Simple Linear Regression

Chapter 13:

3.a.

= 4(100 - 87) 2 + 6(85 - 87) 2 + 5(79 - 87) 2 = 1,020

MSTR = SSB /(k - 1) = 1,020/2 = 510

b. = 3(35.33) + 5(35.60) + 4(43.50) = 458

MSE = SSE /(nT - k) = 458/(15 - 3) = 38.17

c.F = MSTR /MSE = 510/38.17 = 13.36

Using F table (2 degrees of freedom numerator and 12 denominator), p-value is less than .01

Actual p-value = .0009

Because p-value = .05, we reject the null hypothesis that the means of the three populations are equal.

d.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 1020 / 2 / 510 / 13.36
Error / 458 / 12 / 38.17
Total / 1478 / 14

5.a.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 120 / 2 / 60 / 20
Error / 216 / 72 / 3
Total / 336 / 74

b.Using F table (2 numerator degrees of freedom and 72 denominator), p-value is less than .01

Actual p-value = .0000

Because p-value = .05, we reject the null hypothesis that the 3 population means are equal.

7.

Superior / Peer / Subordinate
Sample Mean / 5.75 / 5.5 / 5.25
Sample Variance / 1.64 / 2.00 / 1.93

= (5.75 + 5.5 + 5.25)/3 = 5.5

= 8(5.75 - 5.5) 2 + 8(5.5 - 5.5) 2 + 8(5.25 - 5.5) 2 = 1

MSTR = SSTR /(k - 1) = 1/2 = .5

= 7(1.64) + 7(2.00) + 7(1.93) = 38.99

MSE = SSE /(nT - k) = 38.99/21 = 1.86

F = MSTR /MSE = 0.5/1.86 = 0.27

Using F table (2 degrees of freedom numerator and 21 denominator), p-value is greater than .10

Actual p-value = .7660

Because p-value > = .05, we cannot reject the null hypothesis that the means of the three populations are equal; thus, the source of information does not significantly affect the dissemination of the information.

9.

Real Estate Agent / Architect / Stockbroker
Sample Mean / 67.73 / 61.13 / 65.80
Sample Variance / 117.72 / 180.10 / 137.12

= (67.73 + 61.13 + 65.80)/3 = 64.89

= 15(67.73 - 64.89) 2 + 15(61.13 - 64.89) 2 + 15(65.80 - 64.89) 2 = 345.47

MSTR = SSTR /(k - 1) = 345.47/2 = 172.74

= 14(117.72) + 14(180.10) + 14(137.12) = 6089.16

MSE = SSE /(nT - k) = 6089.16/(45-3) = 144.98

F = MSTR /MSE = 172.74/144.98 = 1.19

Using F table (2 degrees of freedom numerator and 42 denominator), p-value is greater than .10

Actual p-value = .3143

Because p-value > = .05, we cannot reject the null hypothesis that the job stress ratings are the same for the three occupations.

21.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 300 / 4 / 75 / 14.07
Error / 160 / 30 / 5.33
Total / 460 / 34

22.a.H0: u1 = u2 = u3 = u4 = u5

Ha: Not all the population means are equal

b.Using F table (4 degrees of freedom numerator and 30 denominator), p-value is less than .01

Actual p-value = .0000

Because p-value = .05, we reject H0

23.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 150 / 2 / 75 / 4.80
Error / 250 / 16 / 15.63
Total / 400 / 18

Using F table (2 degrees of freedom numerator and 16 denominator), p-value is between .01 and .025

Actual p-value = .0233

Because p-value = .05, we reject the null hypothesis that the means of the three treatments are equal.

25.

A / B / C
Sample Mean / 119 / 107 / 100
Sample Variance / 146.89 / 96.43 / 173.78

= 8(119 - 107.93) 2 + 10(107 - 107.93) 2 + 10(100 - 107.93) 2 = 1617.9

MSTR = SSTR /(k - 1) = 1617.9 /2 = 809.95

= 7(146.86) + 9(96.44) + 9(173.78) = 3,460

MSE = SSE /(nT - k) = 3,460 /(28 - 3) = 138.4

F = MSTR /MSE = 809.95 /138.4 = 5.85

Using F table (2 degrees of freedom numerator and 25 denominator), p-value is less than .01

Actual p-value = .0082

Because p-value = .05, we reject the null hypothesis that the means of the three treatments are equal.

27.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Between / 61.64 / 3 / 20.55 / 17.56
Error / 23.41 / 20 / 1.17
Total / 85.05 / 23

Using F table (3 degrees of freedom numerator and 20 denominator), p-value is less than .01

Actual p-value = .0000

Because p-value = .05, we reject the null hypothesis that the mean breaking strength of the four cables is the same.

31.

A / B / C
Sample Mean / 20 / 21 / 25
Sample Variance / 1 / 25 / 2.5

= (20 + 21 + 25)/3 = 22

= 5(20 - 22) 2 + 5(21 - 22) 2 + 5(25 - 22) 2 = 70

MSTR = SSTR /(k - 1) = 70 /2 = 35

= 4(1) + 4(2.5) + 4(2.5) = 24

MSE = SSE /(nT - k) = 24 /(15 - 3) = 2

F = MSTR /MSE = 35 /2 = 17.5

Using F table (2 degrees of freedom numerator and 12 denominator), p-value is less than .01

Actual p-value = .0003

Because p-value = .05, we reject the null hypothesis that the mean miles per gallon ratings are the same for the three automobiles.

35.

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 310 / 4 / 77.5 / 17.69
Blocks / 85 / 2 / 42.5
Error / 35 / 8 / 4.38
Total / 430 / 14

Using F table (4 degrees of freedom numerator and 8 denominator), p-value is less than .01

Actual p-value = .0005

Because p-value = .05, we reject the null hypothesis that the means of the treatments are equal.

37.Treatment Means:

= 56 = 44

Block Means:

= 46 = 49.5 = 54.5

Overall Mean:

= 300/6 = 50

Step 1

= (50 - 50) 2 + (42 - 50) 2 + · · · + (46 - 50) 2 = 310

Step 2

= 3 [ (56 - 50) 2 + (44 - 50) 2 ] = 216

Step 3

= 2 [ (46 - 50) 2 + (49.5 - 50) 2 + (54.5 - 50) 2 ] = 73

Step 4

SSE = SST - SSTR - SSBL = 310 - 216 - 73 = 21

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 216 / 1 / 216 / 20.57
Blocks / 73 / 2 / 36.5
Error / 21 / 2 / 10.5
Total / 310 / 5

Using F table (1 degree of freedom numerator and 2 denominator), p-value is between .025 and .05

Actual p-value = .0453

Because p-value = .05, we reject the null hypothesis that the mean tune-up times are the same for both analyzers.

39.Treatment Means:

= 16 = 15 = 21

Block Means:

= 18.67 = 19.33 = 15.33 = 14.33 = 19

Overall Mean:

= 260/15 = 17.33

Step 1

= (16 - 17.33) 2 + (16 - 17.33) 2 + · · · + (22 - 17.33) 2 = 175.33

Step 2

= 5 [ (16 - 17.33) 2 + (15 - 17.33) 2 + (21 - 17.33) 2 ] = 103.33

Step 3

= 3 [ (18.67 - 17.33) 2 + (19.33 - 17.33) 2 + · · · + (19 - 17.33) 2 ] = 64.75

Step 4

SSE = SST - SSTR - SSBL = 175.33 - 103.33 - 64.75 = 7.25

Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 100.33 / 2 / 51.67 / 56.78
Blocks / 64.75 / 4 / 16.19
Error / 7.25 / 8 / .91
Total / 175.33 / 14

Using F table (2 degrees of freedom numerator and 8 denominator), p-value is less than .01

Actual p-value = .0000

Because p-value = .05, we reject the null hypothesis that the mean times for the three systems are equal.

Chapter 14:

3.a.


b.Summations needed to compute the slope and y-intercept are:

c.

5.a.

b.Let x = baggage capacity and y = price ($).

There appears to be a linear relationship between x and y.

c.Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part (d) we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion.

d.Summations needed to compute the slope and y-intercept are:

e.A one point increase in the baggage capacity rating will increase the price by approximately $639.

f.

9.a.


b.Summations needed to compute the slope and y-intercept are:

c.

17.The estimated regression equation and the mean for the dependent variable are:

The sum of squares due to error and the total sum of squares are

Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9

r2 = SSR/SST = 5.9/11.2 = .527

We see that 52.7% of the variability in y has been explained by the least squares line.

21.a.The summations needed in this problem are:

b.$7.60

c.The sum of squares due to error and the total sum of squares are:

Thus, SSR = SST - SSE = 5,648,333.33 - 233,333.33 = 5,415,000

r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587

We see that 95.87% of the variability in y has been explained by the estimated regression equation.

d.

14 - 1