eating vegetarian leads to a more active lifestyle.” Not a valid conclusion.

Correlation is not causation. Those who already lead a more active lifestyle might choose to be vegetarians.

12. “Study finds an important dietary contributor to physical activity.” Not a valid conclusion. The actual correlation might have been very low and yet still have been statistically significant if the sample size were large. Importance is a judgment, and is best assessed with a measure of effect size, not with a measure of statistical significance.

Module 37. Linear Prediction 2. a. Y’ = .45 (.57/180) X- .45 (.57/180) 1050 + 2.40 = .45 (.00317) X - .45 (.00317) 1050 + 2.40 = .001426X - .001426 (1050) + 2.40

Solution Manual For All Chapters = .001426X – 1.4973 + 2.40 = .001426 X + .9027 b. Y’ = .001426 (700) + .9027 = .9982 + .9027 = 1.9009 c. Y’ = .001426 (1300) + .9027 = 1.8538 + .9027 = 2.76 GPA 4. a.

 3.496   3.496  Y   .83  X  ( .83)  (108.25)  23.667   12.063   12.063   .83(.28981) X  ( .83)(.28981)(108.25)  23.667 .

 .24054 X  ( .24054)(108.25)  23.667  .24054 X  ( 26.039)  23.667  .24054 X  26.039  23.667  .24054 X  49.706

b. Y′ = −.24054(115) + 49.706 = −27.662 + 49.706 = 22.044 6. a.

Solution Manual For All Chapters

 1.66   1.66  Y   .46  X  .46  (292.90)  25.20   9.52   9.52   .46(.17437) X  .46(.17437)(292.90)  25.20  .08021X  .08021(292.90)  25.20  .08021X  23.494  25.20  .08021X  1.706 b. Y′ = .08021(290) + 1.706 = 23.261 + 1.706 = 24.967 8. a. Y’ = .239 (197.325/10.113) X - .239 (197.325/10.113) 27.583 + 332.083 = .239 (19.512014) X - .239 (19.512014) (27.583) + 332.083 = 4.66337 X – 4.66337 (27.583) + 332.083 = 4.66337 X – 128.62973 + 332.083 = 4.66337 X + 203.45327 b. Y’ = 4.66337 (26) + 203.45327 = 121.24762 + 203.45327 = 324.70 miles

10. a. Y’ = -.834 (6.288/1.352) X - .834 (6.288/1.352) (1.850) + 85.083 = -.834 (4.65089) X – (-.834) (4.65089) (1.850) + 85.083 = -3.87884 X – (-3.87884) (1.850) + 85.083 = -3.87884 X – (-7.175854) + 85.083 = -3.87884 X +7.175854 + 85.083 = -3.87884 X + 92.258854 b. Y’ = -3.87884 (2.3) + 92.258854

Solution Manual For All Chapters = -8.921332 + 92.258854 = 83.34 percent academic average

Module 38. Standard Error of Prediction 2.

CI = 355  (1) 3.4 = 355  3.4 = 351.6 and 358.4

4. a.

SYX  3.496 1  ( .83) 2  3.496 1  .689  3.496 .311  (3.496)(.558)  1.95

b. 22.044 ± (2)(1.95) = 22.044 ± 3.90 = 18.14 and 25.94 6. a.

SYX  1.66 1  (.46)2  1.66 1  .212  1.66 .788  (1.66)(.888)  1.47 b. 24.967 ± (1)(1.47) = 24.967 ± 1.47

Solution Manual For All Chapters = 23.497 and 26.437 8. a. b.

sYX = 197.325 [SQR (1-(.239)2)] = 197.325 [SQR (1-.0571)] = 197.325 [SQR (.9429)] = (197.325) (.97108) = 191.29 68% CI = 324.70  (1) 191.29 = 324.70  191.29 = 133.41 and 515.99 miles

10. a. b.

sYX = 6.288 [SQR (1-(-.834)2)] = 6.288 [SQR (1-.6956)] = 6.288 [SQR (.3044)] = (6.288) (.5517) = 3.47 95% CI = 83.34 (2) (3.47) = 83.34  6.94 = 76.4 and 90.28 percent academic average

.

12. The heterogeneity will increase the correlation, which will in turn decrease the size of the standard error of prediction. .

14. The decreased range of sentences will decrease the width of the confidence interval.

Module 39. Introduction to Multiple Regression 2. 1-.19 = .71 4. They are the same. R2 of .73 = residual of .27.

6. It should go up. An additional predictor should predict an additional portion of the criterion.

8. a. .709 or 70.9% b. the Regression SS divided by the Total SS c. 1-.709 = .291 or 29.1% d. something more than .842. because an additional predictor can account for a

Solution Manual For All Chapters portion of the criterion variance that is not already accounted for by the three existing predictors 10. a. same sex parent’s longevity b. Minutes of exercise daily might be highly correlated with cigarettes smoked daily or with same sex parents’ longevity c. Y’ = (1.018) (77) + (-.269) (20) + (.112) (30) + .448 = 76.809 12. a. 378.55/589/22 = .6425 b. SQR [.6425] = .8015 c. 3 d. N=30. There wer 10 subjects per predictor, which is minimally acceptable. e. yes. F = 15.572, with an alpha of less than 1/10 of 1%.

Even Solutions to Textbook Exercises

Module 40. Selecting the Appropriate Analysis 2. a. χ2 b. χ2 c. Pearson r d. t test e. F test