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 (wg/g): Note: If you want to use this data has been set up so you may copy it directly into R. Wheat=c( 5.3, 4.5, 6.0, 6.2, 6.8, 5.8) Barley=c( 6.5, 7.9, 6.2, 7.6, 5.9, 5.6) Maize=c( 5.8, 4.8, 6.4, 4.8, 6.0, 5.1) Oats = [ 8.2, 6.0, 7.7, 7.0, 5.4, 7.2) In this example I = and J = What are the individual group means, xi.? * Wheat = x Barley = * Maize = x Oats What is the grand mean ? Calculate the mean squared treatment. (Round your answer to two decimal places.) MST = Ś 3.-L. - Calculate the variance for each group. Note: These can be done in R using the var() command. sa wheat = s_Barley = smaize = [ sOats = Calculate the mean squared error. (Round your answer to two decimal places.) MSE - 521+522+ ... +52:- Does this data suggest that at least two of the grains differ with respect to true average thiamin content? Use a level a = 0.05 test based on the P-value method. State the appropriate hypotheses. . Ho: ui = 42 = 43 = 44 Ha: at least two ui's are unequal HO: 41 = 42 = u3 = 44 Ha: all four ui's are unequal . HO: 41 * 2 * 3 * 44 Ha: at least two ui's are equal Ho: u1 u2 43 44 Ha: all four ui's are equal Compute the test statistic value.(Round your answer to two decimal places.) F-MSTE

Answer 1

1) Group means

a) Wheat mean=5.767

b) BArley mean=6.617

c) maize mean=5.483

d) Oats mean=6.917

2) Grand mean=

Grand mean=6.1958

3)Variance of each group

a) wheat variance=0.627

b) Barley variance=0.870

c) Maize variance=0.458

d)Oats variance=1.098

4) Mean square treatment

MSTr=8.3313/3=2.7771

5)Mean square error

MSE=15.2583/20=0.7629

6) Hypothesis

a)H0=$\mu 1=\mu 2=\mu 3=\mu 4$

H1= atleast two $\mu$i are unequal

7)Test statistic

F=MSTr/MSE=2.7771/0.7629=3.6400