Question

In: Statistics and Probability

8. A pet food company has a business objective of expanding its product line beyond its...

8. 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 5-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?

Kidney Shrimp Chicken Liver Salmon Beef
3.00 0.25 2.61 0.63 2.97
3.00 0.34 3.00 1.37 2.85
1.47 0.16 3.00 3.00 0.44
2.62 0.72 2.43 0.54 0.90
1.51 3.00 3.00 3.00 3.00
3.00 3.00 3.00 2.55 3.00
3.00 0.29 2.85 2.93 0.48
3.00 3.00 0.18 2.73 0.45
0.74 1.99 0.98 3.00 3.00
2.21 2.30 1.51 0.96 0.05

Solutions

Expert Solution

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?

The hypothesis being tested is:

H0: µ1 = µ2 = µ3 = µ4 = µ5

Ha: At least one means is not equal

Mean n Std. Dev
2.355 10 0.8351 Kidney
1.505 10 1.2652 Shrimp
2.256 10 1.0120 Chicken Liver
2.071 10 1.0613 Salmon
1.714 10 1.3335 Beef
1.980 50 1.1181 Total
ANOVA table
Source SS    df MS F    p-value
Treatment 5.2146 4 1.30366 1.05 .3938
Error 56.0413 45 1.24536
Total 61.2559 49

The p-value is 0.3938.

Since the p-value (0.3938) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we cannot conclude that there is a difference 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?

No, because there is no difference in the mean amount of food eaten among the various products.

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