Chapter 24, Problem 24.28E

(a)

To determine

To plot the data and report the group means and standard deviations and explain what do the means suggest about the effect of logging on the number of tree species and explain are the conditions for inference with ANOVA met.

Answer to Problem 24.28E

The group means is $14.36$ and standard deviations are $3.53,4.37,4.5$ and yes, all the conditions for inference with ANOVA are met.

Explanation of Solution

In the question, it is given that the exercise $24.5$ described a study of rainforest recovery after logging so, we have to use a software to analyze the effect of logging on the number of species per plot provided in table $24.2$ . Thus, now the plot for the data is as:

By looking at the line plot we can see that most of the data from the three plot is in the middle part and therefore it is approximately normally distributed. And secondly the species are randomly assigned to the plots. Also, by using the software excel we select the data and with the help of data analyses option in data tab we construct the ANOVA test and the result is as:

SUMMARY
Groups	Count	Sum	Average	Variance	Standard deviations
Plot 1	12	210	17.5	12.4545454545455	=SQRT(CE28)
Plot 2	12	141	11.75	19.1136363636364	=SQRT(CE29)
Plot 3	9	123	13.6666666666667	20.25	=SQRT(CE30)
	=SUM(CB28:CB30)	=SUM(CC28:CC30)
	Grand mean=	=CC31/CB31

The results are:

SUMMARY
Groups	Count	Sum	Average	Variance	Standard deviations
Plot 1	12	210	17.5	12.45455	3.5291
Plot 2	12	141	11.75	19.11364	4.371914
Plot 3	9	123	13.66667	20.25	4.5
	33	474
	Grand mean=	14.36364

Thus, the third condition for ANOVA is that the largest sample standard deviation is no more than twice as large as the smallest standard deviation, so we have the ratio as:

  $\begin{array}{l}\frac{\text{Largest }s}{\text{Smallest }s}\\ =\frac{\sqrt{20.25}}{\sqrt{12.45}}\\ =1.28\end{array}$

Thus, it is also satisfied therefore all the inferences for ANOVA are met. And the means suggest about the effect of logging on the number of tree species that all the plots averages are approximately equal and therefore, there is not very much difference in means for the three plot for species.

(b)

To determine

To carry out the ANOVA and report the F statistic and its P-value and conclude in context.

Answer to Problem 24.28E

The F statistic is $6.02$ and its P-valueis $0.0063$ .

Explanation of Solution

In the question, it is given that the exercise $24.5$ described a study of rainforest recovery after logging so, we have to use a software to analyze the effect of logging on the number of species per plot provided in table $24.2$ . The null and alternative hypothesis are defined as:

Null hypothesis: ${\mu }_1={\mu }_2={\mu }_3$ .

Alternative hypothesis: At least one of them is different.

Thus, by using the software excel we select the data and with the help of data analyses option in data tab we construct the ANOVA test and the result is as:

ANOVA: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Plot 1	12	210	17.5	12.45455
Plot 2	12	141	11.75	19.11364
Plot 3	9	123	13.66667	20.25
	33	474
	Grand mean=	14.36364
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	204.3864	2	102.1932	6.020217	0.006336	3.31583
Within Groups	509.25	30	16.975
Total	713.6364	32

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 at least one of the mean is different from the other.