Answers to All Exercises (Hampton & Havel 2005)

Notes:

1) Textbook has answers only to the odd-numbered problems.

2) Rounding errors may cause slight discrepancies between these answers and your answers.

Chapter 2

2.1 a) ratio b) continuous

2.2 a) ratio b) continuous

2.3 a) ratio b) discrete

2.4 ordinal

2.5-2.6 Male Number 1 2 3 4 5 6

Height 181 202 190 185 190 200

Rank 1 6 3.5 2 3.5 5

Category S T S S S T

2.7-2.8 Pond Number 1 2 3 4 5 6 7 8 9 10 11

No. Bullfrogs 34 65 23 34 18 20 15 70 15 18 34

Rank 8.0 10 6 8.0 3.5 5 1.5 11 1.5 3.5 8.0

Category D D S D S S S D S S D

2.9 a) continuous measurement, ratio, continuous, derived

b) discrete measurement, ratio, discrete, simple

c) attribute, nominal

d) continuous measurement, ratio, continuous, transformed

e) continuous measurement, ratio, continuous, simple

2.10 a) 107

b) 6.82 X 10^-2

c) 3.05

d) 7.82 X 10³

e) 2.91

f) 20.2

Chapter 4

4.1

Mean 6.0

Median 6.5

Range 1, 11

IQ range 3.0, 9.0

Variance 8.9655

Standard Deviation 2.9942

CV 49.9%

4.2

Mean 30.3846

Median 23.00

Range 12, 56

IQ range 14.5, 43.0

Variance 261.4231

Standard Deviation 16.1686

CV 53.2%

4.3

Mean 55.6976

Median 56.5

Range 47.0, 62.4

IQ range 52.175, 59.225

Variance 18.4310

Standard Deviation 4.2931

CV 7.7%

4.4

Mean 181.8176

4.5

Mean 274.2100

4.8

299 pits in 100 quadrats; mean = 2.99 pits/quadrat

4.9

Median = 2

Chapter 5

5.1

0.25

5.2

0.25

5.3

0.50

5.4

0.234375

5.5

0.34375

5.6

0.032152

5.7

0.33021

5.8

0.9042

5.9

0.3125

5.10

0.34375

5.11

0.015625

5.12

0.015625

5.13

0.177979

5.15

0.822021

5.16

0.46606

5.17

P(0males) = 0.004577, P(1) = 0.035178, ……, P(8) = 0.003323

5.18

273

5.19

3.323 » 3

5.20

a) 0.07179 b) 0.92821

5.21

mean = 2.99 Sf = 100

x p(x) f(x)

0 0.0503 5.03

1 0.1504 15.04

2 0.2248 22.48

3 0.2240 22.40

4 0.1675 16.75

5 0.1001 10.01

6 0.0499 4.99

7 0.0213 2.13

8 0.0080 0.80

5.23

a) 0.007967 b) 0.179336

5.24

µ = 1.6667 phage/cell; p(0) = 0.1889; p(³1) = 0.8111

5.25

µ = 0.6 phage/cell; p (≥3) = 0.0231

Chapter 6

6.1

0.0021

6.2

0.0455

6.3

0.6331

6.4

23.52 to 45.05

6.5

0.1814

6.6

0.1292

6.7

0.2386

6.8

113.16 to 190.55

6.9

0.0968

6.10

0.0001

6.11

0.3707

6.12

0.0021

6.13

0.8060

6.14

0.9723

Chapter 7

7.1-7.6 [Answers depend on random samples collected]

7.7

90% CI: 6.00 ± 0.93 = 5.07, 6.93

95% CI: 6.00 ± 1.12 = 4.88, 7.12

99% CI: 6.00 ± 1.51 = 4.49, 7.51

7.8

90%; 55.698 ± 1.116

95%; 55.698 ± 1.339

99%; 55.698 ± 1.791

7.9

90%; 19.2556 ± 1.5895

95%; 19.2556 ± 1.9707

99%; 19.2556 ± 2.8672

7.10

90%; 364.2500 ± 56.5054

95%; 364.2500 ± 69.2514

99%; 364.2500 ± 97.7205

7.11

90%; 45.1818 ± 5.4473

95%; 45.1818 ± 6.6967

99%; 45.1818 ± 9.5252

7.12

90%; 35.4000 ± 2.8489

95%; 35.4000 ± 3.4698

99%; 35.4000 ± 4.8159

7.13

90%; 4.9760 ± 0.2877

95%; 4.9760 ± 0.3471

99%; 4.9760 ± 0.4703

7.14

90%; 1934.6429 ± 247.9211

95%; 1934.6429 ± 302.4399

99%; 1934.6429 ± 421.7023

Chapter 8

8.1

H₀: µ ≤ 10 ppb vs. H_a: µ > 10 ppb

t = 2.50, one-tailed p-value: 0.005 < p < 0.01

reject H₀, \exceeds 10 ppb

8.2

H₀: µ ≥ 50 H_a: µ < 50

t = -2.8947; p » 0.005; \ reject H₀; enzyme levels are depressed, suggesting that water is polluted

8.3

H₀: µ = 100 H_a: µ ¹ 100

|t| = 5.367, p < 0.0001, \ reject H₀; mean weight differs from 100; so this hen should be removed from breeding program

8.4

H₀: m ≥ 8.0 vs. H_a: µ < 8.0

mean = 7.94

t = 1.32, p » 0.10, fail to reject, \not less than 8.0

8.5

H₀: m ³ 425 vs. H_a: m < 425

t = 0.78, p > 0.20, fail to reject, \not less than 425 (ponds not lower than lakes)

8.6

H₀: m £ 3.5 vs. H_a: m > 3.5

t = 1.58, p-value 0.10 < p < 0.20, \yes, no evidence of any difference from claim

Chapter 9

9.1

H₀: µ(men) = µ(women) vs. H_a: µ(men) ¹ µ(women)

|t| = 3.83, p-value 0.0001 < p < 0.001, reject H₀, \genders are different

95% CI for μ_w – μ_m = 8.481 ± 4.437 = 4.044, 12.918

We are 95% sure that the resting pulse rate is between 4.0 and 12.9 beats per minute greater in women than in men.

9.2

H₀: µ(men) = µ(women) vs. H_a: µ(men) ¹ µ(women)

|t| = 1.57

9.3

H₀: µ(largemouth) = µ(smallmouth) vs. H_a: µ(largemouth) ¹ µ(smallmouth)

|t| = 11.33

9.4

H₀: µ(A) ≤ µ(B) vs. H_a: µ(A) > µ(B)

t = 0.95

9.5

H₀: µ(down) ≤ µ(up) vs. H_a: µ(down) > µ(up)

t = 5.14

9.6

H₀: µ(exposed) ≥ µ(untreated) vs. H_a: µ(exposed) < µ(untreated)

t = 5.378

9.7

H₀: µ(glucose) = µ(sucrose) vs. H_a: µ(glucose) ¹ µ(sucrose)

t = 0.731

9.8

H₀: µ(treated) ≤ µ(control) vs. H_a: µ(treated) > µ(control)

t = 2.671

9.9

H₀: µ(with) ≤ µ(without) vs. H_a: µ(with) > µ(without)

t = 6.1612

9.10

H₀: µ(A) = µ(B) vs. H_a: µ(A) ¹ µ(B)

t = 2.9414

9.11

H₀: µ(coniferous) = µ(deciduous) vs. H_a: µ(coniferous) ¹ µ(deciduous)

t = 4.1210

9.12

H₀: µ(athletes) ≥ µ(non) vs. H_a: µ(athletes) < µ(non)

t = –2.22

9.13

H₀: µ(smokers) ≤ µ(non) vs. H_a: µ(smokers) > µ(non)

t = –0.67

9.15

H₀: q(hairy) ≤ q(non-hairy) vs. H_a: q(hairy) > q(non-hairy)

U = 91.5, U_crit = 73, p<<0.05, reject H₀, \ hairy appear scarier!

9.16

U = 31, p < 0.05

9.17

U = 123.5, p < 0.05

9.18

U = 45, p < 0.05

9.19

U = 57, U_crit = 71, p > 0.05

9.20

H₀: µ(post – pre) ≤ 0 vs. H_a: µ(post – pre) > 0

t = 10.62, t_{14, 2, (.05)} = 1.761, p < 0.0001, reject H₀, \PHA increases wattle thickness

9.21

H₀: µ(difference) = 0 vs. H_a: µ(difference) ¹ 0

t = 4.27

9.22

H₀: µ(difference) = 0 vs. H_a: µ(difference) ¹ 0

t = 2.139

9.23

H₀: µ(wall – center) ≤ 0 vs. H_a: µ(wall – center) > 0

t = 2.3904

9.24

H₀: q_dif = 0 vs. H_a: q_dif ¹ 0

T = 2.5, T_crit = 4 (note: reject is T is lower than T_crit ), 0.01 < p < 0.05, reject H₀

9.25

T = 0

9.26

k = 8, p =0.5, P(X ≤ 1) = 0.003906 + 0.03125 = 0.035156 (p-value), reject H₀

9.27

p = 0.0625

9.28

p = 0.109375

9.29

U = 90.5, p > 0.05

9.30

H₀: µ(A) = µ(B) vs. H_a: µ(A) ¹ µ(B)

t = 0.5547

9.31

H₀: µ(difference) = 0 vs. H_a: µ(difference) ¹ 0

t = 2.94, 0.01 < p < 0.02

9.32

T = 13.5, T_{14, .05} = 21, \reject H₀, 0.01 < p < 0.05

9.33

H₀: µ(infected) = µ(healthy)

H₁: µ(infected) ¹ µ(healthy)

t = 3.71

Chapter 10

10.1

H₀ all µ’s equal vs. H_a: one or more µ’s different

F_{2,12, (.05)} = 3.89, F_stat = 30.87, p << 0.05, reject H₀

Source Sum of Squares df MS F

Between-Groups 6.2713 2 3.1356 30.87

Error 1.2191 12 0.1016

Total 7.4904 14

10.2

Source Sum of Squares df MS F

Between-Groups 0.09695 2 0.0485 89.81

Error 0.00814 15 0.00054

Total 0.1051 17

10.3

Source Sum of Squares df MS F

Between-Groups 632.95 3 210.98 70.33

Error 48.00 16 3.00

Total 680.95 19

10.4

Source Sum of Squares df MS F

Between-Groups 2365.6 4 591.4 21.05

Error 983.5 35 28.1

Total 3349.1 39

10.5

Source Sum of Squares df MS F

Between-Groups 1740.4 2 870.2 79.11

Error 132.0 12 11.0

Total 1872.4 14

10.6

Source Sum of Squares df MS F

Between-Groups 2112.47 2 1256.23 10.95

Error 3097.00 27 114.70

Total 5609.47 29

10.7

Source Sum of Squares df MS F

Between-Groups 3785.65 4 946.41 72.78

Error 455.13 35 13.00

Total 4240.78 39

10.8

Source Sum of Squares df MS F

Between-Groups 45.31 2 22.65 5.81

Error 89.65 23 3.9

Total 134.96 25

10.9

Source Sum of Squares df MS F

Between-Groups 145.6389 3 48.5463 19.64

Error 79.1111 32 2.4722

Total 224.7500 35

10.10

Source Sum of Squares df MS F

Between-Groups 1442.1333 2 721.0667 16.06

Error 538.8000 12 44.9000

Total 1980.9333 14

10.11 – 10.18

Using MINITAB, homogeneity of variance test suggests all except 10.17 and 10.18 conform to assumptions of equal variance.

10.19

H₀: all q’s equal vs. one or more q’s different

H = 9.39, c² _{2, .05} = 5.99, p < 0.01, reject H₀.

10.20

H = 17.98

10.21

H = 12.42