PROBLEM 1 6 samples of each of 3 types of cereal grain grown in a certain region

Question

PROBLEM 1

6 samples of each of 3 types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data:

The researchers would like to know whether on average there is any difference in thiamin content among different types of cereal.

(1) What statistical analysis procedure should be used in this situation?

(2) What are the basic assumptions for this procedure?

(3) Now assume the basic assumptions are all satisfied. Calculate the SSTr and SSE for this procedure. (You must show your calculation steps to earn credits. Any answer with no work will receive zero credit.)

(4) Complete the ANOVA table.

(5) What should we conclude in this research study? Use a proper approach to make a conclusion and explain why.

Cereal Typ Thiamin Content (ug/g) Wheat 5,2 4,5 6,0 6,1 6,7 5.8 Maize 5.8 4.7 6.4 4.9 6.0 5.2 Oats 8.3 6.1 7.8 7.0 5.5 7.2 d like to kn

Explanation / Answer

(1) One-Way Analysis of Variance (ANOVA)

(2) Assumptions of ANOVA:

(i) All populations involved follow a normal distribution.
(ii) All populations have the same variance (or standard deviation).
(iii) The samples are randomly selected and independent of one another.

(3) Grand Mean = 6.067

SST = (5.2-6.067)^2 + (4.5-6.067)^2 + (6-6.067)^2 + ..... + (7.2-6.067)^2 = 18.32

Mean of Thiamin in wheat = 5.716

Mean of Thiamin in maize = 5.5

Mean of Thiamin in Oats = 6.983

SSTR = [6*(5.716-6.067)^2 + 6*(5.5-6.067)^2 + 6*(6.983-6.067)^2] = 7.702

SSE = SST - SSTR = 18.32 - 7.702 = 10.618

(4) Complete the ANOVA table.

df for SST = N-1 = 18-1 = 17

MST = SST/(N-1) = 18.32/(18-1) = 1.077

df for SSTR = c-1 = 3-1 = 2

MSTR = SSTR/c-1 = 7.702/(3-1) = 3.851

df for SSE = N-c = 18-3 = 15

MSE = SSE/(N-c) = 10.618/(18-3) = 0.7078667

(5) What should we conclude in this research study? Use a proper approach to make a conclusion and explain why.

F = MSTR/MSE = 3.851/0.7078667 = 5.44

Critical value of F for alpha = 0.05 and numerator df = 2, denominator df = 15 is 0.9514

As, F (observed value) > F (critical value) , we reject the null hypothesis and conclude that the thiamine content is not statistically equal for wheat, maize and oats.

F at numerator for alpha = 0.05 and df = 1, denominator df = 15 is 0.826

Number of observations for each type = 6

Least Significant Difference (LSD)  = sqrt(2*MSE*0.826/6) = sqrt(2*0.7078667*0.826/6) = 0.4414

Mean difference between Wheat and Maize = 5.716 - 5.5 = 0.216

Mean difference between Wheat and Oats = 6.983 - 5.716 = 1.267

Mean difference between Maize and Oats = 6.983 - 5.5 = 1.483

So, the mean difference for Oats and Maize; Oats and Wheat is greater than 0.4414, so Thiamine content in Oats is different from wheat and maize. And Thiamine content in wheat is same as that of maize.

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