In: Statistics and Probability
A pet food company has a business objective of expanding its product line beyond its current kidney and shrimp-based cat foods. The company developed two new products, one based on chicken liver and the other based on salmon. The company conducted an experiment to compare the two new products with its two existing ones, as well as a generic beef-based product sold at a supermarket chain. For the experiment, a sample of 50 cats from the population at a local animal shelter was selected. Ten cats were randomly assigned to each of the five products being tested. Each of the cats was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 10-minute time interval that began when the filled dish was presented. The results are summarized in the dataset (CatFood.xlsx).
Where CatFood.xlsx Data:
Kidney Shrimp Chicken Liver Salmon Beef
2.37 2.26 2.29 1.79 2.09
2.62 2.69 2.23 2.33 1.87
2.31 2.25 2.41 1.96 1.67
2.47 2.45 2.68 2.05 1.64
2.59 2.34 2.25 2.26 2.16
2.62 2.37 2.17 2.24 1.75
2.34 2.22 2.37 1.96 1.18
2.47 2.56 2.26 1.58 1.92
2.45 2.36 2.45 2.18 1.32
2.32 2.59 2.57 1.93 1.94
To test that there is evidence of a difference in the mean amount of food eaten among the various products, fill in all the values in the following One-way ANOVA summary table.
Source of Variation | SS | df | MS | F |
Among Groups | ||||
Within Groups | - | |||
Total | - | - |
At the 0.01 level of significance, is there evidence of a difference in the mean amount of food eaten among the various products? Test the hypothesis using F test based on the result of (a).
At the 0.01 level of significance, determine which products appear to differ significantly in the mean amount of food eaten using the Tukey-Kramer method.
At the 0.01 level of significance, is there evidence of a difference in the variation in the amount of food eaten among the various products? Test the homogeneity of the variances using the Levene’s test.
Excel Addon PHStat used
At the 0.01 level of significance, is there evidence of a difference in the mean amount of food eaten among the various products? Test the hypothesis using F test based on the result of (a).
ANOVA: Single Factor |
||||||
SUMMARY |
||||||
Groups |
Count |
Sum |
Average |
Variance |
||
Kidney |
10 |
24.56 |
2.456 |
0.0148 |
||
Shrimp |
10 |
24.09 |
2.409 |
0.0253 |
||
Chicken Liver |
10 |
23.68 |
2.368 |
0.0263 |
||
Salmon |
10 |
20.28 |
2.028 |
0.0544 |
||
Beef |
10 |
17.54 |
1.754 |
0.0990 |
||
ANOVA |
||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
Between Groups |
3.6590 |
4 |
0.9147 |
20.8054 |
0.0000 |
3.7674 |
Within Groups |
1.9785 |
45 |
0.0440 |
|||
Total |
5.6375 |
49 |
||||
Level of significance |
0.01 |
Calculated F= 20.8054, P=0.0000 which is < 0.01 level of significance, Ho is rejected.
There is enough evidence of a difference in the mean amount of food eaten among the various products.
At the 0.01 level of significance, determine which products appear to differ significantly in the mean amount of food eaten using the Tukey-Kramer method.
Tukey-Kramer Multiple Comparisons |
||||||||||
Sample |
Sample |
Absolute |
Std. Error |
Critical |
||||||
Group |
Mean |
Size |
Comparison |
Difference |
of Difference |
Range |
Results |
|||
1: Kidney |
2.456 |
10 |
Group 1 to Group 2 |
0.047 |
0.066307198 |
0.327 |
Means are not different |
|||
2: Shrimp |
2.409 |
10 |
Group 1 to Group 3 |
0.088 |
0.066307198 |
0.327 |
Means are not different |
|||
3: Chicken Liver |
2.368 |
10 |
Group 1 to Group 4 |
0.428 |
0.066307198 |
0.327 |
Means are different |
|||
4: Salmon |
2.028 |
10 |
Group 1 to Group 5 |
0.702 |
0.066307198 |
0.327 |
Means are different |
|||
5: Beef |
1.754 |
10 |
Group 2 to Group 3 |
0.041 |
0.066307198 |
0.327 |
Means are not different |
|||
Group 2 to Group 4 |
0.381 |
0.066307198 |
0.327 |
Means are different |
||||||
Other Data |
Group 2 to Group 5 |
0.655 |
0.066307198 |
0.327 |
Means are different |
|||||
Level of significance |
0.01 |
Group 3 to Group 4 |
0.34 |
0.066307198 |
0.327 |
Means are different |
||||
Numerator d.f. |
5 |
Group 3 to Group 5 |
0.614 |
0.066307198 |
0.327 |
Means are different |
||||
Denominator d.f. |
45 |
Group 4 to Group 5 |
0.274 |
0.066307198 |
0.327 |
Means are not different |
||||
MSW |
0.043966 |
|||||||||
Q Statistic |
4.93 |
At the 0.01 level of significance, is there evidence of a difference in the variation in the amount of food eaten among the various products? Test the homogeneity of the variances using the Levene’s test.
ANOVA: Levene Test |
||||||
SUMMARY |
||||||
Groups |
Count |
Sum |
Average |
Variance |
||
Kidney |
10 |
0.98 |
0.098 |
0.0041 |
||
Shrimp |
10 |
1.23 |
0.123 |
0.0107 |
||
Chicken Liver |
10 |
1.28 |
0.128 |
0.0097 |
||
Salmon |
10 |
1.84 |
0.184 |
0.0174 |
||
Beef |
10 |
2.42 |
0.242 |
0.0374 |
||
ANOVA |
||||||
Source of Variation |
SS |
df |
MS |
F |
P-value |
F crit |
Between Groups |
0.1341 |
4 |
0.0335 |
2.1140 |
0.0947 |
3.7674 |
Within Groups |
0.7137 |
45 |
0.0159 |
|||
Total |
0.8478 |
49 |
||||
Level of significance |
0.01 |
Calculated F= 2.114, P=0.0947 which is > 0.01 level of significance, Ho is not rejected.
There is not enough evidence of a difference in the variation in the amount of food eaten among the various products.