Chapter 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

Answer to Problem 26.6AYK

The population contrast Lis $(1)({\mu }_C)+(-\frac{1}{2})({\mu }_R)+(-\frac{1}{2})({\mu }_E)$ .

Explanation of Solution

In the question, it is given that figure $26.1$ gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

  $\begin{array}{l}L=(c_1)({\mu }_C)+(c_2)({\mu }_R)+(c_3)({\mu }_E)\\ =(1)({\mu }_C)+(-\frac{1}{2})({\mu }_R)+(-\frac{1}{2})({\mu }_E)\end{array}$

(b)

To determine

To give the sample contrast that estimates L and its standard error.

Answer to Problem 26.6AYK

The sample contrast that estimates Lis $-68.59$ and its standard erroris $15.859$ .

Explanation of Solution

In the question, it is given that figure $26.1$ gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

  $\begin{array}{l}L=(c_1)({\mu }_C)+(c_2)({\mu }_R)+(c_3)({\mu }_E)\\ =(1)({\mu }_C)+(-\frac{1}{2})({\mu }_R)+(-\frac{1}{2})({\mu }_E)\end{array}$

Then the sample contrast that estimates L is calculated as:

  $\begin{array}{l}\Rightarrow \widehat{L}=(1)({\overline{x}}_C)+(-\frac{1}{2})({\overline{x}}_R)+(-\frac{1}{2})({\overline{x}}_E)\\ =(1)(605.63)+(-\frac{1}{2})(657.31)+(-\frac{1}{2})(691.13)\\ =-68.59\end{array}$

And the standard error is calculated as:

  $\begin{array}{l}\text{SE}=(54.42)\times \sqrt{\frac{1^2}{19}+\frac{{(-\frac{1}{2})}^2}{16}+\frac{{(-\frac{1}{2})}^2}{15}}\\ =15.859\end{array}$

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure $26.1$ gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is $-68.59$ and its standard error is $15.859$ .

Now, we will define the hypotheses as:

  $\begin{array}{l}{H}_0:L=0\\ H_a:L<0\end{array}$

The value of test statistics is as:

  $\begin{array}{l}t=\frac{\widehat{L}}{\text{SE}}\\ =\frac{-68.59}{15.859}\\ =-4.32\end{array}$

And P-value is calculated as:

  $P<0.005$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a $95\%$ confidence interval.

Answer to Problem 26.6AYK

We are $95\%$ confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about $-100.64\text{ and}-36.54$ grams.

Explanation of Solution

In the question, it is given that figure $26.1$ gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is $-68.59$ and its standard error is $15.859$ .

Thus, the $95\%$ confidence interval can be calculated as:

  $\begin{array}{l}\widehat{L}\pm t*\text{SE}\\ =(-68.59)\pm 2.021(15.859)\\ =(-100.64,-36.54)\end{array}$

Thus, we are $95\%$ confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about $-100.64\text{ and}-36.54$ grams.