Six samples of each of four types of cereal grain grown in a certain region were

Question

Six samples of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data (µg/g): Note: If you want to use R this data has been set up so you may copy it directly into R.

Wheat=c( 5.2, 4.4, 6.0, 6.0, 6.7, 5.8) Barley=c( 6.4, 8.0, 6.1, 7.4, 5.8, 5.7) Maize=c( 5.7, 4.7, 6.5, 4.9, 5.9, 5.2) Oats = c( 8.2, 6.0, 7.9, 6.9, 5.4, 7.1) Six samples of each of four types of cereal grain grown in a certain region were analyzed to determine thiamin content, resulting in the following data (ug/g): Note: If you want to use R this data has been set up so you may copy it directly into R. Wheat-c( 5.2, 4.4, 6.0, 6.0, 6.7, 5.8) Barley c( 6.4, 8.0, 6.1 7.4, 5.8, 5.7) Maize-c( 5.7, 4.7 6.5 4.9 5.9 5.2) Oats 8.2 6.0 7.9 6.9 5.4 7.1) v and J 6 In this example I 4 What are the individual group means, xi.? x wheat 5.68 X Barley 6.57 X Maize 5.48 X oats 6.92 What is the grand mean x. X.. 6.16 Calculate the mean squared treatment. Round your answer to two decimal places.) x )2* 2.88 MSTr I-1 Calculate the variance for each group. Note: These can be done in R using the var() command. wheat 0.5214 s Barley 0.577 Maize 0.38

Explanation / Answer

We calculate the varianc of each group. The results are

In R,

For MSE

in R,

F = MSTr/ MSE = 2.88 / 0.9835 = 2.9283

wheat=c(5.2,4.4,6,6.7,5.8)
Barley=c(6.4,8,6.1,7.4,5.8,5.7)
maize=c(5.7,4.7,6.5,4.9,5.9,5.2)
oats=c(8.2,6.0,7.9,6.9,5.4,7.1)
var(wheat)
var(Barley)
var(oats)
var(oats) > var(wheat)
[1] 0.752
> var(Barley)
[1] 0.8666667
> var(oats)
[1] 1.157667
> var(oats)
[1] 1.157667

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