In: Math
A pet food company has a business objective of expanding its product line beyond its current kidney- and shrimp-based cat foots. 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 in supermarket chains.
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 period that began when the filled dish was presented to the cat. The results for this experiment are summarized in CatFood.
a. At the 0.05 level of significance, is there evidence of a differences in the mean amount of food eaten among the various products?
b. Does the result in (a) give you statistical permission to probe for individual differences between the food products?
Please show me how to do this in excel using the data analysis tab
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 |
data -> data analysis -> Anova: Single factor
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Kidney | 10 | 24.56 | 2.456 | 0.01476 | ||
Shrimp | 10 | 24.09 | 2.409 | 0.025343 | ||
Chicken Liver | 10 | 23.68 | 2.368 | 0.026284 | ||
Salmon | 10 | 20.28 | 2.028 | 0.054418 | ||
Beef | 10 | 17.54 | 1.754 | 0.099027 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 3.65896 | 4 | 0.91474 | 20.80541 | 9.149E-10 | 2.578739184 |
Within Groups | 1.97849 | 45 | 0.043966 | |||
Total | 5.63745 | 49 |
p-value = 9.149*10^(-10) << alpha
hence
there is evidence of a differences in the mean amount of food eaten among the various products
b)
yes
as there is difference