Please help! A researcher wants to determine the impact of soil type on the grow

ID: 3225266 • Letter: P

Question

Please help!

A researcher wants to determine the impact of soil type on the growth of a certain type of plant. She grows three plants in the growth in inches for each plant after one month resulting in the data below. Use this data: 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. Soil 3 provides a lower mean growth for the plant than the other types 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. 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. The mean growth of the plant is not the same for all soil types. Soil 1 has a higher mean growth for the plant than the other types of soil.

Explanation / Answer

Solution

Back-up Theory

Let xij represent the jth observation in the ith row,(i.e., the growth plant j in soil i)

k = 1,2,…,n; i = 1,2,……,r

Then the ANOVA model is: xij = µ + i + ij,…………………………………………..(1)

where µ = common effect, i = effect of ith row (soil), and ij is the error component which is assumed to be Normally Distributed with mean 0 and variance 2.

Null hypothesis: H0: 1 = 2 = 3 = ….. = r Vs HA: H0 is false.……………………..(2)

Now, to work out the solution,

Terminology:

Row total = xi.= sum over j of xij

Grand total = G = sum over i of xi.

Correction Factor = C = G2/N, where N = total number of observations = r x n =

Total Sum of Squares: SST = (sum over i,j of xij2) – C

Row Sum of Squares: SSR = {(sum over i of xi.2)/(n)} – C

Error Sum of Squares: SSE = SST – SSR

Mean Sum of Squares = Sum of squares/Degrees of Freedom

Degrees of Freedom:

Total: N (i.e., rn) – 1;

Error: DF for Total – DF for Row;

Rows: (r - 1);

Fobs: MSSR/MSSE;

Fcrit: upper % point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for Row and n2 is the DF for Error

Significance: Fobs is significant if Fobs > Fcrit

Now, to work out solution,

ANOVA table is given below Part (e) solution

Part (a)

[vide (2) under Back-up Theory],

Null hypothesis: H0: 1 = 2 = 3 = 4 (i.e., mean growth rate of plants is the same for all 4 types of soil.) Vs HA: H0 is false.

Option 4 ANSWER

Part (b)

SStrt = Sum of squares for treatment = SSR = 21.403 (see ANOVA table) ANSWER

Part (c)

DF for Error = DF for Total - DF for Row = (n - 1) - (r - 1) = n – r = 8. ANSWER

Part (d)

Value of F-statistic = 9.276 (see ANOVA table) ANSWER

Part (e)

Since Fcal > Fcrit, null hypothesis is rejected. => mean growth rate of plants is not the same for all soil types option 3 ANSWER

ANOVA

0.1

Source

Fcrit

Row

21.40333

7.134444

9.275551

2.924

Error

6.153333

0.769167

Total

27.55667