Nonparametric Statistics

Chapter 16 Nonparametric Statistics

16.1 The hypotheses

H 0 :˜ µ = 20 minutes

H 1 :˜ µ > 20 minutes.

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: Subtracting 20 from each observation and discarding the zeroes. We obtain the signs

for which n = 10 and x = 7. Therefore, the P -value is

X 10

P = P (X ≥ 7 | p = 1/2) =

b(x; 10, 1/2)

x=7

b(x; 10, 1/2) = 1 − 0.8281 = 0.1719 > 0.05.

x=0

Decision: Do not reject H 0 .

16.2 The hypotheses

H 0 :˜ µ = 12

H 1 :˜ µ 6= 12.

α = 0.02. Test statistic: binomial variable X with p = 1/2. Computations: Replacing each value above and below 12 by the symbol “+” and “−”, respectively, and discarding the two values which equal to 12. We obtain the sequence

258 Chapter 16 Nonparametric Statistics for which n = 16, x = 10 and n/2 = 8. Therefore, the P -value is

X 16

P = 2P (X ≥ 10 | p = 1/2) = 2

b(x; 16, 1/2)

x=10

b(x; 16, 1/2)) = 2(1 − 0.7728) = 0.4544 > 0.02.

x=0

Decision: Do not reject H 0 .

16.3 The hypotheses

H 0 :˜ µ = 2.5

H 1 :˜ µ 6= 2.5.

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: Replacing each value above and below 2.5 by the symbol “+” and “−”, respectively. We obtain the sequence

p for which n = 16, x = 3. Therefore, µ = np = (16)(0.5) = 8 and σ = (16)(0.5)(0.5) =

2. Hence z = (3.5 − 8)/2 = −2.25, and then P = 2P (X ≤ 3 | p = 1/2) ≈ 2P (Z < −2.25) = (2)(0.0122) = 0.0244 < 0.05.

Decision: Reject H 0 .

16.4 The hypotheses

H 0 :˜ µ 1 =˜ µ 2

H 1 :˜ µ 1 <˜ µ 2 .

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: After replacing each positive difference by a “+” symbol and negative difference by a “−” symbol, respectively, and discarding the two zero differences, we have n = 10 and x = 2. Therefore, the P -value is

P = P (X ≤ 2 | p = 1/2) = b(x; 10, 1/2) = 0.0547 > 0.05.

x=0

Decision: Do not reject H 0 .

Solutions for Exercises in Chapter 16 259

16.5 The hypotheses

H 0 :˜ µ 1 − ˜µ 2 = 4.5

H 1 :˜ µ 1 − ˜µ 2 < 4.5.

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: We have n = 10 and x = 4 plus signs. Therefore, the P -value is

P = P (X ≤ 4 | p = 1/2) = b(x; 10, 1/2) = 0.3770 > 0.05.

x=0

Decision: Do not reject H 0 .

16.6 The hypotheses

H 0 :˜ µ A =˜ µ B

H 1 :˜ µ A 6= ˜µ B .

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: We have n = 14 and x = 12. Therefore, µ = np = (14)(1/2) = 7 and p σ= (14)(1/2)(1/2) = 1.8708. Hence, z = (11.5 − 7)/1.8708 = 2.41, and then

P = 2P (X ≥ 12 | p = 1/2) = 2P (Z > 2.41) = (2)(0.0080) = 0.0160 < 0.05.

Decision: Reject H 0 .

16.7 The hypotheses

H 0 :˜ µ 2 − ˜µ 1 =8

H 1 :˜ µ 2 − ˜µ 1 < 8.

α = 0.05. Test statistic: binomial variable X with p = 1/2. Computations: We have n = 13 and x = 4. Therefore, µ = np = (13)(1/2) = 6.5 and p σ= (13)(1/2)(1/2) = 1.803. Hence, z = (4.5 − 6.5)/1.803 = −1.11, and then

P = P (X ≥ 4 | p = 1/2) = P (Z < −1.11) = 0.1335 > 0.05.

Decision: Do not reject H 0 .

16.8 The hypotheses

H 0 :˜ µ = 20

H 1 :˜ µ > 20.

α = 0.05. Critical region: w ≤ 11 for n = 10. Computations:

260 Chapter 16 Nonparametric Statistics

1 9 4 4 7.5 4 7.5 10 6 2 Therefore, w = 12.5.

Rank

Decision: Do not reject H 0 .

16.9 The hypotheses

H 0 :˜ µ = 12

H 1 :˜ µ 6= 12.

α = 0.02. Critical region: w ≤ 20 for n = 15. Computations:

12 3.5 8.5 3.5 8.5 Now, w = 43 and w + = 77, so that w = 43.

Decision: Do not reject H 0 .

16.10 The hypotheses

H 0 :˜ µ 1 − ˜µ 2 =0

H 1 :˜ µ 1 − ˜µ 2 < 0.

α = 0.02. Critiral region: w+ ≤ 1 for n = 5. Computations:

Therefore, w + = 3.5.

Decision: Do not reject H 0 .

16.11 The hypotheses

H 0 :˜ µ 1 − ˜µ 2 = 4.5

H 1 :˜ µ 1 − ˜µ 2 < 4.5.

α = 0.05. Critiral region: w+ ≤ 11. Computations:

Solutions for Exercises in Chapter 16 261 Woman

d i −d 0 −6.0 0.9 −0.9 2.4 1.0 −1.8 −2.2 −1.1 1.4 −3.8 Rank

1.5 8 3 6 7 4 5 9 Therefore, w + = 17.5.

Decision: Do not reject H 0 .

16.12 The hypotheses

H 0 :˜ µ A − ˜µ B =0

H 1 :˜ µ A − ˜µ B > 0.

α = 0.01. Critiral region: z > 2.575. Computations:

Now w = 180, n = 20, µ W + = (20)(21)/4 = 105, and σ W + = (20)(21)(41)/24 = 26.786. Therefore, z = (180 − 105)/26.786 = 2.80

Decision: Reject H 0 ; on average, Pharmacy A fills more prescriptions than Pharmacy

B.

16.13 The hypotheses

H 0 :˜ µ 1 − ˜µ 2 =8

H 1 :˜ µ 1 − ˜µ 2 < 8.

α = 0.05. Critiral region: z < −1.645. Computations:

d i −d 0 −2

Rank 4.5 1.5 10.5

d i −d 0 0 −6

Rank

262 Chapter 16 Nonparametric Statistics Discarding zero differences, we have w + = 15, n = 13, µ W + = (13)(14)/4 = 45, 5, and

p σ W + = (13)(14)(27)/24 = 15.309. Therefore, z = (15 − 45.5)/14.309 = −2.13

Decision: Reject H 0 ; the average increase is less than 8 points.

16.14 The hypotheses

H 0 :˜ µ A − ˜µ B =0

H 1 :˜ µ A − ˜µ B 6= 0.

α = 0.05. Critiral region: w ≤ 21 for n = 14. Computations:

8 10 1 5.5 14 4 13 Hence, w + = 101.5, w − = 3.5, so w = 3.5.

Rank

Decision: Reject H 0 ; the different instruments lead to different results.

16.15 The hypotheses

Critiral region: n 1 = 3, n 2 = 6 so u 1 ≤ 2.

Computations:

Original data 1 7 8 9 10

Now w 1 = 10 and hence u 1 = 10 − (3)(4)/2 = 4

Decision: Do not reject H 0 ; the claim that the tar content of brand B cigarettes is lower than that of brand A is not statistically supported.

16.16 The hypotheses

Critiral region: u 1 ≤ 2.

Computations:

Solutions for Exercises in Chapter 16 263 Original data 0.5 0.9 1.4

Now w 1 = 18 and hence u 1 = 18 − (4)(5)/2 = 8 Decision: Do not reject H 0 .

16.17 The hypotheses

Critiral region: u 2 ≤ 14.

Computations: Original data

7 8 ∗ 9 ∗ Original Data 5.0 5.1 5.2 5.3

Now w 2 = 50 and hence u 2 = 50 − (9)(10)/2 = 5 Decision: Reject H 0 ; calculator A operates longer.

16.18 The hypotheses

H 0 :˜ µ 1 =˜ µ 2

H 1 :˜ µ 1 6= ˜µ 2 .

α = 0.01. Critiral region: u ≤ 27.

Computations: Original data

8.7 9.3 9.5 9.6 9.8 9.8 9.8 9.9 9.9 10.0 Rank

8.5 10 Original Data

19 20 Here “ ∗ ” is for process 2. Now w 1 = 111.5 for process 1 and w 2 = 98.5 for process 2.

Therefore, u 1 = 111.5 − (10)(11)/2 = 56.5 and u 2 = 98.5 − (10)(11)/2 = 43.5, so that u = 43.5.

Decision: Do not reject H 0 .

16.19 The hypotheses

H 0 :˜ µ 1 =˜ µ 2

H 1 :˜ µ 1 6= ˜µ 2 .

264 Chapter 16 Nonparametric Statistics α = 0.05.

Critiral region: u ≤ 5. Computations:

Original data 64 67

Now w 1 = 35 and w 2 = 43. Therefore, u 1 = 35−(5)(6)/2 = 20 and u 2 = 43−(7)(8)/2 =

15, so that u = 15.

Decision: Do not reject H 0 .

16.20 The hypotheses

H 0 :˜ µ 1 =˜ µ 2

H 1 :˜ µ 1 6= ˜µ 2 .

α = 0.05. Critiral region: Z < −1.96 or z > 1.96.

Computations:

Observation 12.7 13.2 13.6 13.6 14.1 14.1 14.5 14.8 15.0 15.0 15.4 Rank

9.5 11.5 ∗ Observation 15.4 15.6 15.9 15.9 16.3 16.3 16.3 16.3 16.5 16.8 17.2 Rank

22 Observation 17.4 17.7 17.7 18.1 18.1 18.3 18.6 18.6 18.6 19.1 20.0 Rank

Now w 1 = 181.5 and u p 1 = 181.5 − (12)(13)/2 = 103.5. Then with µ U 1 = (21)(12)/2 = 126 and σ U 1 = (21)(12)(34)/12 = 26.721, we find z = (103.5 − 126)/26.721 = −0.84.

Decision: Do not reject H 0 .

16.21 The hypotheses

H 0 : Operating times for all three calculators are equal.

H 1 : Operating times are not all equal.

Critiral region: h > χ 2 0.01 = 9.210 with v = 2 degrees of freedom.

Computations:

Solutions for Exercises in Chapter 16 265

Ranks for Calculators

Now h = (18)(19) 5 + 7 + 6 − (3)(19) = 10.47.

Decision: Reject H 0 ; the operating times for all three calculators are not equal.

16.22 Kruskal-Wallis test (Chi-square approximation)

0.05 = 5.991 with 2 degrees of freedom. So, we reject H 0 and claim that the mean sorptions are not the same for all three solvents.

16.23 The hypotheses

H 0 : Sample is random.

H 1 : Sample is not random.

α = 0.1. Test statistics: V , the total number of runs.

Computations: for the given sequence we obtain n 1 = 5, n 2 = 10, and v = 7. Therefore, from Table A.18, the P -value is

P = 2P (V ≤ 7 when H 0 is true) = (2)(0.455) = 0.910 > 0.1

Decision: Do not reject H 0 ; the sample is random.

16.24 The hypotheses

H 0 : Fluctuations are random.

H 1 : Fluctuations are not random.

α = 0.05. Test statistics: V , the total number of runs. Computations: for the given sequence we find ˜ x = 0.021. Replacing each measurement

266 Chapter 16 Nonparametric Statistics by the symbol “+” if it falls above 0.021 and by the symbol “−” if it falls below 0.021

and omitting the two measurements that equal 0.021, we obtain the sequence

− − − − − + + + ++ for which n 1 = 5, n 2 = 5, and v = 2. Therefore, the P -value is

P = 2P (V ≤ 2 when H 0 is true) = (2)(0.008) = 0.016 < 0.05

Decision: Reject H 0 ; the fluctuations are not random.

16.25 The hypotheses

H 0 :µ A =µ B

H 1 :µ A >µ B .

α = 0.01. Test statistics: V , the total number of runs. Computations: from Exercise 16.17 we can write the sequence

B B B B B B AB B AAB AAAAAA

for which n 1 = 9, n 2 = 9, and v = 6. Therefore, the P -value is

P = P (V ≤ 6 when H 0 is true) = 0.044 > 0.01

Decision: Do not reject H 0 .

16.26 The hypotheses

H 0 : Defectives occur at random.

H 1 : Defectives do not occur at random.

α = 0.05. Critical region: z < −1.96 or z > 1.96.

Computations: n 1 = 11, n 2 = 17, and v = 13. Therefore, (2)(11)(17)

(28 2 )(27) and hence σ V = 2.472. Finally,

z = (13 − 14.357)/2.472 = −0.55. Decision: Do not reject H 0 .

Solutions for Exercises in Chapter 16 267

16.27 The hypotheses

H 0 : Sample is random.

H 1 : Sample is not random.

α = 0.05. Critical region: z < −1.96 or z > 1.96. Computations: we find ¯ x = 2.15. Assigning “+” and “−” signs for observations above

and below the median, respectively, we obtain n 1 = 15, n 2 = 15, and v = 19. Hence,

which yields σ V = 2.691. Therefore, z = (19 − 16)/2.691 = 1.11.

Decision: Do not reject H 0 .

16.28 1 − γ = 0.95, 1 − α = 0.85. From Table A.20, n = 30.

16.29 n = 24, 1 − α = 0.90. From Table A.20, 1 − γ = 0.70.

16.30 1 − γ = 0.99, 1 − α = 0.80. From Table A.21, n = 21.

16.31 n = 135, 1 − α = 0.95. From Table A.21, 1 − γ = 0.995.

16.32 (a) Using the computations, we have

Student Test Exam

L.S.A.

W.P.B.

R.W.K.

J.R.L.

J.K.L.

D.L.P.

B.L.P.

D.W.M.

M.N.M.

R.H.S.

r S =1−

268 Chapter 16 Nonparametric Statistics (b) The hypotheses

H 0 :ρ=0

H 1 :ρ>0

α = 0.025. Critical region: r S > 0.648.

Decision: Do not reject H 0 .

16.33 (a) Using the following

we obtain r (6)(1586.5)

S =1− (25)(625−1) = 0.39.

(b) The hypotheses

H 0 :ρ=0

H 1 : ρ 6= 0

α = 0.05. Critical region: r S < −0.400 or r s > 0.400.

Decision: Do not reject H 0 .

16.34 The numbers come up as follows

Ranks

Ranks

Ranks

dx

Solutions for Exercises in Chapter 16 269

X 2 (6)(238.5)

d = 238.5, r S =1−

16.35 (a) We have the following table: Weight Chest Size d i Weight Chest Size d i Weight Chest Size d i

(b) The hypotheses

H 0 :ρ=0

H 1 :ρ>0

α = 0.025. Critical region: r S > 0.683.

Decision: Reject H 0 and claim ρ > 0.

16.36 The hypotheses

H 0 :ρ=0

H 1 : ρ 6= 0

α = 0.05. Critical region: r S < −0.683 or r S > 0.683. Computations:

Manufacture A B C D E F GH I Panel rating 6 9 2 8 5 1 7 4 3

Price rank

0 −1 −6 5 0 0 Therefore, r (6)(176)

S =1− (9)(80) = −0.47.

Decision: Do not reject H 0 .

16.37 (a) (6)(24) d 2 = 24, r

S =1− (8)(63) = 0.71.

(b) The hypotheses

H 0 :ρ=0

H 1 :ρ>0

270 Chapter 16 Nonparametric Statistics α = 0.05.

Critical region: r S > 0.643. Computations: r S = 0.71.

Decision: Reject H 0 , ρ > 0.

16.38 (a) (6)(1828) d 2 = 1828, r

S =1− (30)(899) = 0.59.

(b) The hypotheses

H 0 :ρ=0

H 1 : ρ 6= 0

α = 0.05. Critical region: r S < −0.364 or r S > 0.364. Computations: r S = 0.59.

Decision: Reject H 0 , ρ 6= 0.

16.39 (a) The hypotheses

H 0 :µ A =µ B

H 1 :µ A 6= µ B

Test statistic: binomial variable X with p = 1/2. Computations: n = 9, omitting the identical pair, so x = 3 and P -value is P = P (X ≤ 3) = 0.2539.

Decision: Do not reject H 0 .

(b) w + = 15.5, n = 9.

Decision: Do not reject H 0 .

16.40 The hypotheses:

H 0 :µ 1 =µ 2 =µ 3 =µ 4 .

H 1 : At least two of the means are not equal.

Critical region: h > χ 2 0.05 = 7.815 with 3 degrees of freedom.

Computaions:

Ranks for the Laboratories

6 10.5 5 17 r 1 = 50 r 2 = 76.5 r 3 = 15 r 4 = 68.5

Solutions for Exercises in Chapter 16 271 Now

Decision: Reject H 0 .

16.41 The hypotheses:

H 0 :µ 29 =µ 54 =µ 84 .

H 1 : At least two of the means are not equal.

Kruskal-Wallis test (Chi-squared approximation)

(12)(13) 3 5 4 with 2 degrees of freedom. χ 2 0.05 = 5.991.

Decision: reject H 0 . Mean nitrogen loss is different for different levels of dietary protein.