Chapter 13 Analysis of Variance and Experimental Design

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Chapter 3 Analyss of Varance and Expermental Desgn Learnng Obectves. Understand how the analyss of varance procedure can be used to determne f the means of more than two populatons are equal.. Know the assumptons necessary to use the analyss of varance procedure. 3. Understand the use of the F dstrbuton n performng the analyss of varance procedure. 4. Know how to set up an ANOVA table and nterpret the entres n the table. 5. Be able to use output from computer software pacages to solve analyss of varance problems. 6. Know how to use Fsher s least sgnfcant dfference (LSD) procedure and Fsher s LSD wth the Bonferron adustment to conduct statstcal comparsons between pars of populatons means. 7. Understand the dfference between a completely randomzed desgn, a randomzed bloc desgn, and factoral experments. 8. Know the defnton of the followng terms: comparsonwse Type I error rate expermentwse Type I error rate factor level treatment replcaton parttonng blocng man effect nteracton 3 -

Chapter 3 Solutons:. a. x = (30 + 45 + 36)/3 = 37 = SSTR = n = 5(30-37) + 5(45-37) + 5(36-37) = 570 MSTR = SSTR /( - ) = 570/ = 85 b. n s = 4(6) + 4(4) + 4(6.5) = 66 = SSE = ( ) MSE = SSE /(n T - ) = 66/(5-3) = 5.5 c. F = MSTR /MSE = 85/5.5 = 5.8 F.05 = 3.89 ( degrees of freedom numerator and denomnator) Snce F = 5.8 > F.05 = 3.89, we reect the null hypothess that the means of the three populatons are equal. d. Treatments 570 85 5.8 Error 66 5.5 Total 636 4. a. x = (53 + 69 + 58)/3 = 60 = SSTR = n = 4(53-60) + 4(69-60) + 4(58-60) = 536 MSTR = SSTR /( - ) = 536/ = 68 b. n s = 3(96.67) + 3(97.33) +3(8.00) = 88.00 = SSE = ( ) MSE = SSE /(n T - ) = 88.00 /( - 3) = 9.00 c. F = MSTR /MSE = 68/9 =.9 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F =.9 < F.05 = 4.6, we cannot reect the null hypothess. d. Treatments 536 68.9 Error 88 9 9 Total 364 3 -

Analyss of Varance and Expermental Desgn 3. a. 4(00) + 6(85) + 5(79) x = = 87 5 = SSTR = n = 4(00-87) + 6(85-87) + 5(79-87) =,00 MSTR = SSB /( - ) =,00/ = 50 b. n s = 3(35.33) + 5(35.60) + 4(43.50) = 458 = SSE = ( ) MSE = SSE /(n T - ) = 458/(5-3) = 38.7 c. F = MSTR /MSE = 50/38.7 = 3.36 F.05 = 3.89 ( degrees of freedom numerator and denomnator) Snce F = 3.36 > F.05 = 3.89 we reect the null hypothess that the means of the three populatons are equal. d. 4. a. Treatments 00 50 3.36 Error 458 38.7 Total 478 4 Treatments 00 3 400 80 Error 300 60 5 Total 500 63 b. F.05 =.76 (3 degrees of freedom numerator and 60 denomnator) Snce F = 80 > F.05 =.76 we reect the null hypothess that the means of the 4 populatons are equal. 5. a. Treatments 0 60 0 Error 6 7 3 Total 336 74 b. F.05 = 3.5 ( numerator degrees of freedom and 60 denomnator) F.05 = 3.07 ( numerator degrees of freedom and 0 denomnator) The crtcal value s between 3.07 and 3.5 Snce F = 0 must exceed the crtcal value, no matter what ts actual value, we reect the null hypothess that the 3 populaton means are equal. 3-3

Chapter 3 6. Manufacturer Manufacturer Manufacturer 3 Sample Mean 3 8 Sample Varance 6.67 4.67 3.33 x = (3 + 8 + )/3 = 4 = SSTR = n = 4(3-4) + 4(8-4) + 4( - 4) = 04 MSTR = SSTR /( - ) = 04/ = 5 n s = 3(6.67) + 3(4.67) + 3(3.33) = 44.0 = SSE = ( ) MSE = SSE /(n T - ) = 44.0/( - 3) = 4.89 F = MSTR /MSE = 5/4.89 = 0.63 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F = 0.63 > F.05 = 4.6 we reect the null hypothess that the mean tme needed to mx a batch of materal s the same for each manufacturer. 7. Superor Peer Subordnate Sample Mean 5.75 5.5 5.5 Sample Varance.64.00.93 x = (5.75 + 5.5 + 5.5)/3 = 5.5 = SSTR = n = 8(5.75-5.5) + 8(5.5-5.5) + 8(5.5-5.5) = MSTR = SSTR /( - ) = / =.5 n s = 7(.64) + 7(.00) + 7(.93) = 38.99 = SSE = ( ) MSE = SSE /(n T - ) = 38.99/ =.86 F = MSTR /MSE = 0.5/.86 = 0.7 F.05 = 3.47 ( degrees of freedom numerator and denomnator) Snce F = 0.7 < F.05 = 3.47, we cannot reect the null hypothess that the means of the three populatons are equal; thus, the source of nformaton does not sgnfcantly affect the dssemnaton of the nformaton. 3-4

Analyss of Varance and Expermental Desgn 8. Maretng Managers Maretng Research Advertsng Sample Mean 5 4.5 6 Sample Varance.8.3.4 x = (5 + 4.5 + 6)/3 = 5.7 = SSTR = n = 6(5-5.7) + 6(4.5-5.7) + 6(6-5.7) = 7.00 MSTR = SSTR /( - ) = 7.00/ = 3.5 n s = 5(.8) + 5(.3) + 5(.4) = 7.50 = SSE = ( ) MSE = SSE /(n T - ) = 7.50/(8-3) =.5 F = MSTR /MSE = 3.5/.50 = 7.00 F.05 = 3.68 ( degrees of freedom numerator and 5 denomnator) Snce F = 7.00 > F.05 = 3.68, we reect the null hypothess that the mean percepton score s the same for the three groups of specalsts. 9. Real Estate Agent Archtect Stocbroer Sample Mean 67.73 6.3 65.80 Sample Varance 7.7 80.0 37. x = (67.73 + 6.3 + 65.80)/3 = 64.89 = SSTR = n = 5(67.73-64.89) + 5(6.3-64.89) + 5(65.80-64.89) = 345.47 MSTR = SSTR /( - ) = 345.47/ = 7.74 n s = 4(7.7) + 4(80.0) + 4(37.) = 6089.6 = SSE = ( ) MSE = SSE /(n T - ) = 6089.6/(45-3) = 44.98 F = MSTR /MSE = 7.74/44.98 =.9 F.05 = 3. ( degrees of freedom numerator and 4 denomnator) Note: Table 4 does not show a value for degrees of freedom numerator and 4 denomnator. However, the value of 3.3 correspondng to degrees of freedom numerator and 40 denomnator can be used as an approxmaton. 3-5

Chapter 3 Snce F =.9 < F.05 = 3.3, we cannot reect the null hypothess that the ob stress ratngs are the same for the three occupatons. 0. The Mntab output s shown below: ANALYSIS OF VARIANCE ON P/E SOURCE DF SS MS F p Industry 40.8 0.4 0.94 0.403 ERROR 6 563.8.7 TOTAL 8 604.6 INDIVIDUAL 95 PCT CI'S FOR MEAN BASED ON POOLED STDEV LEVEL N MEAN STDEV ---------+---------+---------+------- 5.50 5.463 (--------*--------) 7 8.86 4.07 (-----------*-----------) 3 0 6.300 3.889 (---------*---------) ---------+---------+---------+------- POOLED STDEV = 4.657 5.0 8.0.0 Snce the p-value = 0.403 > α = 0.05, we cannot reect the null hypothess that that the mean prce/earnngs rato s the same for these three groups of frms.. a LSD= tα / MSE + = t.05 5.5 + =.79. = 3.3 n n 5 5 x x = 30 45= 5> LSD; sgnfcant dfference x x3 = 30 36 = 6> LSD; sgnfcant dfference x x3 = 45 36 = 9> LSD; sgnfcant dfference x x ± t MSE + n n b. α / (30 45).79 5.5 ± + n n -5 ± 3.3 = -8.3 to -.77. a. Sample Sample Sample 3 Sample Mean 5 77 58 Sample Varance 96.67 97.34 8.99 x = (5 + 77 + 58)/3 = 6 = SSTR = n = 4(5-6) +4(77-6) + 4(58-6) =,448 3-6

Analyss of Varance and Expermental Desgn MSTR = SSTR /( - ) =,448/ = 74 n s = 3(96.67) + 3(97.34) + 3(8.99) = 88 = SSE = ( ) MSE = SSE /(n T - ) = 88/( - 3) = 9 F = MSTR /MSE = 74/9 = 7.87 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F = 7.87 > F.05 = 4.6, we reect the null hypothess that the means of the three populatons are equal. LSD= t MSE + = t 9 + =.6 46 = 5.34 4 4 b. α /.05 n n x x = 5 77 = 6 >LSD; sgnfcant dfference x x3 = 5 58 = 7<LSD; no sgnfcant dfference x x3 = 77 58= 9 > LSD; sgnfcant dfference LSD = t MSE + = t 4.89 + =.6.45 = 3.54 4 4 3. α /.05 n n3 Snce x x3 = 3 = < 3.54, there does not appear to be any sgnfcant dfference between the means of populaton and populaton 3. 4. x x ± LSD 3-8 ± 3.54-5 ± 3.54 = -8.54 to -.46 5. Snce there are only 3 possble parwse comparsons we wll use the Bonferron adustment. α =.05/3 =.07 t.07/ = t.0085 whch s approxmately t.0 =.60 BSD=.60 MSE + =.60.5 + =.06 n n 6 6 x x = 5 4.5 =.5 <.06; no sgnfcant dfference x x3= 5 6= <.06; no sgnfcant dfference 3-7

Chapter 3 x x3 = 4.5 6 =.5 >.06; sgnfcant dfference 6. a. Machne Machne Machne 3 Machne 4 Sample Mean 7. 9. 9.9.4 Sample Varance..93.70.0 x = (7. + 9. + 9.9 +.4)/4 = 9.38 = SSTR = n = 6(7. - 9.38) + 6(9. - 9.38) + 6(9.9-9.38) + 6(.4-9.38) = 57.77 MSTR = SSTR /( - ) = 57.77/3 = 9.6 n s = 5(.) + 5(.93) + 5(.70) + 5(.0) = 9.30 = SSE = ( ) MSE = SSE /(n T - ) = 9.30/(4-4) =.97 F = MSTR /MSE = 9.6/.97 = 9.86 F.05 = 3.0 (3 degrees of freedom numerator and 0 denomnator) Snce F = 9.86 > F.05 = 3.0, we reect the null hypothess that the mean tme between breadowns s the same for the four machnes. b. Note: t α/ s based upon 0 degrees of freedom LSD = tα / MSE + = t.05 0.97 + =.086.333=.9 n n 6 6 x x4 = 9..4 =.3 >LSD; sgnfcant dfference 7. C = 6 [(,), (,3), (,4), (,3), (,4), (3,4)] α =.05/6 =.008 and α / =.004 Snce the smallest value for α / n the t table s.005, we wll use t.005 =.845 as an approxmaton for t.004 (0 degrees of freedom) BSD =.845 0.97 + =.6 6 6 Thus, f the absolute value of the dfference between any two sample means exceeds.6, there s suffcent evdence to reect the hypothess that the correspondng populaton means are equal. Means (,) (,3) (,4) (,3) (,4) (3,4) Dfference.8 4.3 0.8.3.5 Sgnfcant? Yes Yes Yes No Yes No 3-8

Analyss of Varance and Expermental Desgn 8. n = n = 8 n 3 = 0 t α/ s based upon 7 degrees of freedom Comparng and LSD = t.05 3 + =.05.7083 = 3.38 8 9.95 4.75 = 4.8 >LSD; sgnfcant dfference Comparng and 3 LSD =.05 3 + =.05.3833 = 3.7 0 9.95-3.5 = 3.55 > LSD; sgnfcant dfference Comparng and 3 LSD =.05 3 + =.05.950 = 3.5 8 0 4.75-3.5 =.5 < LSD; no sgnfcant dfference Usng the Bonferron adustment wth C = 3 α =.05 /3 =.067 α / =.0083 or approxmately.0 t.0 =.473 Means (,) (,3) (,3) BSD 4.07 3.8 4.3 Dfference 4.8 3.55 3.5 Sgnfcant? Yes No No Thus, usng the Bonferron adustment, the dfference between the mean prce/earnngs ratos of banng frms and fnancal servces frms s sgnfcant. 9. a. x = (56 + 4 + 34)/3 = 44 = SSTR = n = 6(56-44) + 6(4-44) + 6(34-44) =,488 b. MSTR = SSTR /( - ) = 488/ = 744 c. s = 64.4 s = 3. s = 0.4 3 3-9

Chapter 3 n s = 5(64.4) + 5(3.) +5(0.4) = 030 = SSE = ( ) d. MSE = SSE /(n T - ) = 030/( - 3) = 35.3 e. F = MSTR /MSE = 744/35.3 = 5.50 F.05 = 3.68 ( degrees of freedom numerator and 5 denomnator) Snce F = 5.50 > F.05 = 3.68, we reect the hypothess that the means for the three treatments are equal. 0. a. Treatments 488 744 5.50 Error 030 5 35.3 Total 358 7 LSD= t MSE + =.3 35.3 + = 4.3 n n 6 6 b. α / 56-4 = 4 < 4.3; no sgnfcant dfference 56-34 = > 4.3; sgnfcant dfference 4-34 = 8 < 4.3; no sgnfcant dfference. Treatments 300 4 75 4.07 Error 60 30 5.33 Total 460 34. a. H 0 : u = u = u 3 = u 4 = u 5 H a : Not all the populaton means are equal b. F.05 =.69 (4 degrees of freedom numerator and 30 denomnator) Snce F = 4.07 >.69 we reect H 0 3. Treatments 50 75 4.80 Error 50 6 5.63 Total 400 8 F.05 = 3.63 ( degrees of freedom numerator and 6 denomnator) Snce F = 4.80 > F.05 = 3.63, we reect the null hypothess that the means of the three treatments are equal. 4. 3-0

Analyss of Varance and Expermental Desgn Treatments 00 600 43.99 Error 600 44 3.64 Total 800 46 F.05 = 3.3 ( degrees of freedom numerator and 40 denomnator) F.05 = 3.5 ( degrees of freedom numerator and 60 denomnator) The crtcal F value s between 3.5 and 3.3. Snce F = 43.99 exceeds the crtcal value, we reect the hypothess that the treatment means are equal. 5. A B C Sample Mean 9 07 00 Sample Varance 46.89 96.43 73.78 8(9) + 0(07) + 0(00) x = = 07.93 8 = SSTR = n = 8(9-07.93) + 0(07-07.93) + 0(00-07.93) = 67.9 MSTR = SSTR /( - ) = 67.9 / = 809.95 n s = 7(46.86) + 9(96.44) + 9(73.78) = 3,460 = SSE = ( ) MSE = SSE /(n T - ) = 3,460 /(8-3) = 38.4 F = MSTR /MSE = 809.95 /38.4 = 5.85 F.05 = 3.39 ( degrees of freedom numerator and 5 denomnator) Snce F = 5.85 > F.05 = 3.39, we reect the null hypothess that the means of the three treatments are equal. 6. a. Treatments 4560 80 9.87 Error 640 7 3. Total 0800 9 b. F.05 = 3.35 ( degrees of freedom numerator and 7 denomnator) Snce F = 9.87 > F.05 = 3.35, we reect the null hypothess that the means of the three assembly methods are equal. 7. 3 -

Chapter 3 Between 6.64 3 0.55 7.56 Error 3.4 0.7 Total 85.05 3 F.05 = 3.0 (3 degrees of freedom numerator and 0 denomnator) Snce F = 7.56 > F.05 = 3.0, we reect the null hypothess that the mean breang strength of the four cables s the same. 8. 50 60 70 Sample Mean 33 9 8 Sample Varance 3 7.5 9.5 x = (33 + 9 + 8)/3 = 30 = SSTR = n = 5(33-30) + 5(9-30) + 5(8-30) = 70 MSTR = SSTR /( - ) = 70 / = 35 n s = 4(3) + 4(7.5) + 4(9.5) = 36 = SSE = ( ) MSE = SSE /(n T - ) = 36 /(5-3) = 9.67 F = MSTR /MSE = 35 /9.67 =.78 F.05 = 3.89 ( degrees of freedom numerator and denomnator) Snce F =.78 < F.05 = 3.89, we cannot reect the null hypothess that the mean yelds for the three temperatures are equal. 9. Drect Experence Indrect Experence Combnaton Sample Mean 7.0 0.4 5.0 Sample Varance 5.0 6.6 4.0 x = (7 + 0.4 + 5)/3 = 0.8 = SSTR = n = 7(7-0.8) + 7(0.4-0.8) + 7(5-0.8) = 5.68 MSTR = SSTR /( - ) = 5.68 / =.84 n s = 6(5.0) + 6(6.6) + 6(4.0) = 9.68 = SSE = ( ) 3 -

Analyss of Varance and Expermental Desgn MSE = SSE /(n T - ) = 9.68 /( - 3) = 5.09 F = MSTR /MSE =.84 /5.09 =.7 F.05 = 3.55 ( degrees of freedom numerator and 8 denomnator) Snce F =.7 > F.05 = 3.55, we reect the null hypothess that the means for the three groups are equal. 30. Pant Pant Pant 3 Pant 4 Sample Mean 3.3 39 36 44 Sample Varance 47.5.50 54.5 x = (33 + 39 + 36 + 44)/3 = 38 = SSTR = n = 5(33-38) + 5(39-38) + 5(36-38) + 5(44-38) = 330 MSTR = SSTR /( - ) = 330 /3 = 0 n s = 4(47.5) + 4(50) + 4() + 4(54.5) = 69 = SSE = ( ) MSE = SSE /(n T - ) = 69 /(0-4) = 43.5 F = MSTR /MSE = 0 /43.5 =.54 F.05 = 3.4 (3 degrees of freedom numerator and 6 denomnator) Snce F =.54 < F.05 = 3.4, we cannot reect the null hypothess that the mean dryng tmes for the four pants are equal. 3. A B C Sample Mean 0 5 Sample Varance 5.5 x = (0 + + 5)/3 = = SSTR = n = 5(0 - ) + 5( - ) + 5(5 - ) = 70 MSTR = SSTR /( - ) = 70 / = 35 n s = 4() + 4(.5) + 4(.5) = 4 = SSE = ( ) MSE = SSE /(n T - ) = 4 /(5-3) = 3-3

Chapter 3 F = MSTR /MSE = 35 / = 7.5 F.05 = 3.89 ( degrees of freedom numerator and denomnator) Snce F = 7.5 > F.05 = 3.89, we reect the null hypothess that the mean mles per gallon ratngs are the same for the three automobles. 3. Note: degrees of freedom for t α/ are 8 LSD = tα / MSE + = t.05 5.09 + =.0.4543 =.53 n n 7 7 x x = 7.0 0.4 = 3.4 >.53; sgnfcant dfference x x3 = 7.0 5.0 = 8>.53; sgnfcant dfference x x3 = 0.4 5 = 4.6 >.53; sgnfcant dfference 33. Note: degrees of freedom for t α/ are LSD = tα / MSE + = t.05 + =.79.8 =.95 n n 5 5 x x = 0 = <.95; no sgnfcant dfference x x3 = 0 5 = 5>.95; sgnfcant dfference x x3 = 5 = 4>.95; sgnfcant dfference 34. Treatment Means: x = 3.6 x =.0 x = 0.6 3 Bloc Means: x = 9 x = 7.67 x = 5.67 x = 8.67 x = 7.67 3 4 5 Overall Mean: x = 76/5 =.73 Step ( x ) x SST = = (0 -.73) + (9 -.73) + + (8 -.73) = 354.93 3-4

Analyss of Varance and Expermental Desgn Step SSTR = b = 5 [ (3.6 -.73) + (.0 -.73) + (0.6 -.73) ] = 6.53 Step 3 SSBL = = 3 [ (9 -.73) + (7.67 -.73) + (5.67 -.73) + Step 4 (8.67 -.73) + (7.67 -.73) ] = 3.3 SSE = SST - SSTR - SSBL = 354.93-6.53-3.3 = 6.08 Treatments 6.53 3.7 6.60 Blocs 3.3 4 78.08 Error 6.08 8.0 Total 354.93 4 F.05 = 4.46 ( degrees of freedom numerator and 8 denomnator) Snce F = 6.60 > F.05 = 4.46, we reect the null hypothess that the means of the three treatments are equal. 35. Treatments 30 4 77.5 7.69 Blocs 85 4.5 Error 35 8 4.38 Total 430 4 F.05 = 3.84 (4 degrees of freedom numerator and 8 denomnator) Snce F = 7.69 > F.05 = 3.84, we reect the null hypothess that the means of the treatments are equal. 36. Treatments 900 3 300.60 Blocs 400 7 57.4 Error 500 3.8 Total 800 3 F.05 = 3.07 (3 degrees of freedom numerator and denomnator) Snce F =.60 > F.05 = 3.07, we reect the null hypothess that the means of the treatments are equal. 37. Treatment Means: x = 56 x = 44 3-5

Chapter 3 Bloc Means: x = 46 x = 49.5 x = 54.5 3 Overall Mean: x = 300/6 = 50 Step ( x ) x SST = = (50-50) + (4-50) + + (46-50) = 30 Step SSTR = b = 3 [ (56-50) + (44-50) ] = 6 Step 3 SSBL = = [ (46-50) + (49.5-50) + (54.5-50) ] = 73 Step 4 SSE = SST - SSTR - SSBL = 30-6 - 73 = Treatments 6 6 0.57 Blocs 73 36.5 Error 0.5 Total 30 5 F.05 = 8.5 ( degree of freedom numerator and denomnator) Snce F = 0.57 > F.05 = 8.5, we reect the null hypothess that the mean tuneup tmes are the same for both analyzers. 38. Treatments 45 4.5 7. Blocs 36 3 Error 9.58 Total 00 9 F.05 = 3.6 (4 degrees of freedom numerator and denomnator) Snce F = 7. > F.05 = 3.6, we reect the null hypothess that the mean total audt tmes for the fve audtng procedures are equal. 39. Treatment Means: 3-6

Analyss of Varance and Expermental Desgn x = 6 x = 5 x = 3 Bloc Means: x = 8.67 x = 9.33 x = 5.33 x = 4.33 x = 9 3 4 5 Overall Mean: x = 60/5 = 7.33 Step ( x ) x SST = = (6-7.33) + (6-7.33) + + ( - 7.33) = 75.33 Step SSTR = b = 5 [ (6-7.33) + (5-7.33) + ( - 7.33) ] = 03.33 Step 3 SSBL = = 3 [ (8.67-7.33) + (9.33-7.33) + + (9-7.33) ] = 64.75 Step 4 SSE = SST - SSTR - SSBL = 75.33-03.33-64.75 = 7.5 Treatments 00.33 5.67 56.78 Blocs 64.75 4 6.9 Error 7.5 8.9 Total 75.33 4 F.05 = 4.46 ( degrees of freedom numerator and 8 denomnator) Snce F = 56.78 > F.05 = 4.46, we reect the null hypothess that the mean tmes for the three systems are equal. 40. The Mntab output for these data s shown below: ANALYSIS OF VARIANCE BPM SOURCE DF SS MS Bloc 9 796 3 Treat 3 9805 660 3-7

Chapter 3 ERROR 7 7949 94 TOTAL 39 30550 Indvdual 95% CI Treat Mean ----+---------+---------+---------+------- 78.0 (-----*-----) 7.0 (-----*----) 3 75.0 (-----*----) 4 3.6 (-----*----) ----+---------+---------+---------+------- 0.0 40.0 60.0 80.0 F.05 =.96 (3 degrees of freedom numerator and 7 denomnator) Snce F = 660 /94 =.46 >.96, we reect the null hypotheses that the mean heart rate for the four methods are equal. 4. Factor B Factor A Level Level Level 3 Means Factor A Level x = 50 x = 78 x = 84 3 x = 04 Level x = 0 x = 6 x = 8 3 x = 8 Factor B Means x = 30 x = 97 x 3 = 06 x = Step ( x ) x SST = = (35 - ) + (65 - ) + + (36 - ) = 9,08 Step SSA = br = 3 () [ (04 - ) + (8 - ) ] = 588 Step 3 SSB = ar = () [ (30 - ) + (97 - ) + (06 - ) ] =,38 Step 4 3-8

Analyss of Varance and Expermental Desgn = [ (50-04 - 30 + ) + (78-04 - 97 + ) + SSAB Step 5 = r x x x + x + (8-8 - 06 + ) ] = 4,39 SSE = SST - SSA - SSB - SSAB = 9,08-588 -,38-4,39 =,70 Factor A 588 588.05 Factor B 38 64 4.06 Interacton 439 96 7.66 Error 70 6 86.67 Total 908 F.05 = 5.99 ( degree of freedom numerator and 6 denomnator) F.05 = 5.4 ( degrees of freedom numerator and 6 denomnator) Snce F =.05 < F.05 = 5.99, Factor A s not sgnfcant. Snce F = 4.06 < F.05 = 5.4, Factor B s not sgnfcant. Snce F = 7.66 > F.05 = 5.4, Interacton s sgnfcant. 4. Factor A 6 3 8.67 3.7 Factor B 3.50 4.94 Interacton 75 6 9.7.5 Error 56 4.33 Total 80 35 F.05 = 3.0 (3 degrees of freedom numerator and 4 denomnator) Snce F = 3.7 > F.05 = 3.0, Factor A s sgnfcant. F.05 = 3.40 ( degrees of freedom numerator and 4 denomnator) Snce F = 4.94 > F.05 = 3.40, Factor B s sgnfcant. F.05 =.5 (6 degrees of freedom numerator and 4 denomnator) Snce F =.5 > F.05 =.5, Interacton s sgnfcant 43. 3-9

Chapter 3 Factor B Factor B Small Large Means A x = 0 x = 0 x = 0 Factor A B x = 8 x = 8 x = 3 C x = 4 x = 6 x = 5 3 3 3 Factor B Means x = 4 x = 8 x = 6 Step ( x ) x SST = = (8-6) + ( - 6) + ( - 6) + + (4-6) = 544 Step SSA = br = () [ (0-6) + (3-6) + (5-6) ] = 344 Step 3 SSB = ar = 3 () [ (4-6) + (8-6) ] = 48 Step 4 = r x x x + x = [ (0-0 - 4 + 6) + + (6-5 - 8 +6) ] = 56 SSAB Step 5 SSE = SST - SSA - SSB - SSAB = 544-344 - 48-56 = 96 Factor A 344 7 7/6 = 0.75 Factor B 48 48 48/6 = 3.00 Interacton 56 8 8/6 =.75 Error 96 6 6 Total 544 F.05 = 5.4 ( degrees of freedom numerator and 6 denomnator) Snce F = 0.75 > F.05 = 5.4, Factor A s sgnfcant, there s a dfference due to the type of advertsement desgn. 3-0

Analyss of Varance and Expermental Desgn F.05 = 5.99 ( degree of freedom numerator and 6 denomnator) Snce F = 3 < F.05 = 5.99, Factor B s not sgnfcant; there s not a sgnfcant dfference due to sze of advertsement. Snce F =.75 < F.05 = 5.4, Interacton s not sgnfcant. 44. Roller Coaster Factor B Screamng Demon Log Flume Factor A Means Factor A Method x = 4 x = 48 x = 48 x = 46 3 Method x = 50 x = 48 x = 46 3 x = 48 Factor B Means x = 46 x = 48 x 3 = 47 x = 47 Step ( x ) x SST = = (4-47) + (43-47) + + (44-47) = 36 Step SSA = br = 3 () [ (46-47) + (48-47) ] = Step 3 SSB = ar = () [ (46-47) + (48-47) + (47-47) ] = 8 Step 4 = r x x x + x = [ (4-46 - 46 + 47) + + (44-48 - 47 + 47) ] = 56 SSAB Step 5 SSE = SST - SSA - SSB - SSAB = 36 - - 8-56 = 60 3 -

Chapter 3 Factor A /0 =. Factor B 8 4 4/0 =.4 Interacton 56 8 8/0 =.8 Error 60 6 0 Total 36 F.05 = 5.99 ( numerator degree of freedom and 6 denomnator) F.05 = 5.4 ( numerator degrees of freedom and 6 denomnator) Snce none of the F values exceed the correspondng crtcal values, there s no sgnfcant effect due to the loadng and unloadng method, the type of rde, or nteracton. 45. Fnancal Manager Factor B Computer Programmer Pharmacst Factor A Means Factor A Male x = 979 x = 797 x = 047 x = 94 3 Female x = 635 x = 74 x = 93 3 x = 769 Factor B Means x = 807 x = 769 x 3 = 989 x = 855 Step ( x ) x SST = = (87-979) + (859-979) + + (87-93) = 864,43 Step SSA = br = 3(5) [(94-855) + (769-855) ] =,880 Step 3 SSB = ar = (5) [(807-855) + (769-855) + (989-855) ] = 76,560 Step 4 = 5[(979-94 - 807-855) + (797-94 - 769 + 855) + SSAB = r x x x + x + (989-769 - 989 + 855) ] = 5,440 Step 5 3 -

Analyss of Varance and Expermental Desgn SSE = SST - SSA - SSB - SSAB = 864,43 -,880-76,560-5,440 = 50,55 Factor A,880,880.5 Factor B 76,560 38,80 3.5 Interacton 5,440 57,70 5.53 Error 50,55 4 0,440 Total 864,43 9 F.05 = 4.6 ( degree of freedom numerator and 4 denomnator) F.05 = 3.40 ( degrees of freedom numerator and 4 denomnator) Snce F =.5 > F.05 = 4.6, Factor A (gender) s not sgnfcant. Snce F = 3.5 > F.05 = 3.40, Factor B (occupaton) s sgnfcant. Snce F = 5.53 > F.05 = 3.40, Interacton s sgnfcant. 46. x = (.3 +.56 +.00)/3 =.563 x = (0.48 +.68 +.86)/3 =.673 x = (.3 + 0.48)/ = 0.805 x = (.56 +.68)/ =.60 x = (.00 +.86)/ =.43 3 x = (.3 +.56 +.00 + 0.48 +.68 +.86)/6 =.68 Step SST = 37.50 (gven n problem statement) Step SSA = br = 3(5)[(.563 -.68) + (.673 -.68) ] = 0.4538 Step 3 SSB = ar = (5)[(0.805 -.68) + (.6 -.68) + (.43 -.68) ] = 66.059 Step 4 3-3

Chapter 3 = r x x x + x = 5[(.3 -.563-0.805 +.68) + (.56 -.563 -.6 SSAB Step 5 +.68) + + (.86 -.673 -.43 +.68) ] = 4.55 SSE = SST - SSA - SSB - SSAB = 37.50-0.4538-66.059-4.55 Factor A 0.4538 0.4538 0.648 Factor B 66.059 33.0080 9.608 Interacton 4.55 7.63 4.583 Error 46.7778 44.737 Total 37.5000 49 F.05 for degree of freedom numerator and 44 degrees of freedom denomnator s between 3.9 and 3.84. F.05 for degrees of freedom numerator and 44 denomnator s between 3.07 and 3.00. Snce 0.648 < F.05 = 3.84, Factor A s not sgnfcant Snce 9.608 > F.05 = 3.07, Factor B s sgnfcant Snce 4.583 > F.05 = 3.07, Interacton s sgnfcant 47. a. Area Area Sample Mean 96 94 Sample Varance 50 40 s + s 50+ 40 pooled estmate = = = 45 estmate of standard devaton of x x = 45 + = 4.74 4 4 x x 96 94 t = = =.4 4.74 4.74 t.05 =.447 (6 degrees of freedom) Snce t =.4 < t.05 =.477, the means are not sgnfcantly dfferent. b. x = (96 + 94)/ = 95 = SSTR = n = 4(96-95) + 4(94-95) = 8 MSTR = SSTR /( - ) = 8 / = 8 3-4

Analyss of Varance and Expermental Desgn n s = 3(50) + 3(40) = 70 = SSE = ( ) MSE = SSE /(n T - ) = 70 /(8 - ) = 45 F = MSTR /MSE = 8 /45 =.8 F.05 = 5.99 ( degree of freedom numerator and 6 denomnator) Snce F =.8 < F.05 = 5.99 the means are not sgnfcantly dfferent. c. Area Area Area 3 Sample Mean 96 94 83 Sample Varance 50 40 4 x = (96 + 94 + 83)/3 = 9 = SSTR = n = 4(96-9) + 4(94-9) + 4(83-9) = 39 MSTR = SSTR /( - ) = 39 / = 96 n s = 3(50) + 3(40) + 3(4) = 396 = SSE = ( ) MSTR = SSE /(n T - ) = 396 /( - 3) = 44 F = MSTR /MSE = 96 /44 = 4.45 F.05 = 4.6 ( degrees of freedom numerator and 6 denomnator) Snce F = 4.45 > F.05 = 4.6 we reect the null hypothess that the mean asng prces for all three areas are equal. 48. The Mntab output for these data s shown below: Analyss of Varance Source DF SS MS F P Factor 753.3 376.6 8.59 0.000 Error 7 546.9 0.3 Total 9 300. Indvdual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---------+---------+---------+------- SUV 0 58.600 4.575 (-----*-----) Small 0 48.800 4. (-----*----) FullSze 0 60.00 4.70 (-----*-----) ---------+---------+---------+------- Pooled StDev = 4.50 50.0 55.0 60.0 3-5

Chapter 3 Because the p-value =.000 < α =.05, we can reect the null hypothess that the mean resale value s the same. It appears that the mean resale value for small pcup trucs s much smaller than the mean resale value for sport utlty vehcles or full-sze pcup trucs. 49. Food Personal Care Retal Sample Mean 5.5 6.5 55.75 Sample Varance.5 5.58 4.9 x = (5.5 + 6.5 + 55.75)/3 = 56.75 = SSTR = n = 4(5.5-56.75) + 4(6.5-56.75) + 4(55.75-56.75) = 06 MSTR = SSTR /( - ) = 06 / = 03 n s = 3(.5) + 3(5.58) + 3(4.9) = 8.5 = SSE = ( ) MSE = SSE /(n T - ) = 8.5 /( - 3) = 4.5 F = MSTR /MSE = 03 /4.5 = 7.3 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F = 7.3 exceeds the crtcal F value, we reect the null hypothess that the mean age of executves s the same n the three categores of companes. 50. Lawyer Physcal Therapst Cabnet Maer Systems Analyst Sample Mean 50.0 63.7 69. 6. Sample Varance 4. 64.68 05.88 36.6 50.0+ 63.7+ 69.+ 6. x = = 6 4 = SSTR = n = 0(50.0-6) + 0(63.7-6) + 0(69. - 6) + 0(6. - 6) = 939.4 MSTR = SSTR /( - ) = 939.4 /3 = 646.47 n s = 9(4.) + 9(64.68) + 9(05.88) + 9(36.6) = 4,78.60 = SSE = ( ) MSE = SSE /(n T - ) = 478.6 /(40-4) = 3.85 F = MSTR /MSE = 646.47 /3.85 = 4.87 F.05 =.84 (3 degrees of numerator and 40 denomnator) 3-6

Analyss of Varance and Expermental Desgn F.05 =.76 (3 degrees of freedom numerator and 60 denomnator) Thus, the crtcal F value s between.76 and.84. Snce F = 4.87 exceeds the crtcal F value, we reect the null hypothess that the mean ob satsfacton ratng s the same for the four professons. 5. The Mntab output for these data s shown below: Analyss of Varance Source DF SS MS F P Factor 4339 69 3.66 0.039 Error 7 599 59 Total 9 0330 Indvdual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+--- West 0 08.00 3.78 (-------*-------) South 0 9.70 9.6 (-------*-------) NE 0.0 8.75 (-------*------) ---+---------+---------+---------+--- Pooled StDev = 4.34 80 00 0 40 Because the p-value =.039 < α =.05, we can reect the null hypothess that the mean rate for the three regons s the same. 5. The Mntab output s shown below: ANALYSIS OF VARIANCE SOURCE DF SS MS F p FACTOR 3 7.0 43.7 8.74 0.000 ERROR 36 744. 48.4 TOTAL 39 305. INDIVIDUAL 95 PCT CI'S FOR MEAN BASED ON POOLED STDEV LEVEL N MEAN STDEV --+---------+---------+---------+---- West 0 60.000 7.8 (------*-----) South 0 45.400 7.60 (------*-----) N.Cent 0 47.300 6.778 (------*-----) N.East 0 5.00 6.5 (-----*------) --+---------+---------+---------+---- POOLED STDEV = 6.96 4.0 49.0 56.0 63.0 Snce the p-value = 0.000 < α = 0.05, we can reect the null hypothess that that the mean base salary for art drectors s the same for each of the four regons. 3-7

Chapter 3 53. The Mntab output for these data s shown below: Analyss of Varance Source DF SS MS F P Factor.40 6.0 9.33 0.00 Error 37 4.596 0.665 Total 39 36.998 Indvdual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ------+---------+---------+---------+ Recever 5 7.433 0.8855 (-------*------) Guard 3 6.077 0.7399 (-------*------) Tacle 7.0583 0.8005 (-------*-------) ------+---------+---------+---------+ Pooled StDev = 0.853 6.00 6.60 7.0 7.80 Because the p-value =.00 < α =.05, we can reect the null hypothess that the mean ratng for the three postons s the same. It appears that wde recevers and tacles have a hgher mean ratng than guards. 54. X Y Z Sample Mean 9 97 84 Sample Varance 30 6 35.33 x = (9 + 97 + 44) /3 = 9 = SSTR = n = 4(9-9) + 4(97-9) + 4(84-9) = 344 MSTR = SSTR /( - ) = 344 / = 7 n s = 3(30) + 3(6) + 3(35.33) = 3.99 = SSE = ( ) MSE = SSE /(n T - ) = 3.99 /( - 3) = 3.78 F = MSTR /MSE = 7 /3.78 = 7.3 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F = 7.3 > F.05 = 4.6, we reect the null hypothess that the mean absorbency ratngs for the three brands are equal. 55. Frst Year Second Year Thrd Year Fourth Year Sample Mean.03-0.99 5.4 9.8 Sample Varance 46.93 343.04 59.3 55.43 x = (.03 -.99 + 5.4 + 9.8) /4 = 6.7 3-8

Analyss of Varance and Expermental Desgn = SSTR = n = 7(.03-6.7) + 7(-.99-6.7) + 7(5.4-6.7) + (9.8-6.7) =,.0 MSTR = SSTR /( - ) =,.0 /3 = 404.03 n s = 6(46.93) + 6(343.04) + 6(59.3) + 6(55.43) = 5,848.6 = SSE = ( ) MSE = SSE /(n T - ) = 5,848.6 /(8-4) = 43.68 F = MSTR /MSE = 404.03 /43.68 =.66 F.05 = 3.0 (3 degrees of freedom numerator and 4 denomnator) Snce F =.66 < F.05 = 3.0, we can not reect the null hypothess that the mean percent changes n each of the four years are equal. 56. Method A Method B Method C Sample Mean 90 84 8 Sample Varance 98.00 68.44 59.78 x = (90 + 84 + 8) /3 = 85 = SSTR = n = 0(90-85) + 0(84-85) + 0(8-85) = 40 MSTR = SSTR /( - ) = 40 / = 0 n s = 9(98.00) + 9(68.44) + 9(59.78) = 3,836 = SSE = ( ) MSE = SSE /(n T - ) = 3,836 /(30-3) = 4.07 F = MSTR /MSE = 0 /4.07 =.48 F.05 = 3.35 ( degrees of freedom numerator and 7 denomnator) Snce F =.48 < F.05 = 3.35, we can not reect the null hypothess that the means are equal. 57. Type A Type B Type C Type D Sample Mean 3,000 7,500 34,00 30,300 Sample Varance,0,500,35,65,7,500,960,000 x = (3,000 + 7,500 + 34,00 + 30,000) /4 = 3,000 3-9

Chapter 3 = SSTR = n = 30(3,000-3,000) + 30(7,500-3,000) + 30(34,00-3,000) + 30(30,300-3,000) = 79,400,000 MSTR = SSTR /( - ) = 79,400,000 /3 = 39,800,000 n s = 9(,0,500) + 9(,35,65) + 9(,7,500) + 9(,960,000) = SSE = ( ) = 64,08,5 MSE = SSE /(n T - ) = 64,08,5 /(0-4) =,77,656.5 F = MSTR /MSE = 39,800,000 /,77,656.5 = 05.8 F.05 s approxmately.68, the table value for 3 degrees of freedom numerator and 0 denomnator; the value we would loo up, f t were avalable, would correspond to 6 denomnator degrees of freedom. Snce F = 05.8 exceeds F.05, whatever ts value actually s, we reect the null hypothess that the populaton means are equal. 58. Desgn A Desgn B Desgn C Sample Mean 90 07 09 Sample Varance 8.67 68.67 00.67 x = (90 + 07 + 09) /3 = 0 = SSTR = n = 4(90-0) +4(07-0) +(09-0) = 87 MSTR = SSTR /( - ) = 87 / = 436 n s = 3(8.67) + 3(68.67) + 3(00.67) = 756.03 = SSE = ( ) MSE = SSE /(n T - ) = 756.03 /( - 3) = 84 F = MSTR /MSE = 436 /84 = 5.9 F.05 = 4.6 ( degrees of freedom numerator and 9 denomnator) Snce F = 5.9 > F.05 = 4.6, we reect the null hypothess that the mean lfetme n hours s the same for the three desgns. 59. a. Nonbrowser Lght Browser Heavy Browser Sample Mean 4.5 5.5 5.75 Sample Varance.07.07.36 x = (4.5 + 5.5 + 5.75) /3 = 5.08 3-30

Analyss of Varance and Expermental Desgn = SSTR = n = 8(4.5-5.08) + 8(5.5-5.08) + 8(5.75-5.08) = 9.33 MSB = SSB /( - ) = 9.33 / = 4.67 n s = 7(.07) + 7(.07) + 7(.36) = 4.5 = SSW = ( ) MSW = SSW /(n T - ) = 4.5 /(4-3) =.7 F = MSB /MSW = 4.67 /.7 = 3.99 F.05 = 3.47 ( degrees of freedom numerator and denomnator) Snce F = 3.99 > F.05 = 3.47, we reect the null hypothess that the mean comfort scores are the same for the three groups. LSD= t MSW + =.080.7 + =. n n 8 8 b. α / Snce the absolute value of the dfference between the sample means for nonbrowsers and lght browsers s 4.5 5.5 =, we cannot reect the null hypothess that the two populaton means are equal. 60. Treatment Means: x =.8 x = 4.8 x = 5.80 3 Bloc Means: x = 9.67 x = 5.67 x3 = 3 x4 = 3.67 x5 =.33 Overall Mean: x = 367 /5 = 4.47 Step ( x ) x SST = = (8-4.47) + ( - 4.47) + + (4-4.47) = 53.73 Step SSTR = b = 5 [ (.8-4.47) + (4.8-4.47) + (5.8-4.47) ] = 3.33 3-3

Chapter 3 Step 3 SSBL = = 3 [ (9.67-4.47) + (5.67-4.47) + + (.33-4.47) ] = 7.0 Step 4 SSE = SST - SSTR - SSBL = 53.73-3.33-7.0 = 3.38 Treatment 3.33.67 6.99 Blocs 7.0 4 54.6 3.49 Error 3.38 8.67 Total 53.73 4 F.05 = 4.46 ( degrees of freedom numerator and 8 denomnator) Snce F = 6.99 > F.05 = 4.46 we reect the null hypothess that the mean mles per gallon ratngs for the three brands of gasolne are equal. 6. I II III Sample Mean.8 4.8 5.8 Sample Varance. 9. 7. x = (.8 + 4.8 + 5.8) /3 = 4.47 = SSTR = n = 5(.8-4.47) + 5(4.8-4.47) + 5(5.8-4.47) = 3.33 MSTR = SSTR /( - ) = 3.33 / =.67 n s = 4(.) + 4(9.) + 4(7.) = 30.4 = SSE = ( ) MSE = SSE /(n T - ) = 30.4 /(5-3) = 9. F = MSTR /MSE =.67 /9. =.6 F.05 = 3.89 ( degrees of freedom numerator and denomnator) Snce F =.6 < F.05 = 3.89, we cannot reect the null hypothess that the mean mles per gallon ratngs for the three brands of gasolne are equal. Thus, we must remove the bloc effect n order to detect a sgnfcant dfference due to the brand of gasolne. The followng table llustrates the relatonshp between the randomzed bloc desgn and the completely randomzed desgn. 3-3

Analyss of Varance and Expermental Desgn Sum of Squares Randomzed Bloc Desgn Completely Randomzed Desgn SST 53.73 53.73 SSTR 3.33 3.33 SSBL 7.0 does not exst SSE 3.38 30.4 Note that SSE for the completely randomzed desgn s the sum of SSBL (7.0) and SSE (3.38) for the randomzed bloc desgn. Ths llustrates that the effect of blocng s to remove the bloc effect from the error sum of squares; thus, the estmate of σ for the randomzed bloc desgn s substantally smaller than t s for the completely randomzed desgn. 6. The Mntab output for these data s shown below: Analyss of Varance Source DF SS MS F P Factor 73.75 365.88 93.6 0.000 Error 63 47.4 3.93 Total 65 979.7 Indvdual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev ---+---------+---------+---------+--- UK.05.393 (--*--) US 4.957.847 (--*--) Europe 0.05.536 (--*--) ---+---------+---------+---------+--- Pooled StDev =.98.0 5.0 8.0.0 Because the p-value =.000 < α =.05, we can reect the null hypothess that the mean download tme s the same for web stes located n the three countres. Note that the mean download tme for web stes located n the Unted Kngdom (.05 seconds) s less than the mean download tme for web stes n the Unted States (4.957) and web stes located n Europe (0.05). 63. Factor B Factor A Spansh French German Means Factor A System x = 0 x = x = 4 x = 3 System x = 8 x = 5 x = 9 3 x = 4 Factor B Means x = 9 x = 3.5 x 3 = 6.5 x = 3 Step ( x ) x SST = = (8-3) + ( - 3) + + ( - 3) = 04 3-33

Chapter 3 Step SSA = br = 3 () [ ( - 3) + (4-3) ] = Step 3 SSB = ar = () [ (9-3) + (3.5-3) + (6.5-3) ] = 4 Step 4 64. = r x x x + x = [(8 - - 9 + 3) + + ( - 4-6.5 +3) ] = 6 SSAB Step 5 SSE = SST - SSA - SSB - SSAB = 04 - - 4-6 = 5 Factor A.38 Factor B 4 57 6.57 Interacton 6.50 Error 5 6 8.67 Total 04 F.05 = 5.99 ( degree of freedom numerator and 6 denomnator) F.05 = 5.4 ( degrees of freedom numerator and 6 denomnator) Snce F = 6.57 > F.05 = 5.4, Factor B s sgnfcant; that s, there s a sgnfcant dfference due to the language translated. Type of system and nteracton are not sgnfcant snce both F values are less than the crtcal value. Factor B Factor B Manual Automatc Means Factor A Machne x = 3 x = 8 x = 36 Machne x = = 6 x = 3.5 Factor B Means x = 6.5 x = 7 x = 6.75 3-34

Analyss of Varance and Expermental Desgn Step ( x ) x SST = = (30-6.75) + (34-6.75) + + (8-6.75) = 5.5 Step SSA = br = () [ (30-6.75) + (3.5-6.75) ] = 84.5 Step 3 SSB = ar = () [ (6.5-6.75) + (7-6.75) ] = 0.5 Step 4 = [(30-30 - 6.5 + 6.75) + + (8-3.5-7 + 6.75) ] SSAB Step 5 = r x x x + x = 40.5 SSE = SST - SSA - SSB - SSAB = 5.5-84.5-0.5-40.5 = 6 Factor A 84.5 84.5 3 Factor B.5.5.08 Interacton 40.5 40.5 6.3 Error 6 4 6.5 Total 5.5 7 F.05 = 7.7 ( degree of freedom numerator and 4 denomnator) Snce F = 3 > F.05 = 7.7, Factor A (Type of Machne) s sgnfcant. Type of Loadng System and Interacton are not sgnfcant snce both F values are less than the crtcal value. 3-35

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