286

11: The Analysis of Variance 11.1 In comparing 6 populations, there are 1k − degrees of freedom for treatments and ( )6 10 60n = = . The

ANOVA table is shown below. Source df Treatments 5 Error 54 Total 59

11.2 a Refer to Exercise 11.1. The given sums of squares are inserted and missing entries found by

subtraction. The mean squares are found as MS SS df= . Source df SS MS F Treatments 5 5.2 1.04 3.467 Error 54 16.2 0.30 Total 59 21.4

b The F statistic, MST MSEF = , has 1 5df = and 2 54df = degrees of freedom. c With .05α = and degrees of freedom from b, H0 is rejected if .05 2.37F F> ≈ . d Since 3.467F = falls in the rejection region, the null hypothesis is rejected. There is a difference

among the means. e The critical values of F with 1 5df = and 2 60df ≈ (Table 6) for bounding the p-value for this one-tailed

test are shown below. α .10 .05 .025 .01 .005 Fα 1.95 2.37 2.79 3.34 3.76

Since the observed value 3.467F = falls between .01F and .005F , .005 -value .01p< < and H0 is rejected as in part d. 11.3 Refer to Exercise 11.2.

a 1 .0251

MSE .33.07 1.96 3.07 .33910

x tn

± ⇒ ± ⇒ ±

or 12.731 3.409μ< < .

b ( )

( )

1 2 .0251 2

1 2

1 1MSE

23.07 2.52 1.96 0.310

.55 .480 or .07 1.03

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± < − <

11.4 Similar to Exercise 11.1. With ( )4 6 24n = = and 4k = , the sources of variation and associated df are

shown below. Source df Treatments 3 Error 20 Total 23

11.5 a Refer to Exercise 11.4. The given sums of squares are inserted and missing entries found by

subtraction. The mean squares are found as MS SS df= .

287

Source df SS MS F Treatments 3 339.8 113.267 16.98 Error 20 133.4 6.67 Total 23 473.2

b The F statistic, MST MSEF = , has 1 3df = and 2 20df = degrees of freedom. c With .05α = and degrees of freedom from b, H0 is rejected if .05 3.10F F> = . d Since 16.98F = falls in the rejection region, the null hypothesis is rejected. There is a difference

among the means. e The critical values of F with 1 3df = and 2 20df = (Table 6) for bounding the p-value for this one-tailed

test are shown below. α .10 .05 .025 .01 .005 Fα 2.38 3.10 3.86 4.94 5.82

Since the observed value 16.98F = is greater than .005F , -value .005p < and H0 is rejected as in part d. 11.6 Refer to Exercise 11.5.

a 1 .0251

MSE 6.6788.0 1.725 88.0 1.8196

x tn

± ⇒ ± ⇒ ±

or 186.181 89.819μ< < .

b ( )

( )

1 2 .0251 2

1 2

1 1MSE

288.0 83.9 1.725 6.676

4.1 2.572 or 1.528 6.672

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± < − <

11.7 The following preliminary calculations are necessary: 1 2 314 19 5 38T T T G= = = =

a ( ) ( )2 238

CM 103.14285714

ijx

n

∑= = =

2 2 2 2 2Total SS CM 3 2 2 1 CM 130 103.142857 26.8571ijx= ∑ − = + + + + − = − =

b 2 2 2 214 19 5SST CM CM 117.65 103.142857 14.5071

5 5 4i

i

Tn

= ∑ − = + + − = − =

and SST 14.5071MST 7.25361 2k

= = =−

c By subtraction, SSE Total SS SST 26.8571 14.5071 12.3500= − = − = and the degrees of freedom, by subtraction, are 13 2 11− = . Then

SSE 12.3500MSE 1.122711 11

= = =

d The information obtained in parts a-c is consolidated in an ANOVA table. Source df SS MS Treatments 2 14.5071 7.2536 Error 11 12.3500 1.1227 Total 13 26.8571

e The hypothesis to be tested is 0 1 2 3 aH : versus H : at least one pair of means are differentμ μ μ= =

288

f The rejection region for the test statistic MST 7.2536 6.46MSE 1.1227

F = = = is based on an F-distribution with 2

and 11 degrees of freedom. The critical values of F for bounding the p-value for this one-tailed test are shown below.

α .10 .05 .025 .01 .005 Fα 2.86 3.98 5.26 7.21 8.91

Since the observed value 6.46F = is between .01F and .025F , .01 -value .025p< < and H0 is rejected at the 5% level of significance. There is a difference among the means. 11.8 The hypothesis to be tested is 0 2 3 a 2 3H : versus H :μ μ μ μ= ≠ and the test statistic is

( )2 3

2

2 3

0 3.8 1.25 3.591 11 1 1.12275 4

x xt

sn n

− − −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Notice that the best estimator of 2σ is 2 MSEs = , which is used in the calculation. The rejection region with .05α = and 11 degrees of freedom is .025 2.201t t> = and the null hypothesis is rejected. We conclude that there is a difference between the means.

11.9 a The 90% confidence interval for 1μ is

1 .051

MSE 1.12272.8 1.796 2.8 .855

x tn

± ⇒ ± ⇒ ±

or 11.95 3.65μ< < . b The 90% confidence interval for 1 3μ μ− is

( )

( )

1 3 .051 3

1 3

1 1MSE

1 12.8 1.25 1.796 1.12275 4

1.55 1.28 or .27 2.83

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± < − <

11.10 a The following preliminary calculations are necessary: 1 2 3380 199 261 840T T T G= = = =

( ) ( )2 2840

CM 64,145.454511

ijx

n

∑= = =

2Total SS CM 65, 286 CM 1140.5455ijx= ∑ − = − =

2 2 2 2380 199 261SST CM CM 641.87883

5 3 3i

i

Tn

= ∑ − = + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 2 641.8788 320.939 Error 8 498.6667 62.333 Total 10 1140.5455

b The hypothesis to be tested is 0 1 2 3 aH : versus H : at least one pair of means are differentμ μ μ= =

289

and the F test to detect a difference in mean student response is

MST 5.15MSE

F = = .

The rejection region with .05α = and 2 and 8 df is 4.46F > and H0 is rejected. There is a significant difference in mean response due to the three different methods.

11.11 a The 95% confidence interval for Aμ is

.025MSE 62.33376 2.306 76 8.142

5A Ax t

n± ⇒ ± ⇒ ±

or 67.86 84.14Aμ< < . b The 95% confidence interval for Bμ is

.025MSE 62.33366.33 2.306 66.33 10.51

3B Bx t

n± ⇒ ± ⇒ ±

or 55.82 76.84Bμ< < . c The 95% confidence interval for A Bμ μ− is

( )

( )

.0251 1MSE

1 176 66.33 2.306 62.3335 3

9.667 13.296 or 3.629 22.963

A BA B

A B

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

d Note that these three confidence intervals cannot be jointly valid because all three employ the same value of MSEs = and are dependent.

11.12 a From the computer printout, the test statistic is 5.70F = and the p-value is given on the printout as

P .025= . That is, H0 can be rejected for any value of α greater than .025. The null hypothesis of equality of means is rejected at the 5% (but not at the 1%) level of significance, and we conclude that there is a difference in mean assembly times for the three programs.

b From the printout, MSE 14.9= and the 99% confidence interval for A Bμ μ− is

( )

( )

.0051 1MSE

1 160.5 54.667 3.25 14.94 3

5.83 9.58 or 3.75 15.41

A BA B

A B

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

c The 99% confidence interval for Aμ

.005MSE 14.960.5 3.25 60.5 6.27

4A Ax t

n± ⇒ ± ⇒ ±

or 54.22 66.77Aμ< < . d Since the measurements represent averages of four assembly times and since time itself is a continuous

random variable, the Central Limit Theorem assures us that even for small values of n, the average assembly times will have a fairly mound-shaped distribution.

11.13 a We would be reasonably confident that the data satisfied the normality assumption because each

measurement represents the average of 10 continuous measurements. The Central Limit Theorem assures us that this mean will be approximately normally distributed.

b We have a completely randomized design with four treatments, each containing 6 measurements. The analysis of variance table is given in the Minitab printout. The F test is

290

MST 6.580 57.38MSE .115

F = = =

with p-value = .000 (in the column marked “P”). Since the p-value is very small (less than .01), H0 is rejected. There is a significant difference in the mean leaf length among the four locations with P < .01 or even P < .001.

c The hypothesis to be tested is 0 1 4 a 1 4H : versus H :μ μ μ μ= ≠ and the test statistic is

1 4

1 4

6.0167 3.65 12.091 11 1 .115MSE 6 6

x xt

n n

− −= = =

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The p-value with 20df = is ( )2 12.09P t > is bounded (using Table 4) as ( )-value 2 .005 .01p < = and the null hypothesis is reject. We conclude that there is a difference between the means. d The 99% confidence interval for 1 4μ μ− is

( )

( )

1 4 .0051 4

1 4

1 1MSE

1 16.0167 3.65 2.845 .1156 6

2.367 .557 or 1.810 2.924

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± < − <

e When conducting the t tests, remember that the stated confidence coefficients are based on random sampling. If you looked at the data and only compared the largest and smallest sample means, the randomness assumption would be disturbed.

11.14 a The design is completely randomized with four treatments. The analysis of variance table is given on

the Minitab printout and the test statistic is

MST 2.6120 63.66MSE .0410

F = = =

with p-value = .000. H0 is rejected and the results are declared highly significant. There is a significant difference in the mean dissolved oxygen content for the four locations.

b The 95% confidence interval for 2 3μ μ− is

( )

( )

2 3 .0252 3

2 3

1 1MSE

1 16.44 4.78 2.131 .04105 5

1.66 .273 or 1.387 1.933

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± < − <

11.15 The design is completely randomized with 3 treatments and 5 replications per treatment. The Minitab

printout on the next page shows the analysis of variance for this experiment.

291

One-way ANOVA: Calcium versus Method Source DF SS MS F P Method 2 0.0000041 0.0000021 16.38 0.000 Error 12 0.0000015 0.0000001 Total 14 0.0000056

S = 0.0003545 R-Sq = 73.19% R-Sq(adj) = 68.72%

Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+-------- 1 5 0.027620 0.000421 (------*------) 2 5 0.026780 0.000396 (------*------) 3 5 0.028040 0.000207 (------*------) -+---------+---------+---------+-------- 0.02650 0.02700 0.02750 0.02800 Pooled StDev = 0.000354

The test statistic, 16.38F = with p-value = .000 indicates the results are highly significant; there is a

difference in the mean calcium contents for the three methods. All assumptions appear to have been satisfied.

11.16 a The following preliminary calculations are necessary: 1 2 3 412.55 4.9 10.24 12.62 40.31T T T T G= = = = =

( ) ( )2 240.31

CM 43.9161137

ijx

n

∑= = = 2Total SS CM 51.649 CM 7.73289ijx= ∑ − = − =

2 2 2 2 212.55 4.9 10.24 12.62SST CM CM .92235

14 4 8 11i

i

Tn

= ∑ − = + + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 3 .92235 .30745 Error 33 6.81054 .20638 Total 36 7.73289

The hypothesis to be tested is 0 1 2 3 4 aH : versus H : at least one pair of means are differentμ μ μ μ= = = and the F test to detect a difference in average prices is

MST 1.49MSE

F = = .

The rejection region with .05α = and 3 and 33 df is approximately 2.92F > and H0 is not rejected. There is not enough evidence to indicate a difference in average prices for the four types of tuna.

b The 95% confidence interval for 1 4μ μ− is

( )

( )

1 4 .0251 4

1 4

1 1MSE

1 1.896 1.147 1.96 .2063814 11

.251 .359 or .61 .108

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

− ± − < − <

c The 95% confidence interval for 3 2μ μ− is

292

( )

( )

3 2 .0253 2

3 2

1 1MSE

1 11.28 1.225 1.96 .206388 4

.055 .545 or .49 .60

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± +⎜ ⎟⎝ ⎠

± − < − <

d The researcher might be interested in the difference between white and light tuna! 11.17 a The design is a completely randomized design (four independent samples). b The following preliminary calculations are necessary: 1 2 3 41211 1074 1158 1243 4686T T T T G= = = = =

( ) ( )2 24686

CM 1,097,929.820

ijx

n

∑= = = 2Total SS CM 1,101,862 CM 3932.2ijx= ∑ − = − =

2 2 2 2 21211 1074 1158 1243SST CM CM 3272.2

5 5 5 5i

i

Tn

= ∑ − = + + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 3 3272.2 1090.7333 Error 16 660 41.25 Total 19 3932.2

c The hypothesis to be tested is 0 1 2 3 4 aH : versus H : at least one pair of means are differentμ μ μ μ= = = and the F test to detect a difference in average prices is

MST 26.44MSE

F = = .

The rejection region with .05α = and 3 and 16 df is approximately 3.24F > and H0 is rejected. There is enough evidence to indicate a difference in the average prices for the four states.

11.18 a The design is a completely randomized design (four independent samples). b The following preliminary calculations are necessary: 1 2 3 4514 577 497 567 2155T T T T G= = = = =

( ) ( )2 22155

CM 244, 422.368419

ijx

n

∑= = = 2Total SS CM 248,473 CM 4050.631579ijx= ∑ − = − =

2 2 2 2 2514 577 497 567SST CM CM 1052.6816

5 5 4 5i

i

Tn

= ∑ − = + + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 3 1052.6816 350.8939 Error 15 2997.95 199.8633 Total 18 4050.6316

c The hypothesis to be tested is 0 1 2 3 4 aH : versus H : at least one pair of means are differentμ μ μ μ= = = and the F test to detect a difference in average scores is

293

MST 1.76MSE

F = = .

The rejection region with .05α = and 3 and 15 df is approximately 3.29F > and H0 is not rejected. There is not enough evidence to indicate a difference in the average scores for the four teaching methods.

11.19 Sample means must be independent and based upon samples of equal size. 11.20 Use Tables 11(a) and 11(b). a ( ).05 5,7 5.06q = b ( ).05 3,10 3.88q = c ( ).01 4,8 6.20q = d ( ).01 7,5 9.32q =

11.21 a ( ).05 4,12 4.20 1.8785 5s sq sω = = =

b ( ).01 6,12 6.10 2.15678 8s sq sω = = =

11.22 a ( ).059.12 9.126,18 4.49 6.78

4 4qω = = =

b The ranked means are shown below. A line under two or more means indicates a difference less than ω and hence no differences between that group of means.

4x 2x 1x 5x 3x 6x 92.9 98.4 101.6 104.2 112.3 113.8 11.23 With 4, 20, 6tk df n= = = ,

( ).01MSE .1154,20 5.02 .69

6tq

nω = = =

The ranked means are shown below. 6.0167 5.65 5.35 3.65 1x 2x 3x 4x 11.24 The Minitab printout for pairwise comparisons is reproduced below.

Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Method Individual confidence level = 97.94%

Method = 1 subtracted from: Method Lower Center Upper 2 -0.0014377 -0.0008400 -0.0002423 3 -0.0001777 0.0004200 0.0010177

Method ---------+---------+---------+---------+ 2 (-----*-----) 3 (-----*-----) ---------+---------+---------+---------+ -0.0010 0.0000 0.0010 0.0020

Method = 2 subtracted from: Method Lower Center Upper 3 0.0006623 0.0012600 0.0018577

Method ---------+---------+---------+---------+ 3 (-----*-----) ---------+---------+---------+---------+ -0.0010 0.0000 0.0010 0.0020

294

Minitab adjusts the differences in the sample means, calculating an interval ( )i jx x ω− ±

for all pairs of treatments. If the interval contains zero, the two means are not judged to be significantly different. Hence, methods 1 and 3 are not significantly different. The ranked means are shown below.

.02678 .02762 .02804 2x 1x 3x 11.25 The design is completely randomized with 3 treatments and 5 replications per treatment. The Minitab

printout below shows the analysis of variance for this experiment.

One-way ANOVA: mg/dl versus Lab Source DF SS MS F P Lab 2 42.6 21.3 0.60 0.562 Error 12 422.5 35.2 Total 14 465.0

S = 5.933 R-Sq = 9.15% R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev --+---------+---------+---------+------- 1 5 108.86 7.47 (-------------*--------------) 2 5 105.04 6.01 (--------------*-------------) 3 5 105.60 3.70 (-------------*-------------) --+---------+---------+---------+------- 100.0 104.0 108.0 112.0 Pooled StDev = 5.93

Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Lab Individual confidence level = 97.94%

Lab = 1 subtracted from: Lab Lower Center Upper +---------+---------+---------+--------- 2 -13.824 -3.820 6.184 (--------------*-------------) 3 -13.264 -3.260 6.744 (-------------*--------------) +---------+---------+---------+--------- -14.0 -7.0 0.0 7.0

Lab = 2 subtracted from: Lab Lower Center Upper +---------+---------+---------+--------- 3 -9.444 0.560 10.564 (-------------*-------------) +---------+---------+---------+--------- -14.0 -7.0 0.0 7.0

a The analysis of variance F test for 0 1 2 3H : μ μ μ= = is .60F = with -value .562p = . The results are not significant and H0 is not rejected. There is insufficient evidence to indicate a difference in the treatment means.

b Since the treatment means are not significantly different, there is no need to use Tukey’s test to search for the pairwise differences. Notice that all three intervals generated by Minitab contain zero, indicating that the pairs cannot be judged different.

11.26 With 4, 16, 5tk df n= = = ,

( ).01MSE 41.254,16 5.19 14.91

5tq

nω = = =

The ranked means are shown below. 214.8 231.6 242.2 248.6 2x 3x 1x 4x The second sample mean is different than the other three; means 3 and 4 are different from each other.

295

11.27 a The following preliminary calculations are necessary: 1 2 32835 3300 2724 8859T T T G= = = =

( ) ( )2 28859

CM 5232125.415

ijx

n

∑= = = 2Total SS CM 5,295,693 CM 63567.6ijx= ∑ − = − =

2 2 2 22835 3300 2724SST CM CM 37354.8

5 5 5i

i

Tn

= ∑ − = + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 2 37354.8 18677.4 Error 12 26212.8 2184.4 Total 14 63567.6

The hypothesis to be tested is 0 1 2 3 aH : versus H : at least one pair of means are differentμ μ μ= = and the F test to detect a difference in average scores is

MST 8.55MSE

F = = .

The rejection region with .05α = and 2 and 12 df is approximately 3.89F > and H0 is rejected. There is evidence of a difference in the average scores for the three graduate programs.

b The 95% confidence interval for 1 2μ μ− is

( )1 2 .0251 2

1 2

1 1MSE

2835 3300 1 12.179 2184.45 5 5 5

93 64.41 or 157.41 28.59

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞− ± +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− ± − < − < −

c With 3, 12, 5tk df n= = = ,

( ).05MSE 2184.43,12 3.77 78.80

5tq

nω = = =

The ranked means are shown below. 544.8 567 660 3x 1x 2x There is no significant difference between programs 1 and 3, but programs 1 and 2, 2, and 3 are different

from each other. 11.28 In comparing 3 treatments within 6 blocks, there are 1 2k − = treatment degrees of freedom and

1 5b − = block df. The ANOVA table is shown below. Source df Treatments 2 Blocks 5 Error 10 Total 17

11.29 Refer to Exercise 11.28. The given sums of squares are inserted and missing entries found by subtraction.

The mean squares are found as MS SS df= .

296

Source df SS MS F Treatments 2 11.4 5.70 4.01 Blocks 5 17.1 3.42 2.41 Error 10 14.2 1.42 Total 17 42.7

11.30 To compare the treatment means, the test statistic MST MSE 4.01F = = and the rejection region with 2

and 10 df is .05 4.10F F> = . The null hypothesis is not rejected. There is insufficient evidence to indicate a difference between treatment means.

11.31 The 95% confidence interval for A Bμ μ− is then

( )

( )

.0252MSE

221.9 24.2 2.228 1.426

2.3 1.533 or 3.833 .767

A B

A B

x x tb

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

− ± − < − < −

11.32 To test for differences among block means, the test statistic is MSB MSE 2.41F = = . The critical values

of F from Table 6 with 5 and 10 df are shown below. α .10 .05 .025 .01 .005 Fα 2.52 3.33 4.24 5.64 6.87

Since the observed value 2.41F = is less than .10F , -value .10p > and the null hypothesis is not rejected. There is insufficient evident to indicate differences among block means.

11.33 Use Minitab to obtain an ANOVA printout, or use the following calculations:

( ) ( )2 2113

CM 1064.0833312

ijx

n

∑= = =

2 2 2 2Total SS CM 6 10 14 CM 1213 CM=148.91667ijx= ∑ − = + + + − = −

2 2 2 2 222 34 27 30SST CM CM 25.58333

3 3jT + + += ∑ − = − =

2 2 2 233 25 55SSB CM CM 120.66667

4 4iB + += ∑ − = − = and

SSE Total SS SST SSB 2.6667= − − = Calculate MS SS df= and consolidate the information in an ANOVA table.

Source df SS MS F Treatments 3 25.5833 8.5278 19.19 Blocks 2 120.6667 60.3333 135.75 Error 6 2.6667 0.4444 Total 11 148.9167

a To test the difference among treatment means, the test statistic is

MST 8.528 19.19MSE .4444

F = = =

and the rejection region with .05α = and 3 and 6 df is 4.76F > . There is a significant difference among the treatment means.

b To test the difference among block means, the test statistic is

MSB 60.3333 135.75MSE .4444

F = = =

297

and the rejection region with .05α = and 2 and 6 df is 5.14F > . There is a significant difference among the block means.

c With 4, 6, 3tk df n= = = ,

( ).01MSE .44444,6 7.03 2.71

3tq

nω = = =

The ranked means are shown below. 7.33 9.00 10.00 11.33 1x 3x 4x 2x d The 95% confidence interval is

( )

( )

.0252MSE

27.333 11.333 2.447 .44443

4 1.332 or 5.332 2.668

A B

A B

x x tb

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

− ± − < − < −

e Since there is a significant difference among the block means, blocking has been effective. The variation due to block differences can be isolated using the randomized block design.

11.34 Similar to Exercise 11.33. Use the Minitab printout to analyze the experiment. You should notice that

there are significant differences among treatment means and that there are also significant differences among the block means. Since there is a difference among the block means, blocking has been effective. The variation due to block differences can be isolated using the randomized block design. To determine where the treatment differences lie, use Tukey’s test with

( ).05MSE .0283,8 4.04 .302

5tq

nω = = =

The ranked means are shown below. 2.5 3.52 3.76 1x 3x 2x The significant difference is between treatment A and the other two treatments, B and C. 11.35 a By subtraction, the degrees of freedom for blocks is 1 34 28 6b − = − = . Hence, there are 7b = blocks. b There are always 7b = observations in a treatment total. c There are 4 1 5k = + = observations in a block total. d

Source df SS MS F Treatments 4 14.2 3.55 9.68 Blocks 6 18.9 3.15 8.59 Error 24 8.8 0.3667 Total 34 41.9

e To test the difference among treatment means, the test statistic is

MST 3.55 9.68MSE .3667

F = = =

and the rejection region with .05α = and 4 and 24 df is 2.78F > . There is a significant difference among the treatment means.

f To test the difference among block means, the test statistic is

MSB 3.15 8.59MSE .3667

F = = =

298

and the rejection region with .05α = and 6 and 24 df is 2.51F > . There is a significant difference among the block means.

11.36 Use Minitab to obtain an ANOVA printout, or use the following calculations:

( ) ( )2 2325.2

CM 8812.9212

ijx

n

∑= = =

2Total SS CM 8819.68 CM 6.76ijx= ∑ − = − =

( ) ( ) ( ) ( )2 2 22 106.1 110.9 108.2

SS formulations CM CM 2.8954

jTb

+ += ∑ − = − =

( ) ( ) ( ) ( ) ( )2 2 2 22 79 82.6 82 81.6

SS auto CM CM 2.5203

iBk

+ + += ∑ − = − = and

SSE Total SS SST SSB 1.345= − − = Calculate MS SS df= and consolidate the information in an ANOVA table.

Source df SS MS F Treatments 2 2.895 1.4475 6.46 Blocks 3 2.520 0.8400 3.75 Error 6 1.345 0.2242 Total 11 6.760

a To test the null hypothesis that there is no difference in mean mileage per gallon for the three formulations, the test statistic is

MST 6.46MSE

F = =

The critical values of F from Table 6 with 2 and 6 df are shown below. α .10 .05 .025 .01 .005 Fα 3.46 5.14 7.26 10.92 14.54

Since the observed value 6.46F = is between .025F and .05F , .025 -value .05p< < and the null hypothesis is rejected at the 5% level of significance. There is a significant difference among the treatment means.

b To test the null hypothesis that there is no difference in mean mileage for the four automobiles, the test statistic is

MSB 3.75MSE

F = =

and the p-value with 3 and 6 df is .05 -value .10p< < There is no evidence of a significant difference among the automobiles. c The 90% confidence interval is

( )

( )

.052MSE

226.525 27.725 1.943 .2241674

1.2 .650 or 1.85 .55

A B

A B

x x tb

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

− ± − < − < −

d To determine where the treatment differences lie, use Tukey’s test with

( ).05MSE .2241673,6 4.34 1.027

4tq

nω = = =

The ranked means are shown below. 26.525 27.05 27.725 Ax Cx Bx

299

Only gasoline formulations A and B are significantly different from each other. 11.37 Similar to previous exercises. The Minitab printout for this randomized block experiment is shown below. Two-way ANOVA: Measurements versus Blocks, Chemicals

Source DF SS MS F P Blocks 2 7.1717 3.58583 40.21 0.000 Chemicals 3 5.2000 1.73333 19.44 0.002 Error 6 0.5350 0.08917 Total 11 12.9067 S = 0.2986 R-Sq = 95.85% R-Sq(adj) = 92.40%

Individual 95% CIs For Mean Based on Pooled StDev Blocks Mean +---------+---------+---------+--------- 1 10.875 (----*-----) 2 12.700 (----*-----) 3 12.225 (-----*----) +---------+---------+---------+--------- 10.50 11.20 11.90 12.60

Individual 95% CIs For Mean Based on Pooled StDev Chemicals Mean ------+---------+---------+---------+--- 1 11.4000 (-----*-----) 2 12.3333 (-----*-----) 3 11.2000 (-----*-----) 4 12.8000 (-----*-----) ------+---------+---------+---------+--- 11.20 11.90 12.60 13.30

Both the treatment and block means are significantly different. Since the four chemicals represent the treatments in this experiment, Tukey’s test can be used to determine where the differences lie:

( ).05MSE .089174,6 4.90 .845

3tq

nω = = =

The ranked means are shown below. 11.20 11.40 12.33 12.80 3x 1x 2x 4x The chemical falls into two significantly different groups – A and C versus B and D. 11.38 a Use the properties of additivity to obtain SS. Degrees of freedom are obtained using 40b = and 4k = .

Source df SS MS F Mirrors 3 46.98 15.66 6.96 Drivers 39 328.38 8.42 3.74 Error 117 263.25 2.25 Total 159 638.61

b To test the equality of treatment means, the test statistic is

MST 6.96MSE

F = =

The critical values of F from Table 6 with 3 and 117 ( )120≈ df are shown below. α .10 .05 .025 .01 .005 Fα 2.13 2.68 3.23 3.95 4.50

Since the observed value 6.96F = is greater than .005F , -value .005p < and H0 is rejected. There is a significant difference in mean glare ratings for the four mirrors.

c To test the equality of block means, the test statistic is

MSB 3.74MSE

F = =

The critical values of F from Table 6 with 39 ( )40≈ and 117 ( )120≈ df are shown below.

300

α .10 .05 .025 .01 .005 Fα 1.37 1.50 1.61 1.76 1.87

Since the observed value 3.74F = is greater than .005F , -value .005p < . There is a difference among the drivers.

11.39 The factor of interest is “soil preparation”, and the blocking factor is “locations”. A randomized block

design is used and the analysis of variance table can be obtained using the computer printout. a The F statistic to detect a difference due to soil preparations is

MST 10.06MSE

F = =

with -value .012p = . The null hypothesis can be rejected at the 5% level of significance; there is a significant difference among the treatment means.

b The F statistic to detect a difference due to locations is

MSB 10.88MSE

F = =

with -value .008p = . The null hypothesis can be rejected at the 1% level of significance; there is a highly significant difference among the block means. c Tukey’s test can be used to determine where the differences lie:

( ).01MSE 1.88893,6 6.33 4.35

4tq

nω = = =

The ranked means are shown below. 12.0 12.5 16.0 3x 1x 2x

Preparations 2 and 3 are the only two treatments that can be declared significantly different.

d The 95% confidence interval is

( )

( )

.0252MSE

216.5 12.5 2.447 1.894

3.5 2.38 or 1.12 5.88

B A

B A

x x tb

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± < − <

11.40 A rearrangement of the results gives Dogs

Level of Digitalis 1 2 3 4 A 1342 1140 1029 1150 B 1608 1387 1296 1319 C 1881 1698 1549 1579

The analysis of variance table can be found on the printout. a There are 1 12 3 4 1 6n k b− − + = − − + = degrees of freedom associated with SSE. b The F statistic to detect a difference in mean uptake of calcium for the three levels of digitalis is

MST 258.24MSE

F = =

with -value .000p = . The digitalis levels are very significantly different; at least one of the levels of digitalis causes a different mean uptake of calcium in the heart muscle of dogs. c Tukey’s test can be used to determine where the differences lie:

( ).01MSE 10153,6 6.33 100.83

4tq

nω = = =

301

The ranked means are shown below. 1165.25 1402.50 1676.75 Ax Bx Cx Average responses are different for all three levels of digitalis. d The F statistic to detect a difference in mean uptake in calcium for the four heart muscles is

MSB 56.96MSE

F = =

with -value .000p = . The null hypothesis of no difference should be rejected; there is evidence of a significant difference among the heart muscles. e Tukey’s test can be used to determine where the differences lie:

( ).01MSE 10154,6 7.03 129.3

3tq

nω = = =

The ranked means are shown below. 1291.33 1349.33 1408.33 1610.33 3x 4x 2x 1x

The heart muscle in the first dog is significantly different from the other three dogs. f The standard deviation of the difference between the mean calcium uptake for two levels of digitalis is

2 1 1 2 21015 22.534

s MSEb b b

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

g The 95% confidence interval is

( )

( ) ( )

.0252MSE

1165.25 1402.5 2.447 22.53237.25 55.13 or 292.38 182.12

A B

A B

x x tb

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

− ±

− ± − < − < −

11.41 A randomized block design has been used with “estimators” as treatments and “construction job” as the

block factor. The analysis of variance table is found in the Minitab printout below. Two-way ANOVA: Cost versus Estimator, Job Source DF SS MS F P Estimator 2 10.8617 5.4308 7.20 0.025 Job 3 37.6073 12.5358 16.61 0.003 Error 6 4.5283 0.7547 Total 11 52.9973 S = 0.8687 R-Sq = 91.46% R-Sq(adj) = 84.34%

Individual 95% CIs For Mean Based on Pooled StDev Estimator Mean -------+---------+---------+---------+-- A 32.6125 (--------*--------) B 34.8875 (--------*--------) C 34.1875 (--------*--------) -------+---------+---------+---------+-- 32.4 33.6 34.8 36.0 Both treatments and blocks are significant. The treatment means can be further compared using Tukey’s test with

( ).05MSE .75473,6 4.34 1.885

4tq

nω = = =

The ranked means are shown below. 32.6125 34.1875 34.8875 Ax Cx Bx

302

Estimators A and B show a significant difference in average costs. 11.42 a A randomized block design has been used with “insurance companies” as treatments and “locations” as

the block factor. The analysis of variance table is found in the Minitab printout below. Two-way ANOVA: Cost versus Location, Company Source DF SS MS F P Location 3 4923543 1641181 24.04 0.000 Company 4 2046196 511549 7.49 0.003 Error 12 819230 68269 Total 19 7788969 S = 261.3 R-Sq = 89.48% R-Sq(adj) = 83.35%

b-c There is a significant difference due to companies ( 7.49F = with -value .003p = ) and also due to

location (blocks) ( 24.04F = with -value .000p = ). d The treatment means can be further compared using Tukey’s test with

( ).05MSE 68,2695,12 4.51 589.194

4tq

nω = = =

The ranked means are shown below. 2426.00 2596.50 2731.50 2742.00 3370.75 21st Century AAA Allstate Fireman’s Fund State Farm The companies overlap into groups within which the means are not significantly different. 11.43 a A randomized block design has been used with “stores” as treatments and “items” as the block factor. b The F statistic to detect a difference in mean prices for the five stores is

MST 14.79MSE

F = =

with -value .000p = . The null hypothesis of no difference should be rejected; there is evidence of a significant difference in average prices among the stores.

c The F statistic to detect a difference in mean prices for the 8 items is

MSB 19.39MSE

F = =

with -value .000p = . The null hypothesis of no difference should be rejected; there is evidence of a significant difference in average prices from item to item. That is, blocking has been effective.

d If WinCo has specifically chosen to list the items for which they know their cost is lower than the other stores, the experiment will be biased in their favor. The average cost of items might not be less than the other stores if the items to be priced had been randomly selected.

11.44 a The treatment means can be further compared using Tukey’s test. From Table 11(a), we use a

conservative estimate with 24df = and .05 (5,24) 4.17q = .

b Calculate ( ).05MSE .281654,24 4.17 .782

8tq

nω = = =

c The ranked means are shown below. WinCo Food 4 Less Staters Ralphs Albertsons 2.08 2.741 3.055 3.304 4.041 The average price at WinCo is not significantly lower than Food 4 Less, but it is lower than the other three

stores. 11.45 a-b There are 4 5 20× = treatments and 4 5 3 60× × = total observations.

303

c In a factorial experiment, variation due to the interaction A B× is isolated from SSE. The sources of variation and associated degrees of freedom are given on the next page.

Source df A 3 B 4 A B× 12 Error 40 Total 59

11.46 a The complete ANOVA table is shown below. Since factor A is run at 3 levels, it must have 2 df. Other

entries are found by similar reasoning. Source df SS MS F A 2 5.3 2.6500 1.30 B 3 9.1 3.0333 1.49 A B× 6 4.8 0.8000 0.39 Error 12 24.5 2.0417 Total 23 43.7

b The test statistic is ( )MS AB MSE 0.39F = = and the rejection region is 3.00F > . Hence, H0 is not rejected. There is insufficient evidence to indicate interaction between A and B.

c The test statistic for testing factor A is 1.30F = with .05 3.89F = . The test statistic for factor B is 1.49F = with .05 3.49F = . Neither A nor B are significant.

11.47 Refer to Exercise 11.46. The 95% confidence interval is

( )

( )

1 2 .025

1 2

2MSE

28.3 6.3 2.179 2.1792

2.0 3.11 or 1.11 5.11

x x tr

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± − < − <

11.48 a The nine treatment (cell) totals needed for calculation are shown in the table. Factor A

Factor B 1 2 3 Total 1 12 16 10 38 2 15 25 17 57 3 25 17 27 69 Total 52 58 54 164

2164CM 1492.2222

18= = Total SS 1662 CM 167.7778= − =

2 2 252 58 54SSA CM 3.1111

6+ +

= − = 2 2 238 57 69SSB CM 81.4444

6+ +

= − =

( )2 2 212 16 27SS AB SSA SSB CM 62.2222

2+ + +

= − − − =

Source df SS MS F A 2 3.1111 1.5556 B 2 81.4444 40.7222 A B× 4 62.2222 15.5556 6.67 Error 9 21.0000 2.3333 Total 17 167.7778

304

b-c The test statistic is ( )MS AB MSE 6.67F = = and the rejection region is 3.63F > . There is evidence of a significant interaction. That is, the effect of factor A depends upon the level of factor B at which A is measured.

d Since 6.67F = lies between .01F and .005F , .005 -value .01p< < . e Since the interaction is significant, the differences in the four factor-level combinations should be

explored individually, using an interaction plot such as the one generated by Minitab below.

Factor BFactor B

Mea

n

321

14

13

12

11

10

9

8

7

6

5

Factor A

3

12

Look at the differences between the three levels of factor A when factor B changes from level 1 to level 2.

Levels 2 and 3 behave very similarly while level 1 behaves quite differently. When factor B changes from level 2 to level 3, levels 1 and 3 of factor A behave similarly, and level 2 behaves differently.

11.49 a Similar to Exercise 11.48. Based on the fact that the mean response for the two levels of factor B

behaves very differently depending on the level of factor A under investigation, there is a strong interaction present between factors A and B.

b The test statistic for interaction is ( )MS AB MSE 37.85F = = with -value .000p = from the Minitab printout. There is evidence of a significant interaction. That is, the effect of factor A depends upon the lvel of factor B at which A is measured.

c In light of this type of interaction, the main effect means (averaged over the levels of the other factor) differ only slightly. Hence, a test of the main-effect terms produces a non-significant result.

d No. A significant interaction indicates that the effect of one factor depends upon the level of the other. Each factor-level combination should be investigated individually.

e Answers will vary. 11.50 Use the computing formulas given in this section or a computer software package to generate the ANVOA

table for this 2 3× factorial experiment. The following printout was generated using Minitab. Two-way ANOVA: Percent Gain versus Markup, Location

Source DF SS MS F P Markup 2 835.17 417.583 11.87 0.008 Location 1 280.33 280.333 7.97 0.030 Interaction 2 85.17 42.583 1.21 0.362 Error 6 211.00 35.167 Total 11 1411.67 S = 5.930 R-Sq = 85.05% R-Sq(adj) = 72.60%

a From the printout, 1.21F = with -value .362p = . Hence, at the .05α = level, H0 is not rejected. There is insufficient evidence to indicate interaction. b Since no interaction is found, the effects of A and B can be tested individually. Both A and B are significant.

305

c The interaction plot generated by Minitab is shown on the next page. Notice that the lines, although not exactly parallel, do not indicate a significant difference in the behavior of the mean responses for the two different locations.

MarkupMarkup

Mea

n

321

20

10

0

-10

-20

Location12

d The 95% confidence interval is

( )

( )

31 32 .025

31 32

2MSE

211 217 .5 2.4476 2

16.5 14.51 or 31.01 1.99

x x tr

μ μ

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

⎛ ⎞− + ± ⎜ ⎟⎝ ⎠

− ± − < − < −

11.51 a The total number of participants was sixty, twenty in each of three categories. Hence, the total degrees

of freedom is fifty-nine. Factor T was run at two levels, factor A at three levels, resulting in the given degrees of freedom.

b MST 103.7009 3.66MSE 28.3015

F = = = MSA 760.5889 26.87MSE 28.3015

F = = =

( )MS TA 124.9905 4.42

MSE 28.3015F = = =

c Since interaction is significant, the main effects need not be tested individually. Attention should be focused on the individual cell means.

d The tabled values for the approximate df are shown below. α .10 .05 .025 .01 .005 Fα (1,60) 2.79 4.00 5.29 7.08 8.49 Fα (2,60) 2.39 3.15 3.93 4.98 5.79

For factor T, .005 -value .01p< < . For factor A, -value .005p < , and for the interaction A T× , .01 -value .025p< < .

11.52 Answers will vary from student to student. There is no significant interaction, nor is the main effect for

cities significant. There is a significant difference in the average cost per mile based on the distance traveled, with the cost per mile decreasing as the distance increases. Perhaps a straight line may model the costs as a function of time.

11.53 a The design is a 2 4× factorial experiment with 5r = replications. There are two factors, Gender and

School, one at two levels and one at four levels.

306

b The analysis of variance table can be found using a computer printout or the following calculations: Schools

Gender 1 2 3 4 Total Male 2919 3257 3330 2461 11967 Female 3082 3629 3344 2410 12465 Total 6001 6886 6674 4871 24432

224432CM 14923065.6

40= = Total SS 15281392 CM 358326.4= − =

2 211967 12465SSG CM 6200.120+

= − =

( )2 2 2 26001 6886 6674 4871SS Sc CM 246725.8

10+ + +

= − =

( ) ( )2 2 22919 3257 2410SS G Sc SSG SS Sc CM 10574.9

5+ + +

× = − − − =

Source df SS MS F G 1 6200.1 6200.100 2.09 Sc 3 246725.8 82241.933 27.75 G Sc× 3 10574.9 3524.967 1.19 Error 32 94825.6 2963.300 Total 39 358326.4

c The test statistic is ( )MS GSc MSE 1.19F = = and the rejection region is 2.92F > (with .05α = ). Alternately, you can bound the -value .10p > . Hence, H0 is not rejected. There is insufficient evidence to indicate interaction between gender and schools.

d You can see in the interaction plot that there is a small difference between the average scores for male and female students at schools 1 and 2, but no difference to speak of at the other two schools. The interaction is not significant.

SchoolSchool

Mea

n

4321

750

700

650

600

550

500

Gender12

e The test statistic for testing gender is 2.09F = with .05 4.17F = (or -value .10p > ). The test statistic for

schools is 27.75F = with .05 2.92F = (or -value .005p < ). There is a significant effect due to schools. Using Tukey’s method of paired comparisons with .01α = , calculate

( ).01MSE 2963.34,32 4.80 82.63

10tq

nω = = =

307

The ranked means are shown below.

487.1 600.1 667.4 688.6 4x 1x 3x 2x

11.54 a The experimental units are the supervisors. b The two factors are the training method (trained or untrained) and the situation (standard or

emergency). c There are two levels of each factor. d There are 2 2 4× = treatments. e The design is a 2 2× factorial experiment, with 4 replications per treatment. 11.55 a The analysis of variance table can be found using a computer printout or the following calculations: Training (A)

Situation (B) Trained Not Trained Total Standard 334 185 185 Emergency 296 177 473 Total 630 362 992

2992CM 61504

16= = Total SS 66640 CM 5136= − =

2 2630 362SSA CM 65993 61504 4489

8+

= − = − =

2 2519 473SSB CM 132.25

8+

= − =

( )2 2 2334 296 117SS A B SSA SSB CM 56.25

4+ + +

× = − − − =

Source df SS MS F A 1 4489 4489 117.49 B 1 132.25 132.25 3.46 A B× 1 56.25 56.25 1.47 Error 12 458.5 38.2083 Total 15 5136

b The test statistic is ( )MS A B MSE 1.47F = × = and the rejection region is 4.75F > (with .05α = ). Alternately, you can bound the -value .10p > . Hence, H0 is not rejected. The interaction term is not significant.

c The test statistic is MSB MSE 3.46F = = and the rejection region is 4.75F > (with .05α = ). Alternately, you can bound the .05 -value .10p< < . Hence, H0 is not rejected. Factor B (Situation) is not significant.

d The test statistic is MSA MSE 117.49F = = and the rejection region is 4.75F > (with .05α = ). Alternately, you can bound the -value .005p < . Hence, H0 is rejected. Factor A (Training) is highly significant.

e The interaction plot is shown on the next page. The response is much higher for the supervisors who have been trained. You can see very little change in the response for the two different situations (standard or emergency). The parallel lines indicate that there is no interaction between the two factors.

308

Situation

Mea

n

21

80

70

60

50

40

Training12

11.56 The design is completely randomized with five treatments, containing four, seven, six, five and five

measurements, respectively. The analysis of variance table can be found using the computer printout or the following calculations:

( ) ( )2 220.6

CM 15.71727

ijx

n

∑= = =

2Total SS CM 17.500 CM 1.783ijx= ∑ − = − =

( ) ( ) ( )2 2 22 2.5 4.7 2.4

SST CM CM 1.2124 7 5

i

i

Tn

= ∑ − = + + + − =

SSE Total SS SST .571= − = a The F test is 11.67F = with -value .000p = . The results are highly significant, and H0 is rejected.

There is a difference in mean reaction times due to the five stimuli. b The hypothesis to be tested is 0H : versus H :A D a A Dμ μ μ μ= ≠ and the test statistic is

.625 .920 2.731 11 1 .026MSE 4 5

A D

A D

x xt

n n

− −= = = −

⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

The rejection region with .05α = and 22 degrees of freedom is .025 2.074t t> = and the null hypothesis is rejected. We conclude that there is a difference between the means.

11.57 The intervals provided in the Minitab printout allow you to declare a difference between a pair of means

only when both endpoints have the same sign. Significant differences are observed between treatments A and C, B and C, C and E and D and E. The ranked means are shown below.

E A B D C .48 .62 .67 .92 1.07 11.58 The residuals in the upper tail of the normal probability plot are smaller than expected, but overall, there is

not a problem with normality. The spreads of the residuals when plotted against the fitted values is relatively constant.

11.59 The objective is to determine whether or not mean reaction time differs for the five stimuli. The four

people used in the experiment act as blocks, in an attempt to isolate the variation from person to person. A randomized block design is used, and the analysis of variance table is given in the printout.

a The F statistic to detect a difference due to stimuli is

MST 27.78MSE

F = =

309

with -value .000p = . There is a significant difference in the effect of the five stimuli. b The treatment means can be further compared using Tukey’s test with

( ).05MSE .007085,12 4.51 .190

4tq

nω = = =

The ranked means are shown below. E A B D C .525 .7 .8 1.025 1.05 c The F test for blocks produces 6.59F = with -value .007p = . The block differences are significant;

blocking has been effective. 11.60 A completely randomized design has been used. The analysis of variance table can be found using a

computer program or the following calculations:

( ) ( )2 21161

CM 33,698.02540

ijx

n

∑= = =

2Total SS CM 34,701 CM 1002.975ijx= ∑ − = − =

2 2 2 2 2309 275 295 282SST CM CM 67.475

10 10 10 10i

i

Tn

= ∑ − = + + + − =

Calculate MS SS df= and consolidate the information in an ANOVA table. Source df SS MS Treatments 3 67.475 22.4917 Error 36 935.500 25.9861 Total 39 1002.975

The Minitab computer printout is shown below. One-way ANOVA: 10-19, 20-39, 40-59, 60-69

Source DF SS MS F P Factor 3 67.5 22.5 0.87 0.468 Error 36 935.5 26.0 Total 39 1003.0 S = 5.098 R-Sq = 6.73% R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+------ 10-19 10 30.900 5.195 (------------*------------) 20-39 10 27.500 4.882 (------------*------------) 40-59 10 29.500 4.696 (------------*------------) 60-69 10 28.200 5.574 (------------*------------) ---+---------+---------+---------+------ 25.0 27.5 30.0 32.5 Pooled StDev = 5.098 a The F test for treatments is

MST .87MSE

F = =

with -value .468p = and H0 is not rejected. There is no evidence to suggest a difference among the four groups. b The 90% confidence interval for 1 4μ μ− is

310

( )

( )

1 4 .051 4

1 4

1 1MSE

230.9 28.2 1.645 25.986110

2.7 3.750 or 1.050 6.450

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± − < − <

c The 90% confidence interval for 2μ is

2 .052

MSE 25.986127.5 1.645 27.5 2.65210

x tn

± ⇒ ± ⇒ ±

2or 24.848 30.152μ< < . d With B = 2, 2 MSEσ ≈ and 2t ≈ , the necessary inequality is

MSE2 2 MSE 25.9861 or 25.9861n nn

≤ ⇒ ≥ = ≥

Samples of size 26n = will be required. In this case, the degrees of freedom associated with MSE will be 4 4 100n − = , which is large enough that 2t ≈ is a valid approximation.

11.61 Answers will vary from student to student. A completely randomized design has been used. The analysis

of variance table is shown in the printout. One-way ANOVA: 1, 2, 3, 4

Source DF SS MS F P Factor 3 1385.8 461.9 9.84 0.000 Error 23 1079.4 46.9 Total 26 2465.2 S = 6.851 R-Sq = 56.21% R-Sq(adj) = 50.50%

Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ----+---------+---------+---------+----- 1 6 80.333 8.595 (------*-------) 2 8 91.875 4.912 (-----*-----) 3 5 80.400 4.930 (-------*------) 4 8 73.500 7.964 (-----*-----) ----+---------+---------+---------+----- 72.0 80.0 88.0 96.0 Pooled StDev = 6.851

Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons

Individual confidence level = 98.90% 1 subtracted from:

Lower Center Upper ---------+---------+---------+---------+ 2 1.313 11.542 21.771 (------*------) 3 -11.402 0.067 11.536 (-------*-------) 4 -17.062 -6.833 3.396 (-----*------) ---------+---------+---------+---------+ -15 0 15 30

2 subtracted from: Lower Center Upper ---------+---------+---------+---------+ 3 -22.273 -11.475 -0.677 (------*-------) 4 -27.845 -18.375 -8.905 (------*-----) ---------+---------+---------+---------+ -15 0 15 30

3 subtracted from: Lower Center Upper ---------+---------+---------+---------+ 4 -17.698 -6.900 3.898 (------*-------) ---------+---------+---------+---------+ -15 0 15 30

311

The student should recognize the significant difference in the mean responses for the four training programs, and should further investigate these differences using Tukey’s test with ranked means shown below:

4 1 3 2 73.5 80.33 80.4 91.875 11.62 This is similar to Exercise 11.45. a-b There are 4 2 8× = treatments and 4 2 8r r× × = total observations. c The sources of variation and associated degrees of freedom are given below.

Source df A 3 B 1 A B× 3 Error 8r – 8 Total 8r – 1

11.63 This is similar to previous exercises. The complete ANOVA table is shown below.

Source df SS MS F A 1 1.14 1.14 6.51 B 2 2.58 1.29 7.37 A B× 2 0.49 0.245 1.40 Error 24 4.20 0.175 Total 29 8.41

a The test statistic is ( )MS AB MSE 1.40F = = and the rejection region is 3.40F > . There is insufficient evidence to indicate an interaction.

b Using Table 6 with 1 2df = and 2 24df = , the following critical values are obtained. α .10 .05 .025 .01 .005 Fα 2.54 3.40 4.32 5.61 6.66

The observed value of F is less than .10F , so that -value .10p > . c The test statistic for testing factor A is 6.51F = with .05 4.26F = . There is evidence that factor A

affects the response. d The test statistic for factor B is 7.37F = with .05 3.40F = . Factor B also affects the response. 11.64 Refer to Exercise 11.63. The 95% confidence interval is

( )

( )

1 2 .0251 2

1 2

1 1MSE

23.7 1.4 2.064 .17515

2.3 .315 or 1.985 2.615

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± < − <

11.65 a The experiment is an 2 3× factorial and a two-way analysis of variance is generated. b Using the Minitab printout given in the exercise, the F test for interaction is .45F = with

-value .642p = . There is insufficient evidence to suggest that the effect of temperature is different depending on the type of plant.

c The plot of treatment means for cotton and cucumber as a function of temperature is shown on the next page. The temperature appears to have a quadratic effect on the number of eggs laid in both cotton and cucumber. However, the treatment means are higher overall for the cucumber plants.

312

TempTemp

Mea

n

827770

60

55

50

45

40

35

Plant01

d The 95% confidence interval for Cotton Cucumberμ μ− is

( ) .0251 1MSE

547 760 22.064 12315 15 15

14.2 8.36 or 22.56 5.84

Cotton CucumberCotton Cucumber

Cotton Cucumber

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞− ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− ± − < − < −

11.66 The completely randomized design has been used. The analysis of variance table can be obtained using a

computer program or the computing formulas.

One-way ANOVA: A, B, C, D Source DF SS MS F P Factor 3 0.4649 0.1550 5.20 0.011 Error 16 0.4768 0.0298 Total 19 0.9417

S = 0.1726 R-Sq = 49.37% R-Sq(adj) = 39.87%

Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -+---------+---------+---------+-------- A 5 1.5680 0.1366 (-------*--------) B 5 1.7720 0.2160 (--------*-------) C 5 1.5460 0.1592 (-------*-------) D 5 1.9160 0.1689 (-------*-------) -+---------+---------+---------+-------- 1.40 1.60 1.80 2.00 Pooled StDev = 0.1726

a To test the difference in treatment means, use MST 5.20MSE

F = = with -value .011p = . H0 is rejected at

the 5% level of significance; there is evidence to suggest a difference in mean discharge for the four plants. b The hypothesis to be tested is 0H : 1.5 versus H : 1.5A a Aμ μ= > and the test statistic is

1.568 1.5 .88MSE .0298

5

A A

A

xt

n

μ− −= = =

The rejection region with .05α = and 16 df is .05 1.746t t> = and the null hypothesis is not rejected. We cannot conclude that the limit is exceeded at plant A.

c The 95% confidence interval for A Dμ μ− is

313

( )

( )

.0251 1MSE

21.568 1.916 2.12 .02985

.348 .231 or .579 .117

A DA D

A D

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

− ± − < − < −

11.67 a The design is a randomized block design, with weeks representing blocks and stores as treatments. b The Minitab computer printout is shown below. Two-way ANOVA: Total versus Week, Store

Source DF SS MS F P Week 3 571.71 190.570 8.27 0.003 Store 4 684.64 171.159 7.43 0.003 Error 12 276.38 23.032 Total 19 1532.73 S = 4.799 R-Sq = 81.97% R-Sq(adj) = 71.45% c The F test for treatments is 7.43F = with -value .003p = . The p-value is small enough to allow rejection of H0. There is a significant difference in the average weekly totals for the five supermarkets.

d With 5, 12, 4tk df n= = = ,

( ).05MSE 23.0325,12 4.51 10.82

4tq

nω = = =

The ranked means are shown below. 1 5 4 3 2 240.23 249.19 252.18 254.87 256.99 11.68 Answers will vary from student to student. The students should mention the significance of both block and

treatment effects. There appear to be no violations of the normality and common variance assumptions. Since the treatment means were significantly different, Tukey’s test is used to explore the differences with

( ).05MSE 1.91655,20 4.23 2.39

6tq

nω = = =

The ranked means are shown below. E B A C D 31.20 32.28 34.35 36.30 36.78

11.69 a This is a factorial experiment. The Minitab analysis of variance is shown below. Two-way ANOVA: VO2 versus Gender, Activity Source DF SS MS F P Gender 1 290.510 290.510 205.69 0.000 Activity 2 167.516 83.758 59.30 0.000 Interaction 2 21.501 10.750 7.61 0.004 Error 18 25.422 1.412 Total 23 504.950 S = 1.188 R-Sq = 94.97% R-Sq(adj) = 93.57% b The F-test for interaction is 7.61F = with a p-value of .004. There is evidence of significant interaction

between gender and levels of physical activity. The F-test for gender is 205.69F = with a p-value of .000. There is sufficient evidence to indicate a difference in the average maximum oxygen uptake due to gender.

314

The F-test for levels of physical activity is 59.30F = with a p-value of .000. There is sufficient evidence to indicate a difference in the average maximum oxygen uptake due to levels of physical activity.

c Since the interaction was significant, Tukey’s test is used to explore the differences in the six factor-level means with

( ).05MSE 1.4126,18 4.49 2.67

4tq

nω = ≈ =

The ranked means are shown below. FL FS FM ML MS MM 36.03 38.25 40.17 40.58 45.40 49.35 ____ _____ The males who have higher levels of physical activity have significantly higher average maximum

oxygen uptake. 11.70 a-b This is a 2 3× factorial design with 5r = replications. The Minitab analysis of variance is shown

below. Two-way ANOVA: Salary versus Gender, School Type Source DF SS MS F P Gender 1 153.23 153.228 2.89 0.102 School Type 2 487.25 243.626 4.60 0.020 Interaction 2 42.52 21.261 0.40 0.674 Error 24 1270.48 52.937 Total 29 1953.49 S = 7.276 R-Sq = 34.96% R-Sq(adj) = 21.41% Individual 95% CIs For Mean Based on Pooled StDev Gender Mean ---------+---------+---------+---------+ 1 68.0933 (-----------*----------) 2 63.5733 (----------*----------) ---------+---------+---------+---------+ 63.0 66.5 70.0 73.5 Individual 95% CIs For Mean Based on School Pooled StDev Type Mean -------+---------+---------+---------+-- 1 61.18 (--------*---------) 2 71.01 (--------*---------) 3 65.31 (---------*--------) -------+---------+---------+---------+-- 60.0 65.0 70.0 75.0

b The F-test for interaction is 0.40F = with a p-value of .102. There is no evidence of significant

interaction. The F-test for gender is 2.89F = with a p-value of .102 and the F-test for school type is 4.60F = with a p-value of .020. There is sufficient evidence to indicate a difference in the average salaries

due school type, but not due to gender. c The 95% confidence interval for M Fμ μ− is

( )

( )

.0251 1MSE

268.0933 63.5733 2.064 52.93715

4.52 5.484 or .964 10.004

M FF M

M F

x x tn n

μ μ

⎛ ⎞− ± +⎜ ⎟

⎝ ⎠

⎛ ⎞− ± ⎜ ⎟⎝ ⎠

± − < − <

Since the value 0M Fμ μ− = falls in the interval, there is not enough evidence to indicate a difference in the average salaries for males and females.

d Tukey’s test is used to explore the differences due to school type with

315

( ).01MSE 52.9373,24 4.55 10.47

10tq

nω = ≈ =

The ranked means are shown below. Public Church Private 61.18 65.31 71.01 _____________________ There is a difference in average salary between public and private institutions, but not between the other

two pairs. 11.71 a The design is a completely randomized design with three samples, each having a different number of

measurements. b Use the computing formulas in Section 11.5 or the Minitab printout below.

One-way ANOVA: Iron versus Site Source DF SS MS F P Site 2 132.277 66.139 126.85 0.000 Error 21 10.950 0.521 Total 23 143.227 S = 0.7221 R-Sq = 92.36% R-Sq(adj) = 91.63%

The F test for treatments has a test statistic 126.85F = with p-value = .000. The null hypothesis is rejected

and we conclude that there is a significant difference in the average percentage of iron oxide at the three sites.

c The diagnostic plots are shown below. There appears to be no violation of the normality assumptions; the variances may be unequal, judging by the differing bar widths above and below the center line.

Residual

Perc

ent

210-1-2

99

95

90

80

70

60504030

20

10

5

1

Normal Probability Plot of the Residuals(response is Iron)

Fitted Value

Res

idua

l

7654321

1.0

0.5

0.0

-0.5

-1.0

-1.5

-2.0

-2.5

Residuals Versus the Fitted Values(response is Iron)

11.72 a The experiment is run in a randomized block design, with telephone companies as treatments and cities

as blocks. b Use the computing formulas in Section 11.8 or the Minitab printout below.

Two-way ANOVA: Score versus City, Carrier Source DF SS MS F P City 3 55.688 18.5625 3.88 0.049 Carrier 3 285.688 95.2292 19.90 0.000 Error 9 43.063 4.7847 Total 15 384.438 S = 2.187 R-Sq = 88.80% R-Sq(adj) = 81.33%

c The F test for treatments (carriers) has a test statistic 19.90F = with p-value = .000. The null

hypothesis is rejected and we conclude that there is a significant difference in the average satisfaction scores for the four carriers.

316

d The F test for blocks (cities) has a test statistic 3.88F = with p-value = .049. The null hypothesis is rejected and we conclude that there is a significant difference in the average satisfaction scores for the four cities.

11.73 There is no evidence of non-normality. There may be a slightly larger error variation for the smaller values

compared to the larger values of y. 11.74 a The experiment is a 2 3× factorial experiment, with two factors (rank and gender). There are r = 10

replications per factor-level combination. b Use the computing formulas in Section 11.10 or the Minitab printout below. Two-way ANOVA: Salary versus Gender, Rank Source DF SS MS F P Gender 1 780.5 780.5 58.57 0.000 Rank 2 20014.1 10007.1 751.02 0.000 Interaction 2 74.0 37.0 2.78 0.071 Error 54 719.5 13.3 Total 59 21588.1 S = 3.650 R-Sq = 96.67% R-Sq(adj) = 96.36%

c The F test for interaction has a test statistic 2.78F = with p-value = .071. The null hypothesis is not

rejected and we conclude that there is no significant interaction between rank and gender. d The F test for rank has a test statistic 751.02F = with p-value = .000, and the F test for gender has a

test statistic 58.57F = with p-value = .000. Both factors are highly significant. We conclude that there is a difference in average salary due to both gender and rank.

e The interaction plot is shown below. Notice the differences in salary due to both rank and gender.

MaleFemale

110

100

90

80

70

60

Gender

Mea

n

AssistantAssociateFull

Rank

Interaction Plot for SalaryData Means

Using Tukey’s test is used to explore the differences with

( ).01MSE 13.33,54 4.37 3.56

20tq

nω = ≈ =

The ranked means are shown below. All three of the ranks have significantly different average salaries. Assistant Associate Full 62.075 72.485 104.96

Case Study: “A Fine Mess” 1 The design is a two-way classification, with type of ticket as the treatment and cities as blocks. 2 The Minitab printout for the randomized block design is shown below.

Two-way ANOVA: Fine versus City, Type Source DF SS MS F P

317

City 12 2122.10 176.84 1.08 0.420 Type 2 3015.74 1507.87 9.18 0.001 Error 24 3943.59 164.32 Total 38 9081.44 S = 12.82 R-Sq = 56.58% R-Sq(adj) = 31.24%

Individual 95% CIs For Mean Based on Pooled StDev Type Mean -----+---------+---------+---------+---- 1 21.8462 (------*------) 2 35.2308 (------*-------) 3 43.1538 (------*------) -----+---------+---------+---------+---- 20 30 40 50

Students should notice the significant difference in average ticket prices for the three types of tickets, but not from city to city. It does not appear that blocking has been effective. To explore the differences between the three types of tickets, use Tukey’s procedure with

( ).05MSE 164.323,24 3.53 12.55

13tq

nω = = =

The ranked means are shown below. Overtime Red Zone Hydrant 21.85 35.23 43.16 The red zone and fire hydrant ticket amounts are not significantly different, but the amount for overtime parking appears to be significantly less than the other two types. 3 Answers will vary, but should summarize the above results.