Answer to Question #261955 in Statistics and Probability for LO3

Question

A veterinary nutritionist developed a diet for overweight dogs. The total volume of food

consumed remains the same, but one half of the dog food is replaced with a low-calorie

“filler” such as canned green beans. Six overweight dogs were randomly selected from

her practice and were put on this program. Their initial weights were recorded, and then

they were weighed again after 4 weeks. At the 0.05 level of significance can it be

concluded that the dogs lost weight?

Before 42 53 48 65 40 52

After 39 45 40 58 42 47

a. State the hypothesis and identify the claim of the researcher.

b. Find the critical value(s).

c. Compute for the mean of the differences, standard deviation of the differences

and test value.

d. Make a decision on the null hypothesis.

e. Make a decision on the claim of the researcher.

Answer 1

a.

$H_0: \mu_d = 0 \\ H_1: \mu_d > 0$

The researcher's claim is that the dogs lost weight

b. The critical value

α=0.05

df = n-1 = 5

t_{crit} = 2.015

Since we are performing a one-sided test, we only have one critical value.

c. The mean of the differences:

$\bar{d} = \frac{\sum(d)}{n}= \frac{29}{6} = 4.833$

To find the standard deviation for the differences, we first determine the variance for the differences:

$V(d) = \frac{\sum (d- \bar{d})^2}{n-1} \\ V(d) = \frac{74.833}{5} = 14.9666$

The standard deviation:

$S(d) = \sqrt{V(d)} = \sqrt{14.9666} = 3.8686$

Test-statistic:

$t = \frac{\bar{d} -0}{S(d) / \sqrt{n}} \\ t = \frac{4.833}{3.8686 / \sqrt{36}} = 3.0603$

The hull hypothesis is rejected if $t > t_{crit}$

d. Since $t= 3.0603 > t_{crit} = 2.015$ we reject the null hypothesis.

e. Because the null hypothesis is rejected we conclude that there is sufficient evidence at 5% significance level to show that the dogs lost weight. Thus researchers claim is true.