A farmer is trying to maximize milk production in his cows and is testing the ro

Question

A farmer is trying to maximize milk production in his cows and is testing the role of diet and exercise. He is hypothesizing that grass is the best diet because cows are adapted to eating grasses. He is also hypothesizing that exercise will affect hormone and stress levels and in turn affect milk production. He fed his cows three diets and treated them to three different exercise regimes. What type of statistics should be used to analyze this problem? Include the null and alternative hypothesis. Fully analyze his results and interpret the outcome.

Explanation / Answer

Solution

A two-way ANOVA with equal number of observations per cell will be the most appropriate statistics to be applied.

Back-up Theory

Suppose we have data of a 2-way classification ANOVA, with r rows, c columns and n observations per cell.

Let xijk represent the kth observation (milk production – L/day) in the kth cell (cow) of ith row (diet)-jth column (exercise), k = 1,2,…,n (3); i = 1,2,……,r (3) ; j = 1,2,…..,c. (3)

Then the ANOVA model is: xijk = µ + i + j + ij + ijk, where µ = common effect, i = effect of ith row, j = effect of jth column, ij = row-column interaction and ijk is the error component which is assumed to be Normally Distributed with mean 0 and variance 2.

Now, to work out the solution,

Terminology:

Cell total = xij. = sum over k of xijk

Row total = xi..= sum over j of xij.

Column total = x.j. = sum over i of xij.

Grand total = G = sum over i of xi.. = sum over j of x.j.

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

Total Sum of Squares: SST = (sum over i,j and k of xijk2) – C

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

Column Sum of Squares: SSC = {(sum over j of x.j.2)/(rxn)} – C

Between Sum of Squares: SSB = {(sum over i and jof xij.2)/n} – C

Interaction Sum of Squares: SSI = SSB – SSR – SSC

Error Sum of Squares: SSE = SST – SSB

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

Degrees of Freedom:

Total: N (i.e., rcn) – 1;

Between: rc – 1;

Within(Error): DF for Total – DF for Between;

Rows: (r - 1);

Columns: (c - 1);

Interaction: DF for Between – DF for Rows – DF for Columns;

Fobs:

for Rows: MSSR/MSSE;

for Columns: MSSC/MSSE;

for Interaction: MSSI/MSSE

Fcrit: upper % point of F-Distribution with degrees of freedom n1 and n2, where n1 is the DF for the numerator MSS and n2 is the DF for the denominator MSS of Fobs

Significance: Fobs is significant if Fobs > Fcrit

Now, to work out solution,

Null hypothesis H01: 1 = 2 = 3 i.e., effects of diet is the same for all 3 types.

Null hypothesis H02: 1 = 2 = 3 i.e., effects of exercise is the same for all 3 types.

Null hypothesis H03: ij = 0 i.e., there is no interaction effect of diet and exercise.

Alternatives: H01, H02 and H03 are false

Calculations:

r =

3

c =

3

n =

3

G =

210

C =

1633.333

=

0.05

ANOVA TABLE

Source

DF

SS

MSS

Fobs

Fcrit

Row

2

353.5556

176.7778

86.78182

3.554557

Column

2

46.88889

23.44444

11.50909

3.554557

Interaction

4

7.555556

1.888889

0.927273

2.927744

Between

8

408

51

Error

18

36.66667

2.037037

Total

26

444.6667

17.10256

Since Fobs > Fcrit for both Row and Column, H01 and H02 are rejected.

=> both diet and exercise have an effect on the milk production of cows.

Since Fobs < Fcrit, H03 is accepted.

=> there is no evidence of interaction between diet and exercise so far as milk production is concerned.

DONE

r =

3

c =

3

n =

3

G =

210

C =

1633.333

=

0.05

ANOVA TABLE

Source

DF

SS

MSS

Fobs

Fcrit

Row

2

353.5556

176.7778

86.78182

3.554557

Column

2

46.88889

23.44444

11.50909

3.554557

Interaction

4

7.555556

1.888889

0.927273

2.927744

Between

8

408

51

Error

18

36.66667

2.037037

Total

26

444.6667

17.10256

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.