Chapter 30, Problem 30.31E

a.

To determine

To identify: The population contrast which expresses this comparison and null and alternative hypotheses in term of this contrast.

Answer to Problem 30.31E

Answer:

The population contrast that expresses the comparison is:

$L=\left(1\right){\mu }_{1.00}+\left(-0.25\right){\mu }_{0.00}+\left(-0.25\right){\mu }_{0.50}+\left(-0.25\right){\mu }_{1.25}+\left(-0.25\right){\mu }_{1.50}$.

Null Hypothesis:

$H_0:L=0$

Alternative hypotheses:

$H_a:L>0$.

Explanation of Solution

Explanation:

Given info:

The researcher hypothesized that the healing is fastest under natural condition if the population mean for the group 1 is larger than the average of the population means for the other four groups.

Justification:

Population contrast:

The population contrast is a combination of the means ${\mu }_1,{\mu }_2,\dots ,{\mu }_I$ of $I$ populations.

$L=c_1{\mu }_1+c_2{\mu }_2+\cdots +c_I{\mu }_I$

The sum of numerical coefficient is 0, $c_1+c_2+\cdots +c_I=0$.

The contrast of the mean healing rate for the natural field of the newts (1.00 group) with the non-natural field of the newts (0.00, 0.5, 1.25 and 1.50 group). The sample contrast is written as shown below,

$\begin{array}{c}L=\left(1\right){\mu }_{1.00}-\frac{\left(1\right){\mu }_{0.00}+\left(1\right){\mu }_{0.50}+\left(1\right){\mu }_{1.25}+\left(1\right){\mu }_{1.50}}{4}\\ ={\mu }_{1.00}-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right){\mu }_{0.00}-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right){\mu }_{0.50}-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right){\mu }_{1.25}-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right){\mu }_{1.50}\\ ={\mu }_{1.00}-0.25{\mu }_{0.00}-0.25{\mu }_{0.50}-0.25{\mu }_{1.25}-0.25{\mu }_{1.50}\end{array}$

The sum of the numerical coefficients is $\left(1-0.25-0.25-0.25-0.25\right)=0$.

The null hypothesis is a statement of equality. It represents that there is no significant difference between the groups or variables.

Null Hypothesis:

$H_0:L=0$ .

It means there is mean healing rate for natural electrical field is equal to the combined mean of the other four groups or there is no significant difference between the healing rate of natural electrical field and non-natural electrical field.

Alternative hypotheses:

$H_a:L>0$.

It means there is mean healing rate for natural electrical field is greater than the combined mean of the other four groups or the natural healing rate is fastest.

Therefore, the sample contrast expresses this comparison is:

$L=\left(1\right){\mu }_{1.00}+\left(-0.25\right){\mu }_{0.00}+\left(-0.25\right){\mu }_{0.50}+\left(-0.25\right){\mu }_{1.25}+\left(-0.25\right){\mu }_{1.50}$.

The null and alternative hypothesis are:

Null Hypothesis:

$H_0:L=0$

Alternative hypotheses:

$H_a:L>0$.

b.

To determine

To test: The conclusion using t statistic test.

Answer to Problem 30.31E

Answer:

Explanation of Solution

Explanation:

Calculation:

The researchers conclude that the healing is fastest under natural condition if the population mean for the group 1 is larger than the average of the population means for the other four groups. The effect of electrical Field on healing rate in newts is show in the given table.

Group	Diff	Group	Diff	Group	Diff	Group	Diff	Group	Diff
0	$-10$	0.5	$-1$	1	$-7$	1.25	1	1.5	$-13$
0	$-12$	0.5	10	1	15	1.25	8	1.5	$-49$
0	$-9$	0.5	3	1	$-4$	1.25	$-15$	1.5	$-16$
0	$-11$	0.5	$-3$	1	$-16$	1.25	14	1.5	$-8$
0	$-1$	0.5	$-31$	1	$-2$	1.25	$-7$	1.5	$-2$
0	6	0.5	4	1	$-13$	1.25	$-1$	1.5	$-35$
0	$-31$	0.5	$-12$	1	5	1.25	11	1.5	$-11$
0	$-5$	0.5	$-3$	1	$-4$	1.25	8	1.5	$-46$
0	13	0.5	$-7$	1	$-2$	1.25	11	1.5	$-22$
0	$-2$	0.5	$-10$	1	$-14$	1.25	$-4$	1.5	2
0	$-7$	0.5	$-22$	1	5	1.25	7	1.5	10
0	$-8$	0.5	$-4$	1	11	1.25	$-14$	1.5	$-4$
		0.5	$-1$	1	10	1.25	0	1.5	$-10$
		0.5	$-3$	1	3	1.25	5	1.5	2
				1	6	1.25	$-2$	1.5	$-5$
				1	$-1$
				1	13
				1	$-8$

Software procedure:

Step by step procedure to obtain ANOVA using the MINITAB software:

Choose Stat > ANOVA > One-Way.

In Response, select the column of Diff.

In Factor, select the column of Group.

Click OK.

Output using the MINITAB software is given below,

From output it is noticed that the mean of group 0.00 is $-6.42$, mean of group 0.50 is $-5.71$, mean of group 1.00 is $-0.17$, mean of group 1.25 is 1.47 and mean of group 1.50 is $-13.80$.

The sample contrast is written as shown below,

$L=\left(1\right){\mu }_{1.00}+\left(-0.25\right){\mu }_{0.00}+\left(-0.25\right){\mu }_{0.50}+\left(-0.25\right){\mu }_{1.25}+\left(-0.25\right){\mu }_{1.50}$

Substitute $-6.42$ for ${\mu }_{0.00}$, $-5.71$ for ${\mu }_{0.50}$, $-0.17$ for ${\mu }_{1.00}$, $1.47$ for ${\mu }_{1.25}$ and $-13.8$ for ${\mu }_{1.50}$ in the above equation.

$\begin{array}{c}\stackrel{⌢}{L}=\left\{\begin{array}{c}\left(1\right)\left(-0.17\right)+\left(-0.25\right)\left(-6.42\right)+\left(-0.25\right)\left(-5.71\right)\\ +\left(-0.25\right)\left(1.47\right)+\left(-0.25\right)\left(-13.8\right)\end{array}\right\}\\ =-0.17+1.605+1.4275-0.3675+3.45\\ =5.945\end{array}$

The sample contrast is 5.945.

The formula for standard error of the sample contrast is given below,

$S{E}_{\widehat{L}}=s_p\sqrt{\frac{c_1^2}{n_1}+\frac{c_2^2}{n_2}+\frac{c_3^2}{n_3}+\frac{c_4^2}{n_4}+\frac{c_5^2}{n_5}}$

From Fig (1), it is noticed that the value of pooled standard deviation is approximately 11.75.

Substitute 11.75 for $s_p$, 1 for $c_1$, 18 for $n_1$, $-0.25$ for $c_2$, $-0.25$ for $c_3$, $-0.25$ for $c_4$, $-0.25$ for $c_5$, 12 for $n_2$, 14 for $n_3$, 15 for $n_4$ and 15 for $n_5$ in the above equation.

$\begin{array}{c}S{E}_{\widehat{L}}=\left(11.75\right)\sqrt{\frac{{\left(1\right)}^2}{18}+\frac{{\left(-0.25\right)}^2}{12}+\frac{{\left(-0.25\right)}^2}{14}+\frac{{\left(-0.25\right)}^2}{15}+\frac{{\left(-0.25\right)}^2}{15}}\\ =\left(11.75\right)\sqrt{\frac{1}{18}+\frac{0.0625}{12}+\frac{0.0625}{14}+\frac{0.0625}{15}+\frac{0.0625}{15}}\\ =\left(11.75\right)\sqrt{0.056+0.0052+0.0044+0.0042+0.0042}\end{array}$

Further solve the above equation.

$\begin{array}{c}S{E}_{\widehat{L}}=\left(11.75\right)\sqrt{\left(0.074\right)}\\ =\left(11.75\right)\left(0.272\right)\\ =3.196\end{array}$

The formula for t test statistic is,

$\begin{array}{c}t=\frac{\widehat{L}}{S{E}_{\widehat{L}}}\\ =\frac{5.945}{3.196}\\ =1.86\end{array}$

Thus, $\underset{\_}{t\text{-statistic}=1.86}$.

Decision rule:

Reject $H_0$ only if $t\text{-statistic value}>\text{critical value}$.

Critical value:

The value of t at 5% level of significance and 73 degree of freedom is approximately 1.66. Thus, $\text{critical value}=1.66$.

Since $t\text{-statistic}\left(=1.86\right)>\text{critical value}\left(=1.66\right)$ the null hypothesis is rejected and it can be concluded that healing is fastest under natural conditions.

c.

To determine

To explain: The reason it is not legitimate to contrast the average for the 1 and 1.25 groups with the average for the other three groups.

Explanation of Solution

Explanation:

Justification:

From part (b), it is noticed that the mean of group 0.00 is $-6.42$, mean of group 0.50 is $-5.71$, mean of group 1.00 is $-0.17$, mean of group 1.25 is 1.47 and mean of group 1.50 is $-13.80$.

From the above information it is noticed that the mean of groups 1.00 and 1.25 is higher that the other three groups. So, it is tempting to contrast the average for the 1 and 1.25 groups with the average for the other three groups.

However, it is not legitimate because researcher hypothesized that the healing is fastest under natural condition if the population mean for the group 1 is larger than the average of the population means for the other four groups.

Therefore, the contrast the average for the 1 and 1.25 groups with the average for the other three groups is tempting but it is not legitimate because according to the researcher the mean of group 1.00 is better than the combined mean of the others, which is the hypothesis of interest. The contrast would not serve the interest of the researcher’s hypothesis.