1. Since σ1 = σ2 and the two samples are independent, we apply the pooled t test.The test statistic is

|T | = |5.5 − 4.3|√(20 − 1)(1.52) + (30 − 1)(1.62)

20 + 30 − 2

(1

20+

130

)= 2.66267473.

t0.05/2,20+30−2 = t0.025,48 = 2.011 < 2.66267473, so we can conclude that the meandriving times via Route 1 and Route 2 are different (µ1 , µ2) at level 0.05.

2. The p-value is 2(P(t(48) > 2.66267473). From the table “Quantiles for t distri-butions”, we have

t0.01,48 < 2.66267473 < t0.005,48,

so0.005 < P(t(48) > 2.66267473) < 0.01

and2 × 0.005︸ ︷︷ ︸

=0.01

< 2P(t(48) > 2.66267473)︸ ︷︷ ︸p-value

< 2 × 0.01︸ ︷︷ ︸=0.02

.

Since the p-value is less than 0.02 and greater than 0.01, we can conclude thatthe mean driving times via Route 1 and Route 2 are different (µ1 , µ2) at level awhen a ≥ 0.02, but we cannot conclude µ1 , µ2 at level a when a ≤ 0.01. Theanswers: (a) Yes; (b) Yes; (c) Yes; (d) No; (e) No.

3. When σ1 and σ2 are known, we can use the test statistic

|Z| = |5.5 − 4.3|√1.52

20+

1.62

30

= 2.697935184.

From the table “Quantiles for t distributions”, we have z0.01/2 = z0.005 = t0.005,∞ =2.576 < 2.697935184. Therefore, we can conclude that the mean driving timesvia Route 1 and Route 2 are different (µ1 , µ2) at level 0.01.

4. Since the two samples of weights can be dependent, we apply the paired t test.The sample of weight increment amounts is

(77 − 80, 54 − 55, 53 − 63, 48 − 48, 51 − 50) = (−3,−1,−10, 0, 1).

The sample mean and sample standard deviation are

−3 − 1 − 10 + 0 + 15

= −2.6

and√(−3)2 + (−1)2 + (−10)2 + (0)2 + (1)2 − 5 × (2.6)2

5 − 1 =√

19.3 = 4.393177

respectively. The test statistic is

|T | =√

5 × | − 2.6|√

19.3= 1.323365.

Since t0.05/2,5−1 = t0.025,4 = 2.776 > 1.323365, we cannot conclude that par-ticipating in the program has a significant effect on weight change at the 0.05level.

1

5. Let

T1 =X̄ − Ȳ − 1√

σ̂2(1/n1 + 1/n2),

where

σ̂ =

√(n1 − 1)S 2X + (n2 − 1)S 2Y

n1 + n2 − 2.

The proposed test rejects H0 : µ1 ≤ µ2 + 1 at level a if T1 > ta,n1+n2−2.Below we will verify that the proposed test is of size a. Since the size of a test isthe largest Type I error probability under H0, to find the size of the test, we needto find the Type I error probability of the proposed test, which is

P(T1 > ta,n1+n2−2)

when µ1 ≤ µ2 + 1. Note that when µ1 ≤ µ2 + 1,

P(T1 > ta,n1+n2−2)

= P

X̄ − Ȳ − 1√σ̂2(1/n1 + 1/n2)

> ta,n1+n2−2

= P

X̄ − Ȳ − (µ1 − µ2)√σ̂2(1/n1 + 1/n2)

+µ1 − µ2 − 1√σ̂2(1/n1 + 1/n2)

> ta,n1+n2−2

≤ P

X̄ − Ȳ − (µ1 − µ2)√σ̂2(1/n1 + 1/n2)

> ta,n1+n2−2

(1)since

µ1 − µ2 − 1√σ̂2(1/n1 + 1/n2)

≤ 0

when µ1 ≤ µ2 + 1. Since

X̄ − Ȳ − (µ1 − µ2)√σ̂2(1/n1 + 1/n2)

∼ t(n1 + n2 − 2),

(1) implies that

P(T1 > ta,n1+n2−2) ≤ P(t(n1 + n2 − 2) > ta,n1+n2−2) = a (2)

when µ1 ≤ µ2 + 1. Moreover, when µ1 = µ2 + 1,

P(T1 > ta,n1+n2−2)

= P

X̄ − Ȳ − (µ1 − µ2)√σ̂2(1/n1 + 1/n2)

> ta,n1+n2−2

= P(t(n1 + n2 − 2) > ta,n1+n2−2) = a. (3)

From (2) and (3), the largest Type I error probability of the proposed test underH0 : µ1 ≤ µ2 + 1 is a, so the proposed test is of size a.

6. The testing problem is

H0 : σ1 = σ2 v.s. H1 : σ1 , σ2.

Let S i be the sample standard deviation of the driving times for Rout i for i = 1,2 and let F = S 22/S

21. We will use the test that rejects H0 at level 0.1 if F >

2

f0.1/2,31−1,21−1 = f0.05,30,20 or 1/F > f0.1/2,21−1,31−1 = f0.05,20,30. From the table“0.95 quantiles for F distributions”, f0.05,20,30 = 1.93 and f0.05,30,20 = 2.04, so

observed F =1.62

1.52= 1.137778 < 2.04 = f0.05,30,20

and

observed1F

=1.52

1.62< 1 < 1.93 = f0.05,20,30.

Therefore, we cannot conclude σ1 , σ2 at level 0.1 based on the test.

7. Consider two random variables X and Y that are independent, where X ∼ χ2(m)and Y ∼ χ2(n). Let F = (X/m)/(Y/n), then F ∼ F(m, n) and 1/F ∼ F(n,m). Wewill show that f1−a,n,m = 1/ fa,m,n by verifying

P(F(n,m) > 1/ fa,m,n) = 1 − a. (4)

Note that

P(F(n,m) > 1/ fa,m,n) = P(1/F > 1/ fa,m,n)= P(F < fa,m,n) = P(F(m, n) < fa,m,n)= 1 − P(F(m, n) ≥ fa,m,n)= 1 − P(F(m, n) > fa,m,n) = 1 − a,

so (4) holds and we have shown that f1−a,n,m = 1/ fa,m,n, which implies fa,m,n =1/ f1−a,n,m.

8. The p-value for the F test for testing

H0 : σ1 = σ2 v.s. H1 : σ1 , σ2

ismin(1.655241, 0.3447595) = 0.3447595 > 0.1,

so we cannot even conclude that there is some evidence for σ1 , σ2. Thus wewill apply the two-sample t test (the pooled t test) for testing H0: µ1 = µ2 versusH1 : µ1 , µ2, where σ1 = σ2 is assumed. The answers for questions in (a) and(b) are given below.

(a) The p-value is 0.3447595.

(b) Yes, we can conclude µ1 , µ2 at level 0.002.

9. The observed test statistic is∣∣∣∣∣ 102000 − 121000∣∣∣∣∣√(

10 + 122000 + 1000

) (1 − 10 + 12

2000 + 1000

) (1

2000+

11000

) = 2.118361.

From the table “Quantiles for t distributions”, we have z0.01/2 = z0.005 = t0.005,∞ =2.576, so 2.118361 < z0.005 and we cannot conclude that the defective rates forthe two lines are different at level 0.01.

3

10. Consider the differenced sample (X1−Y1, . . . , Xn−Yn). Let X̄ and Ȳ be the samplemeans for (X1, . . . , Xn) and (Y1, . . . ,Yn) respectively. Then the sample mean forthe differenced sample is X̄ − Ȳ . Suppose that the differenced data are IID, thenfrom the central limit theorem, the distribution of

√n(X̄ − Ȳ − E(X1 − Y1))√

Var(X1 − Y1)

is approximately N(0, 1) for large n. Let S D be the sample standard deviation forthe differenced sample, then S D ≈

√Var(X1 − Y1) for large n, so the distribution

of √n(X̄ − Ȳ − E(X1 − Y1))

S Dis approximately N(0, 1) for large n. Under H0: p1 = p2, E(X1 −Y1) = p1 − p2 =0, so the distribution of √

n(X̄ − Ȳ)S D

is approximately N(0, 1) for large n under H0: p1 = p2. Therefore, we canpropose the test that rejects H0 : p1 = p2 at level a if∣∣∣∣∣∣

√n(X̄ − Ȳ)

S D

∣∣∣∣∣∣ > za/2.Note that the proposed test is approximately level a for large n if the differenceddata are IID.

11. The grand mean is

5(350/5) + 10(695/10) + 4(288/4)5 + 10 + 4

= 1333/19,

so

SST = 5(350/5 − 1333/19)2 + 10(695/10 − 1333/19)2 + 4(288/4 − 1333/19)2

= 6507.5/361

with degrees of freedom 3 − 1 = 2.

SSE = (5 − 1)(√

6/4)2 + (10 − 1)(√

8.5/9)2 + (4 − 1)(√

6/3)2 = 20.5

with degrees of freedom 5 + 10 + 4 − 3 = 16. The F test statistic is

SST/2SSE/16

=(6507.5/361)/2

20.5/16=

10412014801

= 7.03466.

Since f0.05,2,16 = 3.63 < 7.03466, we can conclude that the means of the scoredistributions for the three classes are not all the same at the 0.05 significant level.

12. (a) Yes. The p-value is 4.156 × 10−5 < 0.001.(b) The degrees of freedom for SST is 2, so the total number of classes is

2 + 1 = 3.

(c) The degrees of freedom for SSE is 96 and the degrees of freedom for SSTis 2, so the total number of students is 96 + 2 + 1 = 99.

4

13. Let X̄ and S X be the sample mean and sample standard deviation of (X1, . . . , Xn1 )respectively, and let Ȳ and S Y be the sample mean and sample standard deviationof (Y1, . . . ,Yn2 ) respectively. Then the two sample t test (pooled t test) rejectsH0 : µ1 = µ2 at level a if

|T | > ta/2,n1+n2−2,where

T =X̄ − Ȳ√

(1/n1 + 1/n2)((n1 − 1)S 2X + (n2 − 1)S 2Y )/(n1 + n2 − 2). (5)

For the ANOVA test, note that the grand mean is (n1X̄ + n2Ȳ)/(n1 + n2), so

SST = n1

(X̄ − n1X̄ + n2Ȳ

n1 + n2

)2+ n2

(Ȳ − n1X̄ + n2Ȳ

n1 + n2

)2= n1

(n2(X̄ − Ȳ)

n1 + n2

)2+ n2

(n1(Ȳ − X̄)

n1 + n2

)2= n1n2(X̄ − Ȳ)2/(n1 + n2) =

(X̄ − Ȳ)21/n1 + 1/n2

with degree of freedom 2 − 1 = 1. Also,

SSE = (n1 − 1)S 2X + (n2 − 1)S 2Y

with degrees of freedom n1 + n2 − 2. Thus the F statistic is

F =SST/1

SSE/(n1 + n2 − 2)

=(n1 + n2 − 2)(X̄ − Ȳ)2

(1/n1 + 1/n2)((n1 − 1)S 2X + (n2 − 1)S 2Y ). (6)

From (5) and (6), we have T 2 = F.

14. Suppose that the treatment has k levels and the group with level i has ni observa-tions Xi,1, . . ., Xi,ni for i = 1, . . ., k. Let X̄G be the grand mean and for i = 1, . . .,k, let

Di = (Xi,1, . . . , Xi,ni )

be the sample of data in the level i group and let X̄i and S i be the sample meanand sample standard deviation of Di respectively. Apply Fact 1 given in theproblem with W = Di and c = X̄G, then we have

ni∑j=1

(Xi, j − X̄G)2 =ni∑j=1

(Xi, j − X̄i)2 + ni(X̄i − X̄G)2

for i = 1, . . ., k. Take the sum over i and we have

k∑i=1

ni∑j=1

(Xi, j − X̄G)2 =k∑

i=1

ni∑j=1

(Xi, j − X̄i)2 +k∑

i=1

ni(X̄i − X̄G)2. (7)

Note that

SStotal =k∑

i=1

ni∑j=1

(Xi, j − X̄G)2,

5

SSE =k∑

i=1

(ni − 1)S 2i =k∑

i=1

ni∑j=1

(Xi, j − X̄i)2,

and

SST =k∑

i=1

ni(X̄i − X̄G)2,

so (7) implies thatSStotal = SST + SSE.

15. Let µ̄ denote the average of all µi, j’s, then

µ = µ̄ =5 + 7 + 3 + 4 + 5 + 3

6= 4.5.

The αi’s are computed using

αi =µi,1 + µi,2 + µi,3

3− µ,

soα1 =

5 + 7 + 33

− µ = 5 − 4.5 = 0.5,

andα2 =

4 + 5 + 33

− µ = 4 − 4.5 = −0.5.

The β j’s are computed using

β j =µ1, j + µ2, j

2− µ,

soβ1 =

5 + 42− µ = 4.5 − 4.5 = 0,

β2 =7 + 5

2− µ = 6 − 4.5 = 1.5,

andβ3 =

3 + 32− µ = 3 − 4.5 = −1.5.

The γi, j’s are computed using

γi, j = µi, j − µ − αi − β j,

soγ1,1 = 5 − (4.5 + 0.5 + 0) = 0,

γ1,2 = 7 − (4.5 + 0.5 + 1.5) = 0.5,γ1,3 = 3 − (4.5 + 0.5 − 1.5) = −0.5,γ2,1 = 4 − (4.5 − 0.5 + 0) = 0,

γ2,2 = 5 − (4.5 − 0.5 + 1.5) = −0.5,and

γ2,3 = 3 − (4.5 − 0.5 − 1.5) = 0.5.

6

16. Let ā = (a1 + a2)/2 and b̄ =∑3

j=1 b j/3. Then

µ =

∑2i=1

∑3j=1 µi, j

6

=

∑2i=1

∑3j=1(ai + b j)

6

=3∑2

i=1 ai + 2∑3

j=1 b j6

= ā + b̄,

αi =µi,1 + µi,2 + µi,3

3− µ

=

∑3j=1(ai + b j)

3− (ā + b̄)

= ai − ā

for i = 1, 2,

β j =µ1, j + µ2, j

2− µ

=

∑2i=1(ai + b j)

2− (ā + b̄)

= b j − b̄

for j = 1, 2, 3, and

γi, j = µi, j − µ − αi − β j= ai + b j − (ā + b̄) − (ai − ā) − (b j − b̄) = 0

for i = 1, 2, j = 1, 2, 3.

17. (a) The sum of squares due to error (SSE) is

(6 − 1) × (0.3 + 0.3 + 0.3 + 0.4 + 0.5 + 0.5 + 0.3 + 0.4 + 0.4 + 0.4 + 0.3 + 0.3)= 22,

(b) The grand mean (total mean) X̄G is

2.8 + 3.0 + 2.7 + 2.5 + 2.6 + 2.8 + 2.7 + 2.8 + 3.0 + 2.8 + 2.7 + 2.512

=32.912≈ 2.7417,

and the sample means for the four major groups are

2.8 + 3.0 + 2.73

=8.53≈ 2.8333,

2.5 + 2.6 + 2.83

=7.93≈ 2.6333,

2.7 + 2.8 + 3.03

=8.53≈ 2.8333,

and2.8 + 2.7 + 2.5

3=

83≈ 2.6667.

7

The sum of squares due to major, denoted by SSmajor, is

3 × 6 × [(8.5/3 − 32.9/12)2 + ((7.9/3 − 32.9/12)2

+(8.5/3 − 32.9/12)2 + (8/3 − 32.9/12)2]= 0.615.

(c) The sum of squares due to interaction (SSI) will be computed using

SSI = SStotal − SSE − SSmajor − SSyear,

where SSyear denotes the sum of squares due to year. Since SSmajor hasbeen computed in Part (b), we will first compute SSyear and then computeSStotal − SSE to obatin SSI.To compute SSyear, note that the sample means for the three year groupsare

2.8 + 2.5 + 2.7 + 2.84

= 2.7,

3.0 + 2.6 + 2.8 + 2.74

= 2.775,

and2.7 + 2.8 + 3.0 + 2.5

4= 2.75,

so SSyear is

4 × 6 × [(2.7 − 32.9/12)2 + (2.775 − 32.9/12)2

+(2.75 − 32.9/12)2]= 0.07.

Next, we compute SStotal − SSE. Let X̄i, j be the sample mean of the datain the group of the i-th major and the j-th graduation year for i = 1, . . ., 4,j = 1, 2, 3. Then

SStotal − SSE =4∑

i=1

3∑j=1

6(X̄i, j − X̄G)2

= 64∑

i=1

3∑j=1

X̄2i, j − 72X̄2G

= 6 × [2.82 + 3.02 + 2.72 + 2.52 + 2.62 + 2.82

+2.72 + 2.82 + 3.02 + 2.82 + 2.72 + 2.52]−72 × (32.9/12)2

= 1.735,

so the sum of squares due to interaction (SSI) is

SStotal − SSE − SSmajor − SSyear= 1.735 − 0.615 − 0.07 = 1.05.

(d) The degree of freedom for the sum of squares due to interaction is (4 −1)(3 − 1) = 6.

8

(e)

F =SSmajor/(4 − 1)

SSE/(4 × 3 × 6 − 4 × 3)

=0.615/322/60

≈ 0.5591 < f0.05,3,60 = 2.76,

so we cannot conclude that the initial mean salaries for the four majors arenot all the same at level 0.05.

(f) The degrees of freedom for SSI and SSE are 6 and 4 × 3 × 6 − 4 × 3 = 60respectively, so

F =SSI/6

SSE/60

=1.05/622/60

≈ 0.4773 < f0.05,6,60 = 2.25.

We cannot conclude that there is a major-year interaction effect on salaryat the 0.05 level.

18. We will show that

SSI = SStotal − SSE − SSA − SSB. (8)

For i = 1, . . ., k and j = 1, . . ., b, let Gi, j denote the group with Factor A level= i and Factor B level = j, and let X̄i, j denote the sample mean for data in Gi, j,X̄i· denote the sample mean for data in ∪bj=1Gi, j, X̄· j denote the sample mean fordata in ∪ki=1Gi, j and X̄G denote the grand mean. Suppose that each Gi, j has `observations and the m-th observation in Gi, j is denoted by Xi, j,m, m = 1, . . ., `.Then

SSI =k∑

i=1

b∑j=1

∑̀m=1

(X̄i, j − X̄i· − X̄· j + X̄G)2

= `

k∑i=1

b∑j=1

(X̄i, j − X̄i· − X̄· j + X̄G)2

= `

k∑i=1

b∑j=1

((X̄i, j − X̄G) − (X̄i· − X̄G) − (X̄· j − X̄G))2

= `

k∑i=1

b∑j=1

(X̄i, j − X̄G)2 + `k∑

i=1

b∑j=1

(X̄i· − X̄G)2︸ ︷︷ ︸SSA

+ `

k∑i=1

b∑j=1

(X̄· j − X̄G)2︸ ︷︷ ︸SSB

−2`k∑

i=1

b∑j=1

(X̄i, j − X̄G)(X̄i· − X̄G) − 2`k∑

i=1

b∑j=1

(X̄i, j − X̄G)(X̄· j − X̄G)

+2`k∑

i=1

b∑j=1

(X̄i· − X̄G)(X̄· j − X̄G). (9)

Some expressions in (9) can be simplified. Note that

−2`k∑

i=1

b∑j=1

(X̄i, j − X̄G)(X̄i· − X̄G) = −2`k∑

i=1

b∑j=1

(X̄i, j − X̄i· + X̄i· − X̄G)(X̄i· − X̄G)

9

= −2`k∑

i=1

b∑j=1

(X̄i, j − X̄i·)(X̄i· − X̄G) − 2`k∑

i=1

b∑j=1

(X̄i· − X̄G)(X̄i· − X̄G)

= −2`k∑

i=1

(X̄i· − X̄G)b∑

j=1

(X̄i, j − X̄i·) − 2 · SSA

= −2 · SSA (10)

since

b∑j=1

(X̄i, j − X̄i·) =b∑

j=1

1` ∑̀m=1

Xi, j,m

− b 1b`

b∑j=1

∑̀m=1

Xi, j,m

= 0.Similarly,

−2`k∑

i=1

b∑j=1

(X̄i, j − X̄G)(X̄· j − X̄G) = −2`k∑

i=1

b∑j=1

(X̄i, j − X̄· j + X̄· j − X̄G)(X̄· j − X̄G)

= −2`k∑

i=1

b∑j=1

(X̄i, j − X̄· j)(X̄· j − X̄G) − 2`k∑

i=1

b∑j=1

(X̄· j − X̄G)(X̄· j − X̄G)

= −2`b∑

j=1

(X̄· j − X̄G)k∑

i=1

(X̄i, j − X̄· j) − 2 · SSB

= −2 · SSB (11)

since

k∑i=1

(X̄i, j − X̄· j) =k∑

i=1

1` ∑̀m=1

Xi, j,m

− k 1k`

k∑i=1

∑̀m=1

Xi, j,m

= 0.Moreover,

b∑j=1

(X̄· j − X̄G) =b∑

j=1

1k`k∑

i=1

∑̀m=1

Xi, j,m

− b 1kb`

k∑i=1

b∑j=1

∑̀m=1

Xi, j,m

= 0,so

k∑i=1

b∑j=1

(X̄i· − X̄G)(X̄· j − X̄G) =k∑

i=1

(X̄i· − X̄G)b∑

j=1

(X̄· j − X̄G)︸ ︷︷ ︸=0

= 0. (12)

From (10), (11), (12), (9) can be simplified as:

SSI = `k∑

i=1

b∑j=1

(X̄i, j − X̄G)2 − SSA − SSB. (13)

To obtain (8) from (13), it remains to verify that

SStotal − SSE = `k∑

i=1

b∑j=1

(X̄i, j − X̄G)2. (14)

To see that (14) holds, note that in one-way ANOVA, the formula

SStotal = SST + SSE

10

can be written as follows:∑all observations

( observation − grand mean)2

=∑

all groups(group size) × ( group mean − grand mean)2

+∑

all groups

∑all observations in the group

( observation − group mean)2 .

Apply the above formula with observations Xi, j,ms and groups Gi, js, then we have

k∑i=1

b∑j=1

∑̀m=1

(Xi, j,m − X̄G)2︸ ︷︷ ︸SStotal

=

k∑i=1

b∑j=1

`(X̄i, j − X̄G)2 +k∑

i=1

b∑j=1

∑̀m=1

(Xi, j,m − X̄i, j)2︸ ︷︷ ︸SSE

,

so (14) holds and (8) follows from (13) and (14).

19. We will use the following results in the solution to Problem 17 for this problem.

(i) The SSE in the two-way ANOVA is 22 with 60 degrees of freedom.

(ii) SStotal−SSE = 1.735 in the two-way ANOVA, so the total sum of squaresin the two-way ANOVA is

SStotal = 1.735 + SSE(i)= 1.735 + 22 = 23.735

with 71 degrees of freedom.

(iii) The sum of squares due to major in the two-way ANOVA is 0.615 withthree degrees of freedom.

(a) The total sum of squares and the sum of squares due to major in one-wayANOVA are the same as those in two-way ANOVA, which are 23.735 and0.615 with degrees of freedom 71 and 3 respectively (based on (ii) and(iii)). Thus the sum of squares due to error in the one-way ANOVA is

SSE = SStotal − SSmajor = 23.735 − 0.615 = 23.12.

(b) The total sum of squares in the one-way ANOVA is 23.735, as explained inPart (a).

(c) The sum of squares due to major and the sum of squares due to error inone-way ANOVA are 0.615 and 23.12 respectively from Part (a), so theobserved F statistic is

F =0.615/3

23.12/(71 − 3) = 0.6029412.

Since2.68 = f0.05,3,120 < f0.05,3,68 < f0.05,3,60 = 2.76,

0.6029412 < f0.05,3,68. Therefore, we cannot conclude that inital expectedwages for students of different majors are not all the same at the 0.05 levelbased on the one-way ANOVA.

11

20. (a) No. The p-value is 0.55981 > 0.01.(b) No. The p-value is 0.01032 > 0.01.(c) Yes. The p-value is 2.548 × 10−7 < 0.01.(d) The degrees of freedom for the sum of squares due to the drug factor is 4,

so the drug factor has 4 + 1 = 5 levels and there are 5 types of drugs.(e) The degrees of freedom for the sum of squares due to the exercise factor

is 3, so the exercise factor has 3 + 1 = 4 levels and there are 4 types ofexercises.

(f) The total number of participants is 4 + 3 + 12 + 80 + 1 = 100.

21. (a) The total sum of squares in the one-way ANOVA is the same as that in thetwo-way ANOVA in Problem 20), which is

43.784 + 127.562 + 33.027 + 247.244 = 451.617.

(b) The sum of squares due to the drug factor is the same as that in the two-wayANOVA in Problem 20), which is 43.784 with 4 degrees of freedom. Thusthe SSE in the one-way ANOVA is

SStotal − 43.784(a)= 451.617 − 43.784 = 407.833.

(c) From Part (b), the sum of squares due to the drug factor in the one-wayANOVA is 43.784 with 4 degrees of freedom, and the SSE is 407.833 with100 − 1 − 4 = 95 degrees of freedom since the total number of participantsis 100 from Part (f) of Problem 20). Therefore, the observed F test statisticin the one-way ANOVA is

F =43.784/4

407.833/95= 2.549745.

From the table of 0.99 quantiles for F distributions,

3.48 = f0.01,4,120 < f0.01,4,95 < f0.01,4,60 = 3.65,

so 2.549745 < f0.01,4,95 and we cannot conclude that not all drugs have thesame effect on weight loss at level 0.01 based on the one-way ANOVA.

22. Given the two samples (X1, . . . , Xn) and (Y1, . . . ,Yn), let

f (λ) = (n − 1) · ( sample variance of (X1 + λY1, . . . , Xn + λYn))

for λ ∈ (−∞,∞). Let X̄ and Ȳ denote the sample means of (X1, . . . , Xn) and(Y1, . . . ,Yn) respectively, then the sample mean of (X1 + λY1, . . . , Xn + λYn) is

1n

n∑i=1

(Xi + λYi) =1n

n∑i=1

Xi + +1n

n∑i=1

λYi = X̄ + λȲ ,

so

f (λ) =n∑

i=1

(Xi + λYi − (X̄ + λȲ))2

=

n∑i=1

(Xi − X̄ + λ(Yi − Ȳ))2

=

n∑i=1

(Xi − X̄)2 + λ2n∑

i=1

(Yi − Ȳ)2 + 2λn∑

i=1

(Xi − X̄)(Yi − Ȳ).

12

Here we assume that∑n

i=1(Yi−Ȳ)2 > 0 and∑n

i=1(Xi− X̄)2 > 0 (otherwise the sam-ple correlation cannot be computed). It is clear that f (λ) is a quadratic functionof λ that has a global minimum. Solving

ddλ

f (λ) = 0

gives λ = −∑ni=1(Xi − X̄)(Yi − Ȳ)/∑ni=1(Yi − Ȳ)2 and the minimum isf(−∑n

i=1(Xi − X̄)(Yi − Ȳ)∑ni=1(Yi − Ȳ)2

)=

n∑i=1

(Xi − X̄)2

+

(−∑n

i=1(Xi − X̄)(Yi − Ȳ)∑ni=1(Yi − Ȳ)2

)2 n∑i=1

(Yi − Ȳ)2

+2(−∑n

i=1(Xi − X̄)(Yi − Ȳ)∑ni=1(Yi − Ȳ)2

) n∑i=1

(Xi − X̄)(Yi − Ȳ)

=

n∑i=1

(Xi − X̄)2 −

(∑ni=1(Xi − X̄)(Yi − Ȳ)

)2∑ni=1(Yi − Ȳ)2

.

Since f (λ) ≥ 0 for all λ,

the minimum of f =n∑

i=1

(Xi − X̄)2 −

(∑ni=1(Xi − X̄)(Yi − Ȳ)

)2∑ni=1(Yi − Ȳ)2

≥ 0,

which gives n∑i=1

(Xi − X̄)(Yi − Ȳ)2 ≤ n∑

i=1

(Xi − X̄)2 ·n∑

i=1

(Yi − Ȳ)2.

Divide both sides of the above inequalty by (n − 1)2 and we have(sample covariance between (X1, . . . , Xn) and (Y1, . . . ,Yn)

)2 ≤ S 2XS 2Y ,where S X and S Y are the sample standard deviations of (X1, . . . , Xn) and (Y1, . . . ,Yn)respectively. Therefore,∣∣∣sample correlation between (X1, . . . , Xn) and (Y1, . . . ,Yn)∣∣∣

=

∣∣∣∣∣ sample covariance between (X1, . . . , Xn) and (Y1, . . . ,Yn)S XS Y∣∣∣∣∣ ≤ 1.

23. (a) The total sum of squares is

43.784 + 127.562 + 33.027 + 247.244 = 451.617.

(b) The SSE is

451.617 − 43.784 − 127.562(= 33.027 + 247.244) = 280.271.

(c) The degrees of freedom for the SSE is 12 + 80 = 92. The observed Fstatistics is

43.784/4280.271/92

= 3.593065.

13

From the R output, P(F(4, 92) > 3.59) = 0.009128695 < 0.01, so

f0.01,4,92 < 3.59 < 3.593065

and we can conclude that not all drugs have the same effect on weight lossat level 0.01 based on the two-way ANOVA without interaction.Note that the table “0.99 quantiles for F distributions” only gives two rele-vant quantiles f0.01,4,60 = 3.65 and f0.01,4,120 = 3.48, which implies

3.48 < f0.01,4,92 < 3.65. (15)

However, we cannot tell whether f0.01,4,92 < 3.593065 from (15) and it isnecessary to use the R output for P(F(4, 92) > 3.59) to solve this problem.

24. Direct calculation gives

E(X) = (−1) × 0.3 + 0 × 0.4 + a × 0.3 = 0.3(a − 1),

E(X2) = (−1)2 × 0.3 + 02 × 0.4 + a2 × 0.3 = 0.3(a2 + 1),E(Y) = a × 0.3 + 0 × 0.4 + (−a) × 0.3 = 0,

E(Y2) = a2 × 0.3 + 02 × 0.4 + (−a)2 × 0.3 = 0.6a2,and

E(XY) = (−1) × a × 0.3 + 0 × 0 × 0.4 + a × (−a) × 0.3 = −0.3a(a + 1)

so

Var(X) = E(X2)−(E(X))2 = 0.3(a2 +1)−(0.3)2(a−1)2 = 0.3(0.7a2 +0.6a+0.7),

Var(Y) = E(Y2) − (E(Y))2 = 0.6a2 − 02 = 0.6a2,and

Cov(X,Y) = E(XY) − E(X)E(Y) = −0.3a(a − 1) − E(X) × 0 = −0.3a(a + 1).

Therefore,

Corr(X,Y) =Cov(X,Y)

√Var(X)

√Var(Y)

=−0.3a(a + 1)√

0.3(0.7a2 + 0.6a + 0.7) ·√

0.6a2

= − a|a| ·a + 1

√1.4a2 + 1.2a + 1.4

.

25. (a) The grand mean (total mean) X̄G is

2.8 + 3.0 + 2.7 + 2.5 + 2.6 + 2.8 + 2.7 + 2.8 + 3.0 + 2.8 + 2.7 + 2.512

=32.912≈ 2.7417,

and the sample means for the four major groups are

2.8 + 3.0 + 2.73

=8.53,

14

2.5 + 2.6 + 2.83

=7.93,

2.7 + 2.8 + 3.03

=8.53,

and2.8 + 2.7 + 2.5

3=

83.

The sum of squares due to major, denoted by SSmajor, is

3 × [(8.5/3 − 32.9/12)2 + ((7.9/3 − 32.9/12)2

+(8.5/3 − 32.9/12)2 + (8/3 − 32.9/12)2]= 0.1025.

To compute the sum of squares due to graduation year, note that the samplemeans for the three year groups are

2.8 + 2.5 + 2.7 + 2.84

= 2.7,

3.0 + 2.6 + 2.8 + 2.74

= 2.775,

and2.7 + 2.8 + 3.0 + 2.5

4= 2.75,

so the sum of squares due to graduation year, denoted by SSyear, is

4 × [(2.7 − 32.9/12)2 + (2.775 − 32.9/12)2

+(2.75 − 32.9/12)2]

=0.07

6.

The total sum of squares, denoted by SStotal, is

2.82 + 3.02 + 2.72 + 2.52 + 2.62 + 2.82 + 2.72 + 2.82 + 3.02 + 2.82 + 2.72 + 2.52 − 12(

32.912

)2= 90.49 − 1082.41

12

The SSE is

SStotal − SSmajor − SSyear

= 90.49 − 1082.4112

− 0.1025 − 0.076

= 0.175

with degrees of freedom 12 − 1 − (4 − 1) − (3 − 1) = 6. The observed Fstatistic (for testing the main effect of major) is

SSmajor/(4 − 1)SSE/6

=0.1025/30.175/6

= 1.171429.

From the table “0.95 quantiles for F distributions”, f0.05,3,6 = 4.76 >1.171429, so we cannot conclude that inital expected wages for studentsof different majors are not allthe same at level 0.05.

15

(b) No. There is only one observation for each major-year combination, so theSSE in the two-way ANOVA with interaction is zero and the F statistic fortesting the interaction effect cannot be computed.

26. (a)

Corr(X,Y) =Cov(X,Y)

√Var(X)

√Var(Y)

=0.5

√1√

0.25= 1.

(b)

Var(X − 2Y) = Var(X) + 2Cov(X,−2Y) + Var(−2Y)= Var(X) + 2(−2)Cov(X,Y) + (−2)2Var(Y)= 1 + (−4)(0.5) + (4)(0.25) = 0.

27. Since Y = a − bX, we have

Cov(X,Y) = Cov(X, a − bX)= Cov(X, a) + Cov(X,−bX)= E(X · a) − E(X)E(a) + (−b)Cov(X, X)= 0 − bVar(X) = −bVar(X)

andVar(Y) = Var(a − bX) = Var(−bX) = b2Var(X).

From the definition of correlation and the above calculation, we have

Corr(X,Y) =Cov(X,Y)

√Var(X)Var(Y)

=−bVar(X)√

Var(X)b2Var(X)= − b|b| = −1

since b > 0.

28. (a) The sample correlation is

r =0.2895

√0.383 · 5.273

= 0.2037137.

(b) The test statistic is

T =√

(70 − 2) · r√1 − r2

=68 · 0.2037137√

1 − 0.20371372= 1.715847.

Since t0.05/2,70−2 = t0.025,68 = 1.995 > |1.715847|, we cannot conclude thatCorr(X1,Y1) , 0 at level 0.05.

(c) Since t0.05,68 = 1.668 < 1.715847, we can conclude that Corr(X1,Y1) > 0at level 0.05.

29. In the following calculation, we will use the assumption that the kb random vari-ables X̄i, j: i = 1, . . ., k, j = 1, . . . b are independent with common variance σ2/`,which implies that

Cov(X̄i, j, X̄i′, j′ ) ={σ2/` if (i, j) = (i′, j′);0 otherwise.

16

(a) For i = 1, . . ., k, j = 1, . . ., b,

Cov(X̄i, j, X̄i·) = Cov(X̄i, j,

X̄i,1 + · · · + X̄i,bb

)=

1b

Cov(X̄i, j, X̄i, j) =1b

(σ2

`

)=σ2

b`.

(b) For i = 1, . . ., k, j = 1, . . ., b,

Cov(X̄i, j, X̄G) = Cov

X̄i, j, 1kbk∑

i′=1

b∑j′=1

X̄i′, j′

=

1kb

Cov(X̄i, j, X̄i, j) =1kb

(σ2

`

)=σ2

kb`.

(c) For i = 1, . . ., k,

Cov(X̄i·, X̄G) = Cov

1bb∑

j′=1

X̄i, j′ , X̄G

=

1b

b∑j′=1

Cov(X̄i, j′ , X̄G)

Part (b)=

1b

b∑j′=1

σ2

kb`=σ2

kb`.

(d) For i = 1, . . ., k, j = 1, . . ., b,

Cov(X̄i·, X̄· j) = Cov(

X̄i,1 + · · · + X̄i,bb

,X̄1, j + · · · + X̄k, j

k

)=

1bk

Cov(X̄i, j, X̄i, j) =1bk

(σ2

`

)=σ2

kb`.

(e) Since X̄G =∑k

i′=1∑b

j′=1 X̄i′, j′/(kb) and X̄i′, j′s are independent,

Var(X̄G) =(

1kb

)2 k∑i′=1

b∑j′=1

Var(X̄i′, j′ ) =(

1kb

)2 (σ2

`

)· kb = σ

2

kb`.

From the above result for Var(X̄G) and Equations (5)–(7) given in the prob-lem, we have

Var(X̄G) = Cov(X̄i·, X̄G) = Cov(X̄· j, X̄G) = Cov(X̄i·, X̄· j) =σ2

kb`,

so

Cov(X̄i· − X̄G, X̄· j − X̄G)= Cov(X̄i·, X̄· j) + Cov(X̄i·,−X̄G))

+Cov(−X̄G, X̄· j) + Cov(−X̄G,−X̄G)

=σ2

kb`− σ

2

kb`− σ

2

kb`+σ2

kb`= 0.

17

(f) Since X̄i· =∑b

j′=1 X̄i, j′/b and X̄i, j′s are independent,

Var(X̄i·) =(

1b

)2 b∑j′=1

Var(X̄i, j′ ) =1b2

(σ2

`

)· b = σ

2

b`.

From the result in Part (e), Cov(X̄i· − X̄G, X̄· j − X̄G) = 0, so

Cov(X̄i· − X̄G, X̄i, j − X̄i· − X̄· j + X̄G)= Cov(X̄i· − X̄G, X̄i, j − X̄i·) + Cov(X̄i· − X̄G,−X̄· j + X̄G)︸ ︷︷ ︸

=0

= Cov(X̄i· − X̄G, X̄i, j − X̄i·)= Cov(X̄i·, X̄i, j) + Cov(X̄i·,−X̄i·) + Cov(−X̄G, X̄i, j) + Cov(−X̄G,−X̄i·)

=σ2

b`− σ

2

b`− σ

2

kb`+σ2

kb`= 0.

Here the last second equality follows from Equations (2), (4), (5) given inthe problem and the result Var(X̄i·) = σ2/(b`) computed in this part.

30. (a) Solving Cov(Y − bX, X) = 0 for b gives Cov(Y, X) − bCov(X, X) = 0, so

b =Cov(Y, X)Cov(X, X)

=69

=23,

and Y − (2/3)X is independent of X.(b)

E(U) = E(Y − (2/3)X) = E(Y) − 23

E(X) = 2 − 23· 1 = 4

3.

Var(U) = Var(Y − (2/3)X)= Var(Y) + Var(−(2/3)X) + 2Cov(Y,−(2/3)X)

= 16 +(−2

3

)2Var(X) − 4

3Cov(Y, X)

= 16 +49· 9 − 4

3· 6 = 12.

31. (a)

Cov

ε̄, n∑i=1

(Xi − X̄)εi = n∑

i=1

Cov(ε̄, (Xi − X̄)εi)

=

n∑i=1

(Xi − X̄)Cov(ε1 + · · · + εn

n, εi

)=

n∑i=1

(Xi − X̄)n

Cov (ε1 + · · · + εn, εi)

=

n∑i=1

(Xi − X̄)n

Cov(εi, εi)

=

n∑i=1

(Xi − X̄)n

· σ2ε =σ2εn

n∑i=1

Xi − nX̄ = 0.

18

(b) Consider Equation (8) given in the problem. Dividing both sides of theequation by σ2ε, then

∑ni=1(Yi − â − b̂Xi)2

σ2ε=

n∑i=1

(εiσε

)2−

( √nε̄σε

)2−

∑n

i=1(Xi − X̄)εiσε

√(n − 1)S 2X

2

.

Let Zi = εi/σε for i = 1, . . ., n, then Z1, . . ., Zn are IID N(0, 1) randomvariables and the above equation becomes

∑ni=1(Yi − â − b̂Xi)2

σ2ε=

n∑i=1

Z2i −(√

nZ̄)2 −

∑n

i=1(Xi − X̄)Zi√(n − 1)S 2X

2

, (16)

where Z̄ =∑n

i=1 Zi/n. Let U1 =√

nε̄/σε =√

nZ̄ and

U2 =∑n

i=1(Xi − X̄)εiσε

√(n − 1)S 2X

=

∑ni=1(Xi − X̄)Zi√

(n − 1)S 2X,

then (16) becomes∑ni=1(Yi − â − b̂Xi)2

σ2ε=

n∑i=1

Z2i − U21 − U22 , (17)

where

Cov(U1,U2) =( √

nσε

) 1σε √(n − 1)S 2XCov

ε̄, n∑i=1

(Xi − X̄)εi Part (a)= 0 ,

Var(U1) = nVar(Z̄) = n ·1n

= 1,

and

Var(U2) =∑n

i=1 Var((Xi − X̄)Zi)

(√

(n − 1)S 2X)2=

∑ni=1(Xi − X̄)2

(n − 1)S 2X= 1

Apply Fact 2 in the handout “Sum of squares of IID N(0, 1) random vari-ables”, ∑n

i=1(Yi − â − b̂Xi)2σ2ε

(17)=

n∑i=1

Z2i − U21 − U22 ∼ χ2(n − 2)

since Z1, . . ., Zn are IID N(0, 1) random variables, U1 and U2 are lin-ear combinations of Z1, . . ., Zn such that Var(U1) = 1 = Var(U2) andCov(U1,U2) = 0.

32. (a) Apply the result∑n

i=1 w2i =

∑ni=1(wi− w̄)2 + nw̄2 with wi = Yi−a−bXi, then

we have w̄ = Ȳ − a − bX̄ andn∑

i=1

(Yi − a − bXi)2

=

n∑i=1

(Yi − a − bXi − (Ȳ − a − bX̄))2 + n(Ȳ − a − bX̄)2

19

=

n∑i=1

(Yi − Ȳ − b(Xi − X̄))2 + n(Ȳ − a − bX̄)2

=

n∑i=1

(Yi − Ȳ)2 − 2bn∑

i=1

(Yi − Ȳ)(Xi − X̄) + b2n∑

i=1

(Xi − X̄)2 + n(Ȳ − a − bX̄)2

= (n − 1)(S 2Xb

2 − 2rS XS Yb + S 2Y)

+ n(Ȳ − a − bX̄)2

(b) LetS (b) = S 2Xb

2 − 2rS XS Yb + S 2Y ,then from the result in Part (a),

n∑i=1

(Yi − a − bXi)2 = (n − 1)S (b) + n(Ȳ − a − bX̄)2. (18)

Note that when S X , 0, S (b) is a quadratic function of b with a globalminimum. Solving S ′(b) = 0 and we have

2bS 2X − 2rS XS Y = 0,

which gives b = rS Y/S X = b̂, so S (b) is minimized when b = b̂. From(18), we have

n∑i=1

(Yi − a − bXi)2 ≥ (n − 1)S (b)

≥ (n − 1)S (b̂) (S (b) is minimized when b = b̂)= (n − 1)S (b̂) + n(Ȳ − â − b̂X̄)2︸ ︷︷ ︸

=0 since â=Ȳ−b̂X̄

=

n∑i=1

(Yi − a − bXi)2∣∣∣∣∣∣∣(a,b)=(â,b̂)

,

so∑n

i=1(Yi − a − bXi)2 is minimized when (a, b) = (â, b̂).

33. The simple linear regression model considered here is

price = a + b · age + error,

where price and age represents car price and car age respectively. In the aboveregression model, b is estimated by

b̂ =r · S price

S age=−0.6 · 1.9

2.2= −0.518

and a is estimated by

â = price − b̂ · age = 6.9 −(−0.6 · 1.9

2.2

)· 8.9 = 11.512,

so the estimated regression equation is

p̂rice = 11.512 − 0.518 · age.

20

34. For the proof of

Var(â + b̂x0) = σ2ε

1n + (x0 − X̄)2(n − 1)S 2X ,

we first compute Cov(ε̄, b̂ − b). The result in Part (a) of Problem 31 states thatCov(ε̄,

∑ni=1(Xi − X̄)εi) = 0, so

Cov(ε̄, b̂ − b) = Covε̄, ∑ni=1(Xi − X̄)εi

(n − 1)S 2X

= Cov(ε̄,∑ni=1(Xi − X̄)εi)(n − 1)S 2X

= 0. (19)

Next, we express â + b̂x0 − a − bx0 using ε̄ and b̂ − b:

â + b̂x0 − a − bx0 = Ȳ − b̂X̄ + b̂x0 − a − bx0= a + bX̄ + ε̄ + b̂(x0 − X̄) − a − bx0= ε̄ + (b̂ − b)(x0 − X̄).

Then

Var(â + b̂x0) = Var(â + b̂x0 − a − bx0)= Var(ε̄ + (b̂ − b)(x0 − X̄))= Var(ε̄) + Var((b̂ − b)(x0 − X̄)) + 2(x0 − X̄)Cov(ε̄, b̂ − b)︸ ︷︷ ︸

(19)= 0

=σ2εn

+ (x0 − X̄)2Var(b̂ − b)

= σ2ε

1n + (x0 − X̄)2(n − 1)S 2X .

Here the last equality holds since

Var(b̂ − b) = σ2ε

(n − 1)S 2X,

which has been computed in Page 5 in the handout “Simple linear regression”.

35. Let D̄ and S D be the sample mean and sample standard deviation for the sample(D1, D2, D3, D4), then for a given distance D,

Var(â + b̂d0) = σ2ε

14 + (d0 − D̄)23S 2D .

The average variance of estimated expected signal losses is

13

8∑d0=6

Var(â + b̂d0) =13

8∑d0=6

σ2ε

14 + (d0 − D̄)23S 2D

Apply the fact thatn∑

i=1

w2i =n∑

i=1

(wi − w̄)2 + n(w̄)2

with n = 3 and (w1,w2,w3) = (6 − D̄, 7 − D̄, 8 − D̄), then we have8∑

d0=6

(d0 − D̄)2 =8∑

d0=6

(d0 − D̄ − (7 − D̄))2 + 3(7 − D̄)2

=

8∑d0=6

(d0 − 7)2 + 3(7 − D̄)2 = 2 + 3(7 − D̄)2,

21

so13

8∑d0=6

Var(â + b̂d0) = σ2ε

14 + 2 + 3(7 − D̄)29S 2D .

Let

S (D1,D2,D3,D4) =2 + 3(7 − D̄)2

3S 2D,

then the average variance of estimated expected signal losses is minimized whenS (D1,D2,D3,D4) is minimized. Direct calculation gives

(4 + 6 + 8 + 10)/4 = 7 = (5 + 6 + 8 + 9)/4,

S (4, 6, 8, 10) =2 + 3(7 − 7)2

(4 − 7)2 + (6 − 7)2 + (8 − 7)2 + (10 − 7)2 =2

20,

S (5, 6, 8, 9) =2 + 3(7 − 7)2

(5 − 7)2 + (6 − 7)2 + (8 − 7)2 + (9 − 7)2 =2

10,

It is clear thatS (4, 6, 8, 10) < S (5, 6, 8, 9),

so the (D1,D2,D3,D4) in (a) gives the smallest average variance of estimatedexpected signal losses.

36. (a) For brevity, will use “price” and “age” to denote the car price and car agerespectively. The estimated regression equation is

p̂rice = â + b̂ · age.

whereb̂ =−0.6 · 1.9

2.2= −0.518

and

â = 6.9 − b̂ · 8.9 = 6.9 −(−0.6 · 1.9

2.2

)· 8.9 = 11.512,

so the estimated regression equation is

p̂rice = 11.512 − 0.518 · age.

(b) A 95% confidence interval is given by

b̂ ± t0.05/2,30−2

√σ̂2ε

(30 − 1) · (2.2)2 ,

where b̂ = −0.6 · 1.9/2.2 from Part (a), t0.05/2,30−2 = t0.025,28 = 2.048, and

σ̂2ε =RS S

30 − 2

=(30 − 1)(1.9)2(1 − (−0.6)2)

30 − 2 (using the RS S formula in Problem 37),

so√σ̂2ε

(30 − 1) · (2.2)2 =

√(30 − 1)(1.9)2(1 − (−0.6)2)

30 − 2 ·1

(30 − 1) · (2.2)2

=1.92.2

√1 − (−0.6)2

28.

22

The 95% confidence interval is

−0.6 · 1.92.2

± 2.048 · 1.92.2

√1 − (−0.6)2

28,

which gives [−0.786,−0.251]. Using the open interval (−0.786,−0.251) asa 95% confidence interval is also fine.

(c) The observed T statistic is

√30 − 2

−0.6√1 − (−0.6)2

= −3.968627and t0.05/2,30−2 = t0.025,28 = 2.048 < | − 3.968627|, so we can conclude thereis a linear relation between car price and car age at level 0.05.

37. Let X̄ and Ȳ be the sample means of (X1, . . . , Xn) and (Y1, . . . ,Yn) respectively,and let S X be the sample standard deviations for (X1, . . . , Xn). Let r be the samplecorrelation between (X1, . . . , Xn) and (Y1, . . . ,Yn). Let b̂ = rS Y/S X and â =Ȳ − b̂X̄, then

RS S =n∑

i=1

(Yi − â − b̂Xi)2

=

n∑i=1

(Yi − (Ȳ − b̂X̄) − b̂Xi)2

=

n∑i=1

(Yi − Ȳ − b̂(Xi − X̄))2

=

n∑i=1

(Yi − Ȳ)2 + (b̂)2n∑

i=1

(Xi − X̄)2 − 2b̂n∑

i=1

(Yi − Ȳ)(Xi − X̄). (20)

Since1

n − 1

n∑i=1

(Yi − Ȳ)(Xi − X̄) = rS XS Y ,

dividing both sides of (20) by (n − 1) gives

RS Sn − 1 = S

2Y + (b̂)

2S 2X − 2b̂rS XS Y

= S 2Y +(

rS YS X

)2S 2X − 2

(rS YS X

)rS XS Y

= S 2Y + r2S 2Y − 2r2S 2Y

= S 2Y (1 − r2),

which givesRS S = (n − 1)S 2Y (1 − r2).

38. (a) The residual sum of squares (RS S ) is

(30 − 1)S 2price · (1 − (−0.6)2) = 29 · (1.9)2 · 0.64 = 67.0016.

(b) (−0.6)2 = 0.36, so 36% of variation in car sale price can be explained bycar age.

23

(c) A 95% confidence interval for the expected car price when the car age is 8is given by

â + b̂ · 8 ± t0.05/2,30−2 · σ̂ε√

130

+(8 − age)2

(30 − 1)S 2age,

where

â + b̂ · 8 = 6.9 −(−0.6 · 1.9

2.2

)· 8.9 + −0.6 · 1.9

2.2· 8 = 16.202

2.2,

t0.05/2,30−2 = t0.025,28 = 2.048,

σ̂ε =

√(30 − 1)(1.9)2(1 − (−0.6)2)

30 − 2 (from the solution to Problem 36(b)),

and

σ̂ε

√1

30+

(8 − age)2(30 − 1)S 2age

=

√(30 − 1)(1.9)2(1 − (−0.6)2)

30 − 2

√1

30+

(8 − 8.9)229 · (2.2)2

=1.92.2·√

105.3824840

.

The upper bound for the 95% confidence interval is

16.2062.2

+ 2.048 · 1.92.2·√

105.3824870

≈ 7.993,

and the lower bound for the 95% confidence interval is

16.2062.2

− 2.048 · 1.92.2·√

105.3824870

≈ 6.740,

so the 95% confidence interval is [6.740, 7.993].

(d) A 95% prediction interval for car price when the car age is 8 is given by

â + b̂ · 8 ± t0.05/2,30−2 · σ̂ε√

1 +130

+(8 − age)2

(30 − 1)S 2age

where

â + b̂ · 8 = 6.9 −(−0.6 · 1.9

2.2

)· 8.9 + −0.6 · 1.9

2.2· 8 = 16.202

2.2,

t0.05/2,30−2 = t0.025,28 = 2.048 and

σ̂ε

√1 +

130

+(8 − age)2

(30 − 1)S 2age=

√(30 − 1)(1.9)2(1 − (−0.6)2)

30 − 2 ·

√1 +

130

+(8 − 8.9)229 · (2.2)2

=1.92.2·√

2800.2944840

.

The upper bound for the 95% prediction interval is

16.2062.2

+ 2.048 · 1.92.2·√

2800.2944840

≈ 10.596,

24

and the lower bound for the 95% prediction interval is

16.2062.2

− 2.048 · 1.92.2·√

105.3824840

≈ 4.137,

so the 95% predition interval is [4.137, 10.596].

39. (a) The p-value is 0.0121 > 0.01, so we cannot conclude that there is a linearrelation between height and weight at level 0.01.

(b) The estimated regression equation is

̂weight = −27.2405 + 0.4489 × height.(c) 24.39%.(d) √

RS S23

= 4.314,

so the residual sum of squares (RS S ) is 23 · (4.314)2 = 428.0437.

40. (a) The 95% C.I. for the coefficient of height is 0.4489 ± t0.025,23 × 0.1648,where t0.025,23 = 2.069. The lower bound of the 95% C.I. is

0.4489 − 2.069 × 0.1648 ≈ 0.108

and the upper bound of the 95% C.I. is

0.4489 + 2.069 × 0.1648 ≈ 0.790

The 95% C.I. for the coefficient of height is [0.108, 0.790].

(b) The number of students in the group is 23 + 2 = 25 since the degrees offreedom of RS S is (n − 2) = 23 in this regression model.

(c) The variance of the estimated coefficient of height is

σ2ε(n − 1) · (sample variance of height data) ,

which is estimated by

RS S/(n − 2)(n − 1) · (sample variance of height data) . (21)

From the R output, the estimated variance of the estimated coefficient ofheight is (0.1648)2, so (21) gives

(0.1648)2 =RS S/(n − 2)

24 · (sample variance of height data) .

Note that from the R output, RS S/(n − 2) = RS S/23 = (4.314)2. Plug thevalue of RS S/(n − 2) in the above equation, then we have

(0.1648)2 =(4.314)2

24 · (sample variance of height data) ,

so the sample variance for the heights of the students is

(4.314)2

24 · (0.1648)2 ≈ 28.55187.

25

41. (a) The estimated regression equation is

p̂rice = 160.90813 − 6.77199 · age − 0.13038 ·mileage.

(b) Based on the value of multiple R-squared, 21.81% of the variation in pricethat can be explained by age and mileage.

(c) Yes. The p-value for the F test is 0.01056 < 0.05, so we can conclude thatthe coefficients for age and mileage are not all zero at level 0.05.

(d) No. The p-value is 0.0278 > 0.01, so we cannot conclude that the coeffi-cient for age is not zero at level 0.01.

(e) An observed 95% confidence interval for the coefficient of mileage is

−0.13038 ± t0.05/2,n−1−k · 0.06114,

where n − 1 − k = 37 since the residual standard error has 37 degrees offreedom, so t0.05/2,n−1−k = t0.025,37 = 2.026. Plug in t0.05/2,n−1−k = 2.026 inthe formula for the 95% confidence interval, then the upper bound of theinterval is

−0.13038 + 2.026 · 0.06114 = −0.00651036and the lower bound is

−0.13038 − 2.026 · 0.06114 = −0.2542496

and we have the 95% confidence interval for the coefficient of mileage is[−0.2542496,−0.00651036].

(f) Since n − 1 − k = 37 with k = 2, n = 37 + 1 + 2 = 40.(g)√

RS S/37 = 36.38, so the residual sum of squares is RS S = 37 ·(36.38)2 =48969.66.

(h) The regression sum of squares S S reg satisfies

S S reg/k

RS S/(n − 1 − k) = F value,

where k = 2 and n − 1 − k = 37, so

S S reg/2

37 · (36.38)2/37 = 5.16

and the regression sum of squares S S reg is 5.16 · 2 · 37 · (36.38)2/37 =13658.57.

(i) The total sum of squares is

S S total = RS S +S S reg = 37·(36.38)2+5.16·2·37·(36.38)2/37 = 62628.23.

Note. For the computation of S S total and RS S , one can also use

RS SS S total

= 1 − 0.2181 = 0.7819

and RS S = 48969.66 to find S S total = 48969.66/0.7819 = 62629.06 and thencompute

S S reg = S S total − RS S = 62629.06 − 48969.66 = 13659.4.

26

42. We will apply chi-squared goodness of fit test to solve this problem. We firstcompute the probabilities that a N(47, 402) variable falls in the 5 salary ranges.

P(N(47, 402) < 25) = P(N(0, 1) < (25 − 47)/40)= P(N(0, 1) < −0.55)= 0.5 − 0.2088 = 0.2912.

P(25 < N(47, 402) < 50) = P((25 − 47)/40 < N(0, 1) < (50 − 47)/40)= P(−0.55 < N(0, 1) < 0.075)≈ P(−0.55 < N(0, 1) < 0.08)= 0.2088 + 0.0319 = 0.2407.

P(50 < N(47, 402) < 75) = P((50 − 47)/40 < N(0, 1) < (75 − 47)/40)= P(0.075 < N(0, 1) < 0.7)≈ P(0.08 < N(0, 1) < 0.7)= 0.2580 − 0.0319 = 0.2261.

P(75 < N(47, 402) < 100) = P((75 − 47)/40 < N(0, 1) < (100 − 47)/40)= P(0.7 < N(0, 1) < 1.325)≈ P(0.7 < N(0, 1) < 1.33)= 0.4082 − 0.2580 = 0.1502.

P(N(47, 402) > 100) = P(N(0, 1) > 1.325)≈ P(N(0, 1) > 1.33)= 0.5 − 0.4082 = 0.0918.

The test statistic for the chi-squared goodness of fit test is

(90 − 1000 · 0.2912)21000 · 0.2912 +

(450 − 1000 · 0.2407)21000 · 0.2407

+(190 − 1000 · 0.2261)2

1000 · 0.2261 +(140 − 1000 · 0.1502)2

1000 · 0.1502

+(130 − 1000 · 0.0918)2

1000 · 0.0918= 343.3646.

The degree of freedom of the test statistic is 5−1 = 4. The 0.99 quantile of χ2(4)is 13.277 < 343.3646, so based on the chi-squared goodness of fit test, at level0.01, we can conclude that the distribution of the annal salary (in 10,000 NTD)is not N(47, 402).

43. We will apply the chi-square test of independence to solve this problem. Therow totals are

1758 + 5619 + 1697 = 9074 and 273 + 565 + 88 = 926,

and the column totals are

1758 + 273 = 2031, 5619 + 565 = 6184, and 1697 + 88 = 1785.

27

The test statistic is

(1758 − 10000 · (2031/10000) · (9074/10000))210000 · (2031/10000) · (9074/10000)

+(5619 − 10000 · (6184/10000) · (9074/10000))2

10000 · (6184/10000) · (9074/10000)

+(1697 − 10000 · (1785/10000) · (9074/10000))2

10000 · (1785/10000) · (9074/10000)

+(273 − 10000 · (2031/10000) · (926/10000))2

10000 · (2031/10000) · (926/10000)

+(565 − 10000 · (6184/10000) · (926/10000))2

10000 · (6184/10000) · (926/10000)

+(88 − 10000 · (1785/10000) · (926/10000))2

10000 · (1785/10000) · (926/10000)= 82.20876.

The degree of freedom of the test statistic is (3−1)(2−1) = 2. The 0.99 quantileof χ2(2) is 9.210 < 82.20876, so at level 0.01, we can conclude that the jobfunction and the the annal salary range are associated.

44. We will apply the chi-square goodness of fit test to solve this problem. The teststatistic is

(1758 − 10000 · 0.2)210000 · 0.2 +

(5619 − 10000 · 0.5)210000 · 0.5 +

(1697 − 10000 · 0.2)210000 · 0.2

+(273 − 10000 · 0.03)2

10000 · 0.03 +(565 − 10000 · 0.06)2

10000 · 0.06 +(88 − 10000 · 0.01)2

10000 · 0.01= 157.7304.

The degree of freedom of the test statistic is 3 · 2 − 1 = 5. The 0.95 quantileof χ2(5) is 11.070 < 157.7304, so at level 0.05, we can conclude that the actualproportions are not all the same as the proposed proportions.

45. (a) The multiple R2 in the R output of summary(lm(V4˜V2+V3)) is 0.9946,so 99.46% of variation in V4 can be explained by V2 and V3.

(b) In the summary of the model lm(V1˜V2+V3+V4), the multiple R2 is 0.9191,so the three variables V2, V3 and V4 can explain over 90% of variation inV1. However, none of the coefficients of V2, V3 and V4 is significantlydifferent from zero at level 0.05, which indicates there might be a prob-lem of collinearity. In fact, the multiple R2 for lm(V4˜V2+V3) is closeto one, which shows that there is a serious collinearity problem. Afterremoving the variable V4, the model lm(V1˜V2+V3) still has a multipleR2 that is greater than 0.9, and the estimated coefficients become signif-cantly different from zero at level 0.05. Moreover, the standard errors ofthe estimated coeffiencts of V2 and V3 are much smaller than those in themodel lm(V1˜V2+V3+V4). Therefore, here the model lm(V1˜V2+V3) ispreferred to the model lm(V1˜V2+V3+V4).Based on the model lm(V1˜V2+V3), the estimated regression equation is

V̂1 = 1.0554 + 1.9442 × V2 + 2.9395 × V3,

so the predicted value for V1 when (V2, V3, V4) = (1, 2, 3) is

1.0554 + 1.9442 · 1 + 2.9395 · 2 = 8.8786.

28

46. (a) Sort the sample data and we have 1.5 < 1.9 < 2 < 2.1. The rank of 2 is 3.(b) Sort the sample data and we have 1.5 < 1.9 < 2 = 2 = 2. The rank of 2 is

(3 + 4 + 5)/3 = 4.

47. To obtain the test statistic for the signed-rank test, we first substract −1 from thedata:

(−9.58,−8.68, 10.22, 10.01, 9.75, 10.99)−(−1, . . . ,−1) = (−8.58−7.68, 11.22, 11.01, 10.75, 11.99).

Then, we sort the differences by their absolute values:

| − 7.68| < | − 8.58| < |10.75| < |11.01| < |11.22| < |11.99|.

The rank of | − 7.68| and | − 8.58| are 1 and 2, so the observed test statisticR− = 1 + 2 = 3. From the R output, the p-value for the signed-rank test is

P(DS R(6) ≤ 3) = 0.078125.

Based on the signed-rank test, we can conclude that µ > −1 at level 0.1.

48. The one-sample t test is applicable when the sample is from a normal distribu-tion. The shape of the histogram in Problem 47 shows two peaks, which does notlook like the shape of a normal density. Therefore, it is not reasonable to applythe one-sample t test to the data.

49. Note that

I(Y2 > Y1)I(Y3 > Y1) ={

1 if Y2 > Y1 and Y3 > Y1;0 otherwise,

so

Cov(I(Y2 > Y1), I(Y3 > Y1))= E(I(Y2 > Y1)I(Y3 > Y1)) − E(I(Y2 > Y1))E(I(Y3 > Y1))= P(Y2 > Y1 and Y3 > Y1) − P(Y2 > Y1)P(Y3 > Y1). (22)

Since Y1, Y2, Y3 are IID continuous random variables, for each (i1, i2, i3) that isone of the six permutations of (1, 2, 3),

P(Yi1 < Yi2 < Yi3 ) =16.

Thus

P(Y2 > Y1 and Y3 > Y1) = P(Y1 < Y2 < Y3)+P(Y1 < Y3 < Y2) =16·2 = 1

3. (23)

Moreover,

P(Y2 > Y1) =12

(24)

since P(Y2 > Y1) + P(Y1 > Y2) = 1 and P(Y2 > Y1) = P(Y1 > Y2), and similarly,

P(Y3 > Y1) =12

(25)

since P(Y3 > Y1) + P(Y1 > Y3) = 1 and P(Y3 > Y1) = P(Y1 > Y3). Therefore,from (22), (23), (24) and (25),

Cov(I(Y2 > Y1), I(Y3 > Y1)) = P(Y2 > Y1 and Y3 > Y1) − P(Y2 > Y1)P(Y3 > Y1)

=13−

(12

) (12

)=

112.

29

50. LetΛ = {(i, j) : i ∈ {1, . . . , n1}, j ∈ {1, . . . , n2}},

then

Var(S ) = Cov

∑(i, j)∈Λ

I(Y j > Xi),∑

(i′, j′)∈ΛI(Y j′ > Xi′ )

=

∑(i, j)∈Λ

∑(i′, j′)∈Λ

Cov(I(Y j > Xi), I(Y j′ > Xi′ ))

=

n1∑i=1

n2∑j=1

n1∑i′=1

n2∑j′=1

Cov(I(Y j > Xi), I(Y j′ > Xi′ )),

where

Cov(I(Y j > Xi), I(Y j′ > Xi′ )) =

0 if i , i′ and j , j′;1/12 if i , i′ and j = j′;1/12 if i = i′ and j , j′;1/4 if i = i′ and j = j′.

For i, i′ ∈ {1, . . . , n1} and j, j′ ∈ {1, . . . , n2},

• the number of (i, j, i′, j′)s such that i , i′ and j = j′ is n1(n1 − 1)n2,• the number of (i, j, i′, j′)s such that i = i′ and j , j′ is n2(n2 − 1)n1, and• the number of (i, j, i′, j′)s such that i = i′ and j = j′ is n1n2.,

so

Var(S ) =14· n1n2 +

112· n1(n1 − 1)n2 +

112· n2(n2 − 1)n1

=1

12· n1n2(3 + n1 − 1 + n2 − 1) =

n1n2(n1 + n2 + 1)12

.

51. Here we assume that the 2500 rents can be viewed as IID observations from arent distribution. Let µ be median of the rent distribution. We consider the testingproblem

H0 : µ ≥ 7 versus H1 : µ < 7. (26)Based on the histogram of rents, we assume that the rent distribution is not sym-metric, so the only applicable test for (26) that we have learned is the sign test.We define rent < 7 as a “+” sign event which occurs with probability p+, thenthe testing problem in (26) is equivalent to

H0 : p+ ≤ 0.5 versus H1 : p+ > 0.5.

Three of the observed four rents are less than 7, so S +, the number of “+” signevents is 3. The p-value for the sign test is

P(Bin(4, 0.5) ≥ 3) = 1 − P(Bin(4, 0.5) ≤ 2) = 1 − 0.6875 = 0.3125.

The p-value is greater than 0.07, so we cannot conclude that µ < 7 at level 0.07.

52. We will apply the Wilcoxon rank-sum test for the testing problem

H0 : µ ≤ 0 versus H1 : µ > 0.

30

The sorted combined sample is

(4, 4.1, 4.2, 4.5, 5.1, 5.5, 5.6, 6.3, 6.5)

so the ranks for 5.5, 4.5, 4, 4.1, 4.2, 6.3, 5.1, 5.6, 6.5 are 6,4,1,2,3,8,5,7,9. TheR statistic for the rank-sum test is 8 + 5 + 7 + 9 = 29 and the S statistic is29 − 4 · 5/2 = 29 − 10 = 19. The p-value is

P(Wilcox(5, 4) ≤ 20 − 19) = P(Wilcox(5, 4) ≤ 1).

Let X ∼ Wilcox(5, 4). Since the distirbution for Wilcox(5, 4) is symmetric about5 · 4/2 = 10, 20 − X ∼ X, so

P(Wilcox(5, 4) ≤ 1) = P(X ≤ 1)= P(20 − X ≤ 1)= P(X ≥ 19)= P(Wilcox(5, 4) ≥ 19)= 1 − P(Wilcox(5, 4) ≤ 18)= 1 − 0.9841270 = 0.015873 < 0.05,

so we can conclude that µ > 0 at level 0.05.

53. Let µ and σ2 be the mean and variance for the distribution Wilcox(5, 4), then

µ = 5 · 4/2 = 10

andσ2 =

5 · 4 · (5 + 4 + 1)12

=503.

(a) Using normal approximation with continuity correction gives

P(Wilcox(5, 4) ≤ 15) = P(N(10, 50/3) < 15.5)= P(N(0, 1) < (15.5 − 10)/

√50/3)

≈ P(N(0, 1) < 1.35) = 0.5 + 0.4115 = 0.9115,

and the absolute approximation error is |0.9047619−0.9115| = 0.0067381.(b) Using normal approximation without continuity correction gives

P(Wilcox(5, 4) ≤ 15) = P(N(10, 50/3) < 15)= P(N(0, 1) < (15 − 10)/

√50/3)

≈ P(N(0, 1) < 1.22) = 0.5 + 0.3888 = 0.8888,

and the absolute approximation error is |0.9047619−0.8888| = 0.0159619.

Since 0.0067381 < 0.0159619, the normal approximation with continuity cor-rection is more accurate.

31