Practice of Statistics in the Life Sciences
Practice of Statistics in the Life Sciences
4th Edition
ISBN: 9781319013370
Author: Brigitte Baldi, David S. Moore
Publisher: W. H. Freeman
Question
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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.

(a)

Expert Solution
Check Mark

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:

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.28E

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 11221017.512.4545454545455=SQRT(CE28)
    Plot 21214111.7519.1136363636364=SQRT(CE29)
    Plot 3912313.666666666666720.25=SQRT(CE30)
    =SUM(CB28:CB30)=SUM(CC28:CC30)
    Grand mean==CC31/CB31

The results are:

    SUMMARY
    Groups Count Sum Average Variance Standard deviations
    Plot 11221017.512.454553.5291
    Plot 21214111.7519.113644.371914
    Plot 3912313.6666720.254.5
    33474
    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:

  Largest sSmallest s=20.2512.45=1.28

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.

(b)

Expert Solution
Check Mark

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: μ1=μ2=μ3 .

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 11221017.512.45455
    Plot 21214111.7519.11364
    Plot 3912313.6666720.25
    33474
    Grand mean=14.36364
    ANOVA
    Source of Variation SS df MS F P-value F crit
    Between Groups204.38642102.19326.0202170.0063363.31583
    Within Groups509.253016.975
    Total713.636432

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<0.05Reject H0

Thus, we have sufficient evidence to conclude that at least one of the mean is different from the other.

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