Question

A researcher wants to determine the impact of soil type on the growth of a certain type of plant. She grows three plants in each of four different types of soil and measures the growth in inches for each plant after one month resulting in the data below.

Inches	Soil
12.2	1
12.8	1
11.9	1
10.8	2
12.2	2
12.3	2
9.3	3
9.9	3
10.8	3
13	4
11.8	4
11.9	4

a) What null hypothesis is the researcher testing if she runs an ANOVA with this data?

The mean growth of the plant is different in each type of soil.

The variability in growth of the plant in each type of soil is the same.    

The mean growth of the plant in each type of soil is the same.

One type of soil has a higher mean growth for the plant than the others.

Soil 3 provides a lower mean growth for the plant than the other types of soil.

b) What is the SS_trt for the ANOVA? Give your answer to at least three decimal places.  

c) What is DF_err for the ANOVA?  

d) What is the value of the F statistic for the ANOVA? Give your answer to at least three decimal places.  
e) Using a 0.1 level of significance, what conclusion should the researcher reach?

There is not enough evidence to reject the claim that the mean growth of the plant is the same in each type of soil.

Soil 3 has a lower mean growth for the plant than the other types of soil.    

Soil 1 has a higher mean growth for the plant than the other types of soil.

The mean growth of the plant is not the same for all soil types.

Answer 1

We get the following sample statistics

	Group 1	Group 2	Group 3	Group 4
1	12.2	10.8	9.3	13
2	12.8	12.2	9.9	11.8
3	11.9	12.3	10.8	11.9
Total	36.9	35.3	30	36.7
n	3	3	3	3
Mean	12.3	11.77	10	12.23
Variance	0.21	0.7033	0.57	0.4433
Std. Dev	0.45826	0.83865	0.75498	0.66581

The number of groups = k = 4

The number of Observations = N = 4 * 3 = 12

The overall mean = (12.3 + 11.77 + 10.12.23)/4 = 11.575

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a) The Null Hypothesis (Option 3): The mean growth of the plant in each type of soil is the same.

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b) SS treatment = SUM (n * (Mean - Overall Mean)2 = 3 * (12.3 - 11.575)² + 3 * (11.77 - 11.575)² + 3 * (10 - 11.575)² + 3 * (12.23 - 11.575)² = 10.420

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c) Df error = N - k = 12 - 4 = 8

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d) F Statistic = MS treatment / MS error

MS treatment = SS treatment / df treatment

df Treatment = k - 1 = 4 - 1 = 3

Therefore MS treatment = 10.42/3 = 3.473

MS within = SS error / Df error

SS error = SUM[ (n-1) * Variance)] = (2 * 0.21) + (2 * 0.7033) + (2 * 0.57) + (2 * 4433) = 3.85

Therefore MS error = 3.85/8 = 0.482

F statistic = 3.473 / 0.482 = 7.211

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e) The p value for F = 7.211, df between, df error = 3,8- p value = 0.0116

Since p value is < 90.1), we Reject H0

Option 4: The mean growth of the plant is not the same for all soil types.

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