2. -15 points My Note A researcher wants to determine the impact of soil type on

Question

2. -15 points My Note 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. Use this file: Assignment 9q2 data FOR R a) What null hypothesis is the researcher testing if she runs an ANOVA with this data? O One type of soil has a higher mean growth for the plant than the others. The mean growth of the plant is different in each type of soil O The mean growth of the plant in each type of soil is the same. The variability in growth of the plant in each type of soil is the same. Soil 3 provides a lower mean growth for the plant than the other types of soil. b) What is the SS for the ANOVA? Give your answer to at least three decimal places c) What is DF for the ANOVA? err d) What is the value of the F statistic for the ANOVA? Give your answer to at least three decimal places. e) Using a 0.05 level of significance, what conclusion should the researcher reach? Soil 3 has a lower mean growth for the plant than the other types of soi The mean growth of the plant is not the same for all soil types. O There is not enough evidence to reject the claim that the mean growth ofthe plant is the same in each type of soil. O Soil 1 has a higher mean growth for the plant than the other types of soil.

Explanation / Answer

a)

Null hypothesis in ANOVA is based on mean of those factors are equal, so here null hypothesis based on data is

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

b)

SStrt for ANOVA is 13.740

c)

error degree of freedom for ANOVA is 8

d)

Value of F statistic for ANOVA is 3.800.

e)

Using 0.05 level of significant P-value of this analysis is 0.058 which is greater than 0.05, so we failed to reject Ho.

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

I am solving this problem in MINITAB software and also uploaded the output of the same

MTB > Oneway 'Height in Inches' 'Soil Type'.

One-way ANOVA: Height in Inches versus Soil Type

Source    DF     SS      MS   F      P-value
Soil Type    3    13.74    4.58    3.80     0.058
Error         8    9.63    1.20
Total        11 23.37

S = 1.097   R-Sq = 58.78%   R-Sq(adj) = 43.32%

                        Individual 95% CIs For Mean Based on
                         Pooled StDev
Level N    Mean StDev ---------+---------+---------+---------+
1      3 13.167 0.981                   (---------*---------)
2      3 12.400 0.529              (---------*--------)
3      3 10.633 1.550 (---------*---------)
4      3 10.800 1.082   (---------*---------)
                         ---------+---------+---------+---------+
                               10.5      12.0      13.5      15.0

Pooled StDev = 1.097

Thank you....

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