1.In order to determine the most effective of six different colors of soda on pe
ID: 3361293 • Letter: 1
Question
1.In order to determine the most effective of six different colors of soda on people’s taste, 60 people were randomly divided in to six groups and each group tasted one of the six colors of the soda. The soda was otherwise the same in every case. The people rated their taste reaction and the scores were recorded and the results analyzed. The ANOVA software produced the following output.
SOURCE DF SS MS F
Model ___ 831 ______ ______
Error ___ 7114 ______
Total ___
State the hypotheses being tested in this ANOVA, fill in the table, and draw conclusions at the .05 level. Be sure to include a statement to be tested, describe the random variables involved and assumptions about them, level of significance, test statistic, and the critical region.
Here are the results of a Duncan multiple range test:
Color
N
Subset for alpha = .05
1
4
10
94
6
10
98
1
10
101
5
10
103
3
10
104
2
10
105
What does this tell us about how color affects taste?
* please try to help me understand this probelm in most simplest hand calculation if possible ( note - please write clear so i can clearly read out properly for my own understanding ... thank you very much )....
Color
N
Subset for alpha = .05
1
4
10
94
6
10
98
1
10
101
5
10
103
3
10
104
2
10
105
Explanation / Answer
null hypothesis H0: all colour means are same i.e. equal
alternate hypothesis H1: at least one colour mean differ from others
ther are 6 colour so model df=6-1=5 , total cases =60 so total df=60-1=59 and
error df=total df- model df
MS=SS/df and F=MS(model)/MS(error) with (model df, error df)
SOURCE DF SS MS F
Model 5 831 166.2 1.26
Error 54 7114 131.75
Total 59
critical F(0.05,5,54)=2.39 is more than calculated F=1.26, so we fail to reject H0 and conclude that there is no difference among all colors.
range=highest mean - lowerst mean=105-94=11
q value for DMRT for 6 treatment and 54 error df is given as q(0.05,6,54)=3.814 ( available on textbook or internet)
DMRT critical difference =sqrt(2*mse/t)*q=sqrt(2*131.75/6)*3.814= 25.28
here DMRT critical difference =25.28 is more than range=11 so, non of the pair is significant
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