ExampleNine dogs and ten cats were

Example

Nine dogs and ten cats were tested to determine if there is a difference in the average number of days that the animal can survive without food. The dogs averaged 11 days with a standard deviation of 2 days while the cats averaged 12 days with a standard deviation of 3 days. What can be concluded? (Use a = .05)



Solution

We write:

H0: mdog - mcat = 0

H1: mdog - mcat 0

We have:

n1 = 9, n2 = 10

x1 = 11, x2 = 12

s1 = 2, s2 = 3

so that



and



The t-critical value corresponding to a = .05 with 10 + 9 - 2 = 17 degrees of freedom is 2.11 which is greater than .84. Hence we fail to reject the null hypothesis and conclude that there is not sufficient evidence to suggest that there is a difference between the mean starvation time for cats and dogs.