Answer to Question #259831 in Statistics and Probability for Marcus

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.

The hypothesis tested is,

$H_0:\mu_d=0$

$Against$

$H_1:\mu_d\gt 0$ , as the researcher's claim is that the dogs lost weight

The student's t distribution is applied because the observations are paired.

b.

The critical value is given as,

$t_{\alpha,n-1}=t_{0.05,6-1}=t_{0.05,5}=2.015$

Since we are performing a one sided test, we only have on critical value which is given as,

$t_{0.05,5}=2.015$

c.

The mean of the differences $(\bar{d})$ is given as,

$\bar{d}=\sum(d)/n=29/6=$ 4.833333

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

$V(d)=\sum(d-\bar{d})^2/(n-1)$

$V(d)=74.8333/5=14.96667$

The standard deviation is given as,

$S(d)=\sqrt{V(d)}=\sqrt{14.96667}=3.868678$

Let us now determine the t-test statistic $(t^*)$.

$t^*=\bar{d}/(S(d)/\sqrt{n})=4.833333/(3.868678/\sqrt{6})=4.833333/1.579381=3.0603$

The hull hypothesis is rejected if $t^*\gt t_{0.05,5}$

d.

Since $t^*=3.0603\gt t_{0.05,5}=2.015$ 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.