Chapter 12, Problem 49E

(a)

To determine

To find: The expression of contrast for each of the treatments in terms of their respective means.

Answer to Problem 49E

Solution: The expression of contrast for the experiment including the treatment of Placebo verses average of two low dose treatment is given below:

${\psi }_1={\mu }_1-0.5{\mu }_2-0.5{\mu }_4$

The expression of contrast for the experiment including the treatment of difference between high A and low A verses high B and low B is given below:

${\psi }_2=\left({\mu }_3-{\mu }_2\right)-\left({\mu }_5-{\mu }_4\right)$

Explanation of Solution

Calculation: The average of means of low A and low B treatment is $0.5\left({\mu }_2+{\mu }_4\right)$ and the mean of Placebo treatment is ${\mu }_1$. Therefore, null and alternative hypotheses can be formed as follows:

$\begin{array}{l}{H}_0:{\psi }_1=0\\ H_a:{\psi }_1>0\end{array}$

Null hypotheses represent that there is no difference in the Placebo treatment and average of low A and low B treatment, which is represented as below:

$H_0:{\mu }_1=0.5\left({\mu }_2+{\mu }_4\right)$

Hence, on equating ${\psi }_1=0$, the expression of contrast can be obtained as given below:

${\mu }_1-0.5\left({\mu }_2+{\mu }_4\right)=0$, since ${\psi }_1$ is equal to 0.

The difference of means of low A and high A treatment is $\left({\mu }_3-{\mu }_2\right)$ and the difference of means of high B and low B treatment is $\left({\mu }_5-{\mu }_4\right)$. Therefore, null and alternative hypotheses can be formed as follows:

$\begin{array}{l}{H}_0:{\psi }_2=0\\ H_a:{\psi }_2>0\end{array}$

Null hypotheses represent that there is no difference in the difference of low A and high A treatment and difference of high B and low B treatment, which is represented as below:

$H_0:\left({\mu }_3-{\mu }_2\right)=\left({\mu }_5-{\mu }_4\right)$

Hence, on equating ${\psi }_1=0$, expression of contrast can be obtained as given below:

$\left({\mu }_3-{\mu }_2\right)-\left({\mu }_5-{\mu }_4\right)=0$, since ${\psi }_2=0$

The coefficients of contrast for the contrast ${\psi }_1$ are given below:

$\begin{array}{c}{a}_1=1\\ a_2=-0.5\\ a_3=0\\ a_4=-0.5\end{array}$

$a_5=0$

The coefficients of contrast for the contrast ${\psi }_2$ are given below:

$\begin{array}{c}{a}_1=0\\ a_2=-1\\ a_3=1\\ a_4=1\end{array}$

$a_5=-1$

(b)

To determine

To find: The sample and standard error for each of the contrast.

Answer to Problem 49E

Solution: The sample of contrast $c_1$ for the contrast expression ${\psi }_1$ is $\underset{\_}{-3.9}$ and the sample of contrast $c_2$ for the contrast expression ${\psi }_2$ is 2.35.

The standard error for the contrast $c_1$ is 2.1353 and the standard error for the contrast $c_2$ is 3.487.

Explanation of Solution

Calculation: The value of contrast $c_1$ can be calculated by using the formula given below:

$\begin{array}{c}{c}_1={\displaystyle \sum _i{a}_i{\overline{x}}_i}\\ =\left(1\times 11.80\right)+\left(\left(-0.5\right)\times 15.25\right)+\left(\left(-0.5\right)\times 16.15\right)\\ =-3.9\end{array}$

The value of contrast $c_2$ can be calculated by using the formula given below:

$c_2={\displaystyle \sum _i{a}_i{\overline{x}}_i}$

$\begin{array}{c}{c}_2=\left(\left(-1\right)\times 15.25\right)+\left(1\times 18.55\right)+\left(1\times 16.15\right)+\left(\left(-1\right)\times 17.10\right)\\ =2.35\end{array}$

Standard error can be calculated by using the formula given below:

$S{E}_{c_1}=s_{pooled}\sqrt{{\displaystyle \sum \frac{a_i{}^2}{n_i}}}$

Pooled standard deviation can be calculated as given below:

$\begin{array}{c}{s}_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right)s_2{}^2+\cdots +\left(n_5-1\right)s_2{}^2}{n_1+n_2+n_3+n_4+n_5-k}}\\ =\sqrt{\frac{\left(5-1\right)17.20+\left(5-1\right)13.10+\cdots +\left(5-1\right)12.50}{\left(5+5+5+5+5\right)-5}}\\ =3.487\end{array}$

For the first contrast, the standard error $S{E}_{c_1}$ can be calculated as given below:

$\begin{array}{c}S{E}_{c_1}=3.487\sqrt{\frac{1}{4}+\frac{{\left(-0.5\right)}^2}{4}+\frac{{\left(-0.5\right)}^2}{4}}\\ =3.487\times \sqrt{0.375}\\ =2.1353\end{array}$

For the second contrast, the standard error $S{E}_{c_2}$ can be calculated as given below:

$\begin{array}{c}S{E}_{c_1}=3.487\sqrt{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}\\ =3.487\times 1\\ =3.487\end{array}$

Interpretation: Therefore, it can be concluded that for the first contrast, the estimated value of standard deviation is 2.1353 and for the second contrast, the estimated value of standard deviation is 3.487.

(c)

To determine

To test: The significance test for each of the contrast and summarize the obtained results.

Answer to Problem 49E

Solution: The test statistic for contrast of the experiment including the treatment of Placebo verses average of two low dose treatment is 0.0414 and the test statistics for contrast for the experiment including the treatment of difference between high A and low A verses high B and low B is 0.2540. For the first contrast, the result is significant but for the second contrast the result is insignificant.

Explanation of Solution

Calculation: Test statistic can be calculated by using following formula, which is mentioned below:

$t=\frac{c}{S{E}_c}$

To calculate the test statistic for first contrast, the value of $c_1=-3.9$ and is substituted in the formula mentioned above:

$\begin{array}{l}t=\frac{c}{S{E}_c}\\ =\frac{-3.9}{2.1353}\\ =-1.8264\end{array}$

The total sample size of rats is 25. There are five groups, which are represented by k.

The degree of freedom for contrast can be obtained as shown below:

$\begin{array}{c}N-k=25-5\\ =20\end{array}$

The P-value can also be calculated by the software Microsoft Excel by using the following command:

Insert the values of x and degree of freedom, which are –1.8264 and 20, respectively. After this, press enter and the output is shown below in the snapshot:

Hence, the P- value is obtained as 0.0414, which is less than the significance level 0.05, so the decision is to reject the null hypothesis.

To calculate the test statistic for second contrast, the value of $c_2=2.35$ and $S{E}_{c_2}=3.487$ is substituted in the formula mentioned above:

$\begin{array}{l}t=\frac{c}{S{E}_c}\\ =\frac{2.35}{3.487}\\ =0.6739\end{array}$

The P-value can also be calculated by the software Microsoft Excel by using the following command:

Insert the values of x and degree of freedom, which is 0.6739 and 20, respectively. After this, press enter and the output is shown below in the snapshot:

Hence, the P- value is obtained as 0.2540, which is greater than the significance level 0.05, so the decision is to accept the null hypothesis.

Conclusion: Therefore, it can be concluded that the result is significant for the first contrast and the result is not significant for the second contrast.