Question

My husband and I often travel from a small town called Bay City TX to my sister’s house in Houston TX, we like to take our dogs with us, they are like our children (we don’t have kids yet). Often during our trips, we like to visit dog parks to wear them out. The dog park in Bay city TX there were 3 dogs including ours Needville TX – 2, Wharton TX – 8, Rosengberg TX, - 6, Sugarland TX, - 10, Katy TX, – 15. My husband likes to keep track of these numbers to see the effect that it may have on our dogs... he’s protective. According to him the less dogs there are, the less chances of dogs being aggressive towards our puppies. Help him test claim at the .05 significance level.

Answer 1

Solution:

Null Hypothesis (Ho): The data followed a specific distribution.

Alternative Hypothesis (Ha): The data do not followed a specific distribution.

Expected value = Total x Equal probability

Expected value = 41 x 1/5 = 8.2

observed	expected	O - E	(O - E)² / E
2	8.200	-6.200	4.688
8	8.200	-0.200	0.005
6	8.200	-2.200	0.590
10	8.200	1.800	0.395
15	8.200	6.800	5.639
41	41.000	0.000	11.317

Test Statistics

= 11.32

Degrees of freedom, df = n - 1 = 5 - 1 = 4

Using chi-square tables, the p-value is

P [ (4) > 11.32] = 0.0232

Since p-value is less than 0.05 significance level, we reject Ho.

Hence, we can conclude that the data does not follow a specific distribution.