Practice Problems Anova[ 5911] PDF

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Practice Problems on Analysis of Variance Problem 1 To test whether the mean time needed to mix a batch of material is the same for machines produced by three manufacturers, the Jacobs Chemical Company obtained the following data on the time (in minutes) needed to mix the material.

1 20 26 24 22

Manufacturer 2 28 26 31 27

3 20 19 23 22

(a) Use these data to test whether the population mean times for mixing a batch of material differ for the three manufacturers. Use  = 0.05 . (b) At the  = 0.05 level of significance, use Fisher’s LSD procedure to test for the equality of the means for manufacturers 1 and 3. What conclusion can you draw after carrying out this test? Problem 2 Samples of four salespeople from each of the four regions were asked to predict percentage increases in sales volume for their territories in the next twelve months. A summary of the sample results is given below.

Sample Means Sample SD Sample Size

West 5.35 1.0878 4

Midwest 6.65 0.6191 4

South 4.85 0.6807 4

East 7.95 0.7724 4

(a) Set out the analysis of variance table. (b) Test the null hypothesis that the four population mean predictions are equal. (c) Use Fisher’s LSD procedure to carry out pairwise comparisons of means. Problem 3 Suppose that scores given by judges to competitors in the ski jumping competition events of the Winter Olympics were analyzed. For the men’s ski jumping competition, suppose there were twenty-two contestants and nine judges. Each judge in seven sub-events assessed each contestant. The scores given can be treated as a two-way analysis of variance: contestant X judge. The total sum of squares of scores about the grand mean is 1440.19. The sum of squares between contestant means is 364.50, the sum of squares between judge means is 0.81, and the interaction sum of squares is 4.94.

(a) Set up the two-way analysis of variance table. At the 0.05 level of significance: (b) Test for interaction between contestants and judges. (c) Is there a difference between the judges mean scores? (d) Is there a difference between contestants? Problem 4 A consumer group wishes to determine whether there is a significant difference in the mean cost of various breakfast cereals at four supermarket chains. The data they collected is displayed in the table below. Product Store 1 Store 2 Store 3 Store 4

1 8.49 5.99 7.99 5.99

2 9.49 9.47 9.89 7.97

3 2.66 2.59 1.99 2.69

4 6.19 4.99 4.99 5.99

5 7.69 5.99 5.99 6.29

6 2.19 2.49 3.74 2.69

7 8.99 7.49 6.99 6.99

8 7.49 5.99 6.49 5.47

Mean 6.649 5.625 6.009 5.510

They then used Minitab’s Two-Way ANOVA procedure to test whether the mean prices at the stores were the same using a randomized block design. The stores were the treatments; the products, the blocks. The results are below (with most items missing). Set up the appropriate hypotheses. (a) Complete the parts of the ANOVA table that are required, and use your results to test the hypothesis. Use a 5% significance level. Two-way ANOVA: Price versus Store, Product

Source Store Product Error Total

DF

SS 6.327 149.762

MS

F

P

166.906

(b) Use Fisher’s LSD procedure to determine which means are different. Use  = 0.05 .

Solutions to Practice Problems on Analysis of Variance Problem 1 (a) Manufacturer 1 n1 = 4

Manufacturer 2 n2 = 4

Manufacturer 3 n3 = 4

x1 = 23 s1 = 2.5820

x2 = 28 s2 = 2.1602

x3 = 21

s3 = 1.8257

k = Number of Treatments = 3 nT = n1 + n2 + n3 = 4 + 4 + 4 = 12 n x +n x +n x x + x + x 23 + 28 + 21 = 24 x= 1 1 2 2 3 3= 1 2 3= nT 3 3 k

SSTR =  n j (x j − x )2 = n1 (x1 − x )2 + n2 (x2 − x )2 + n3 (x3 − x ) 2 j =1

= 4(23 − 24)2 + 4( 28 − 24) 2 + 4(21 − 24) 2 = 104 k

SSE =  ( nj − 1) s 2j = ( n1 − 1) s12 + ( n2 −1) s22 + ( n3 −1) s32 j =1

= (4− 1)(2.5820)2 + (4− 1)(2.1602)2 + (4 − 1)(1.8257)2 = 44 SST = SSTR + SSE = 104 + 44 =148 DF for Treatments = k −1 = 3 −1 = 2 DF for Error = nT − k = 12 − 3 = 9 DF for Total = nT − 1 = 12 − 1 = 11

SSTR 104 = = 52 k − 1 3− 1 SSE 44 MSE = = = 4.8889 nT − k 12 − 3

MSTR =

ANOVA Table: Source of Variation (Source) Treatments Error Total

Degrees of Freedom (DF) 104 44 148

Sum of Squares (SS) 2 9 11

Mean Squares (MS) 52 4.8889

F-Ratio (F) 10.6364

P-Value < 0.01

H0 : 1 = 2 = 3 (The population mean times for mixing a batch of material for the three manufacturers are equal.) H a : Not all population means are equal (The population mean times for mixing a batch of material for the three manufacturers are different.)

MSTR 52 = = 10. 6364 MSE 4. 8889

Test Statistic:

F=

P-Value:

p- value = P (F  10.6364)  0.01 with df1 = 2 and df 2 = 9

Decision Rule: Reject H 0 if p-value  0.05 Conclusion:

Since the p-value is less than the 0.05 level of significance, we reject H 0 . We have sufficient evidence to conclude that the population mean times for mixing a batch of material are significantly different for the three manufacturers.

(b) Absolute Difference Approach: 1 1 1 1 LSD = t 2 MSE +  = 2.262 4.8889 +  = 3.5366  4 4  n1 n3 

x1 − x3 = 23− 21 = 2 Since the absolute difference is smaller than the Fisher’s LSD, we cannot conclude that the means for manufacturers 1 and 3 are significantly different.

Hypothesis Testing Approach: H 0 : 1 = 3

H a : 1  3

Test Statistic:

t=

x1 − x 3 1 1 MSE  +   n1 n3 

=

23 − 21 1 1 4.8889  +  4 4

=1.2792

Decision Rule: At the 0.05 level of significance, reject H 0 if t  −2.262 or t  2.262 with 9 degrees of freedom. Conclusion: Since our test statistic does not fall in the rejection region, we fail to reject the null hypothesis that the means for manufacturers 1 and 3 are the same.

Confidence Interval Approach: The 95% confidence interval for 1 −  3 is 1 1  x1 − x3  t 2 MSE  +   n1 n 3  1 1 = (23 − 21)  2.262 4.8889  +  4 4 = 2 3.5366 = ( −1.5366, 5.5366)

Since zero is included in the 95% confidence interval, we cannot conclude that the means for manufacturers 1 and 3 are significantly different.

Problem 2 (a) x=

n1 x1 + n2 x2 + n3 x3 + n4 x4 n1 x1 + n2 x2 + n3 x3 + n4 x4 n( x1 + x2 + x3 + x4 ) = = nT n1 + n2 + n3 + n4 4n

x1 + x2 + x3 + x4 4 5.35 + 6.65 + 4.85 + 7.95 = 4 = 6.20 =

k

SSTR =  n j ( x j − x ) 2 j =1

= n 1(x 1 − x ) 2 + n 2 (x 2 − x )2 + n3 (x3 − x ) 2 + n 4 (x4 − x ) 2 = 4(5.35 − 6.20)2 + 4(6.65 − 6.20)2 + 4(4.85 − 6.20)2 + 4(7.95 − 6.20)2 = 23.24 k

SSE =  ( n j −1) s j2 j =1

= (n1 − 1)s12 + (n2 − 1)s22 + (n3 − 1)s32 + (n4 − 1)s42 = (4 − 1)(1.0878)2 + (4 −1)(0.6191)2 + (4 −1)(0.6807) 2 + (4 −1)(0.7724) 2 = 7.8796

ANOVA Table: Source of Variation (Source) Treatments Error Total

Degrees of Freedom (DF) 3 12 15

Sum of Squares (SS) 23.24 7.8796 31.1196

Mean Squares (MS) 7.7467 0.6566

F-Ratio (F) 11.7975

(b) H 0 : 1 =  2 =  3 =  4 H a : Not all population means are equal Test Statistic:

F=

MSTR 7.7467 = = 11.7975 MSE 0.6566

Decision Rule: At the 0.05 level of significance, reject H 0 if F  3.49 with 3 and 12 degrees of freedom. Conclusion:

Since the test statistic falls in the rejection region, we reject H 0 . We have sufficient evidence to conclude that there is a significant difference in the mean prediction for the four regions.

(c)

1 1  1 1 LSD = t 2 MSE +  = 2.179 * 0.6566 +  = 1.2485 n n   4 4 j   i West vs. Midwest

x1 − x 2 = 5.35 − 6.65 = 1.30

Different

West vs. South

x1 − x3 = 5.35 − 4.85 = 0.50

No Different

West vs. East

x1 − x 4 = 5.35 − 7.95 = 2.60

Different

Midwest vs. South

x 2 − x3 = 6.65 − 4.85 = 1.80

Different

Midwest vs. East

x 2 − x 4 = 6.65 − 7.95 = 1.30

Different

South vs. East

x3 − x 4 = 4.85 − 7.95 = 3.10

Different

All pairwise comparisons of means are significant except West versus South.

Problem 3 (a) ANOVA Table: Source of Variation (Source) Contestant Judge Interaction Error Total

Degrees of Freedom (DF) 21 8 168 1188 1385

Sum of Squares (SS) 364.50 0.81 4.94 1069.94 1440.19

Mean Squares (MS) 17.3571 0.1013 0.0294 0.9006

F-Ratio (F) 19.2724 0.1124 0.0326

(b) H 0 : There is no interactio n between contestant and judge. H a : There is an interactio n between contestant and judge. Test Statistic:

F=

MSAB 0.0294 = = 0.0326 MSE 0.9006

Decision Rule: At the 0.05 level of significance, reject H 0 if F  1.20 with 168 and 1188 degrees of freedom. Conclusion:

Since the test statistic does not fall in the rejection region, we do not reject H 0 . We have insufficient evidence to conclude that there is an interaction between contestants and judges.

(c) H 0 : There is no effect due to judge. H a : There is an effect due to judge. Test Statistic:

F=

MSB 0.1013 = = 0.1124 MSE 0.9006

Decision Rule: At the 0.05 level of significance, reject H 0 if F  1.95 with 8 and 1188 degrees of freedom. Conclusion:

Since the test statistic does not fall in the rejection region, we do not reject H 0 . We have insufficient evidence to conclude that there is an effect due to judge.

(d) H 0 : There is no effect due to contestant . H a : There is an effect due to contestant . Test Statistic:

F=

MSA 17.3571 = = 19.2724 MSE 0.9006

Decision Rule: At the 0.05 level of significance, reject H 0 if F  1.56 with 21 and 1188 degrees of freedom. Conclusion:

Since the test statistic falls in the rejection region, we reject H 0 . We have sufficient evidence to conclude that there is an effect due to contestant.

Problem 4 (a) Two-way ANOVA: Price versus Store, Product Source Store Product Error Total

DF 3 7 21 31

SS 6.327 149.762 10.817 166.906

MS 2.109 21.3946 0.5151

F 4.0944 41.5352

P 0.01 < p < 0.025 p < 0.01

H0 :  1 =  2 =  3 =  4 H a : Not all population means are equal

Test Statistic:

F=

MSTR 2.109 = = 4.0944 MSE 0.5151

Decision Rule: At the 0.05 level of significance, reject H 0 if F  3.07 with 3 and 21 degrees of freedom. Conclusion:

Since our test statistic falls in the rejection region, we have sufficient evidence to conclude that the mean prices at the stores are significantly different.

(b)

 1 1  1 1 LSD = t 2 MSE  +  = 2.080* 0.5151 +  = 0.7464  ni n j   8 8   Store 1 vs. Store 2

x1 − x2 = 6.649 − 5.625 = 1.024  LSD

Different

Store 1 vs. Store 3

x1 − x3 = 6.649 − 6.009 = 0.640  LSD

No Different

Store 1 vs. Store 4

x1 − x4 = 6.649 − 5.510 = 1.139  LSD

Different

Store 2 vs. Store 3

x2 − x3 = 5.625 − 6.009 = 0.384  LSD

No Different

Store 2 vs. Store 4

x2 − x4 = 5.625 − 5.510 = 0.115  LSD

No Different

Store 3 vs. Store 4

x3 − x 4 = 6.009 − 5.510 = 0.499  LSD

No Different...