Assume you are employed by a consumer-products rating service and your assignmen
ID: 3176889 • Letter: A
Question
Assume you are employed by a consumer-products rating service and your assignment is to assess car batteries. For this part of your investigation, you want to determine whether there is a difference in useful life among the top-of-the-line car batteries produced by three manufacturers (A, B, and C). To provide the database for your assessment, you randomly sample four batteries from each manufacturer and run them through laboratory tests that allow you to determine the useful life of each battery. The following are the results given in months of useful battery life: Battery Manufacturer A B C 58 48 42 55 53 53 57 51 51 57 48 50 (a) Use the analysis of variance with = 0.05 to determine whether there is a difference among these three brands of batteries. (i) Fill in the missing values. Source SS df s2 Fobt Between Within Total (ii) What is the value of Fcrit? Fcrit = (iii) What do you conclude? Since Fobt < Fcrit, reject H0 and conclude that the batteries of at least one manufacturer differ regarding useful life. Since Fobt < Fcrit, retain H0. We conclude that the batteries of at least one manufacturer differ regarding useful life. Since Fobt Fcrit, reject H0 and conclude that the batteries of at least one manufacturer differ regarding useful life. Since Fobt Fcrit, retain H0. We conclude that the batteries of at least one manufacturer differ regarding useful life. (b) Suppose you are asked to make a recommendation regarding the batteries based on useful life. (i) Use the HSD test with = 0.052 tail to compare all possible means. (Select all that apply.) Reject H0 for the comparison between batteries of manufacturer A and B. Reject H0 for the comparison between batteries of manufacturer A and C. Reject H0 for the comparison between batteries of manufacturer B and C. (ii) Given the conclusion in part (i), what is your recommendation? I cannot make any recommendation since the results could have been due to chance. I recommend the batteries of manufacturer A because they have significantly longer life than the batteries of manufacturers B and C.
Explanation / Answer
Solution:
(Due to lot of calculations, most of the calculations are carried out by using Excel.)
Part a
The completed table is given as below:
We know that there are three groups. So, degrees of freedom for between groups = 3 – 1 = 2
We are given that the total number of observations = n = 12
So, total degrees of freedom = n – 1 = 12 – 1 = 11
Within degrees of freedom = total df – df between = 11 – 2 = 9
A
B
C
A^2
B^2
C^2
58
48
42
3364
2304
1764
55
53
53
3025
2809
2809
57
51
51
3249
2601
2601
57
48
50
3249
2304
2500
= 227
= 200
= 196
= 12887
= 10018
= 9674
Groups
Count
Sum
Average
Variance
A
4
227
56.75
1.583333
B
4
200
50
6
C
4
196
49
23.33333
ANOVA
Source of Variation
SS
df
MS
F
P-value
Between Groups
142.1667
2
71.08333
6.897574
0.015269
Within Groups
92.75
9
10.30556
Total
234.9167
11
F-crit = 4.256495
Since Fobt Fcrit, reject H0 and conclude that the batteries of at least one manufacturer differ regarding useful life.
Part b
Multiple Comparisons
Life
HSD
(I) Manufacturer
(J) Manufacturer
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
A
B
6.75000*
2.26997
.016
1.6150
11.8850
C
7.75000*
2.26997
.008
2.6150
12.8850
B
A
-6.75000*
2.26997
.016
-11.8850
-1.6150
C
1.00000
2.26997
.670
-4.1350
6.1350
C
A
-7.75000*
2.26997
.008
-12.8850
-2.6150
B
-1.00000
2.26997
.670
-6.1350
4.1350
*. The mean difference is significant at the 0.05 level.
We are given a level of significance or alpha value as 0.052.
Pair B and C does not have significant difference as the p-value for this pair is greater than the alpha value. All remaining pairs have significant difference as the p-values are less than the given level of significance or alpha value.
A
B
C
A^2
B^2
C^2
58
48
42
3364
2304
1764
55
53
53
3025
2809
2809
57
51
51
3249
2601
2601
57
48
50
3249
2304
2500
= 227
= 200
= 196
= 12887
= 10018
= 9674
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