Chapter 24, Problem 24.10AYK

(a)

To determine

To explain do the standard deviations satisfy the rule of thumb for safe use of ANOVA.

Answer to Problem 24.10AYK

Yes, the standard deviations satisfy the rule of thumb for safe use of ANOVA.

Explanation of Solution

In the question, it is given that a study examined the impact of exercise type on visceral and subcutaneous fat. Overweight but otherwise disease free adults were assigned to three exercise regiments for eight months. The study report contains the information given in the question about the visceral fat reduction achieved by the subjects in each group. Thus, the standard deviation rule of thumb is that the largest sample standard deviation is no more than twice as large as the smallest standard deviation. Thus, the standard deviation is already given in the question, so let us find the ratio as:

  $\begin{array}{l}\frac{\text{Largest }s}{\text{Smallest }s}\\ =\frac{34}{19}\\ =1.79\end{array}$

Thus, the standard deviations satisfy the rule of thumb for safe use of ANOVA.

(b)

To determine

To explain why ANOVA is nonetheless safe for these data.

Explanation of Solution

In the question, it is given that a study examined the impact of exercise type on visceral and subcutaneous fat. Overweight but otherwise disease free adults were assigned to three exercise regiments for eight months. The study report contains the information given in the question about the visceral fat reduction achieved by the subjects in each group. Thus, the standard deviation rule of thumb is that the largest sample standard deviation is no more than twice as large as the smallest standard deviation. And the standard deviations satisfy the rule of thumb for safe use of ANOVA. The report does not provide the distributions of visceral fat reduction. But ANOVA nonetheless safe for these data because as we look at the means and standard deviations given then we can assume that they are approximately normally distributed and also all the other conditions are satisfied.

(c)

To determine

To calculate the overall mean response $\overline{x}$ , the mean square for groups (MSG) and the mean square for error (MSE).

Answer to Problem 24.10AYK

The overall mean response $\overline{x}$ is $8.96$ , the mean square for groups (MSG) is $2231.226$ and the mean square for error (MSE) is $852.3738$ .

Explanation of Solution

In the question, it is given that a study examined the impact of exercise type on visceral and subcutaneous fat. Overweight but otherwise disease free adults were assigned to three exercise regiments for eight months. The study report contains the information given in the question about the visceral fat reduction achieved by the subjects in each group. Thus, the standard deviation rule of thumb is that the largest sample standard deviation is no more than twice as large as the smallest standard deviation. And the standard deviations satisfy the rule of thumb for safe use of ANOVA. Thus, the overall mean response $\overline{x}$ , the mean square for groups (MSG) and the mean square for error (MSE) can be calculated as:

The calculations are as:

Treatment	n	x bar	n*x bar	n*(x-x total)^2	s^2	SS=(n-1)*s^2
1	36	15.9	=BO50*BP50	=BO50*(BP50-$BP$54)^2	=34^2	=(BO50-1)*BS50
2	39	0.8	=BO51*BP51	=BO51*(BP51-$BP$54)^2	=19^2	=(BO51-1)*BS51
3	35	10.9	=BO52*BP52	=BO52*(BP52-$BP$54)^2	=33^2	=(BO52-1)*BS52
Total	=SUM(BO50:BO52)		=SUM(BQ50:BQ52)	=SUM(BR50:BR52)		=SUM(BT50:BT52)
	Grand mean	=BQ53/BO53		SStr		SSE

Source of variation	df	SS	MS	F
Groups	=3-1	4462.452	=BP60/BO60	=BQ60/BQ61
Error	=110-3	91204	=BP61/BO61
Total	=BO60+BO61	=SUM(BP60:BP61)

The result will be as:

Treatment	n	x bar	n*x bar	n*(x-x total)^2	s^2	SS=(n-1)*s^2
1	36	15.9	572.4	1736.162	1156	40460
2	39	0.8	31.2	2593.946	361	13718
3	35	10.9	381.5	132.344	1089	37026
Total	110		985.1	4462.452		91204
	Grand mean	8.955455		SStr		SSE

Source of variation	df	SS	MS	F
Groups	2	4462.452	2231.226	2.617661
Error	107	91204	852.3738
Total	109	95666.45

Thus, the overall mean response $\overline{x}$ is $8.96$ , the mean square for groups (MSG) is $2231.226$ and the mean square for error (MSE) is $852.3738$ .

(d)

To determine

To obtain the ANOVA F statistic and the test P-value and explain is there evidence that the mean visceral fat reduction in overweight adults depends on which three exercise programs they follow.

Answer to Problem 24.10AYK

The ANOVA F statistic is $2.62$ and the test P-valueis between $0.05<P<0.10$ and there is no evidence that the mean visceral fat reduction in overweight adults depends on which three exercise programs they follow.

Explanation of Solution

In the question, it is given that a study examined the impact of exercise type on visceral and subcutaneous fat. Overweight but otherwise disease free adults were assigned to three exercise regiments for eight months. The study report contains the information given in the question about the visceral fat reduction achieved by the subjects in each group. Thus, the standard deviation rule of thumb is that the largest sample standard deviation is no more than twice as large as the smallest standard deviation. And the standard deviations satisfy the rule of thumb for safe use of ANOVA. And from part (d) we have the ANOVA table as:

Source of variation	df	SS	MS	F
Groups	2	4462.452	2231.226	2.617661
Error	107	91204	852.3738
Total	109	95666.45

Thus, the P-value is $0.05<P<0.10$ and 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,

  $P>0.05\Rightarrow \text{Fail to Reject }{H}_0$

Thus, we do not have sufficient evidence to conclude that the mean visceral fat reduction in overweight adults depends on which three exercise programs they follow.