Question

Numerator Degrees of Freedom de df, 2 3 4 5 7 8 9 1 2 3 4 5 6 7 3 Denominator Degrees of Freedom 10 11 12 13 14 15 16 17 161.
Question 5 9 pts Calculate the F-test value by completing the following table. df Sum of Squares F Mean Square [Select] [Sele


Calculate the degrees of freedom within groups 9 , the degrees of freedom between 2 net groups , and determine the F critical
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Answer #1

There are three treatments.

The overall means for all 12 observations, that is grand mean

\bar \bar x = \frac{Sum \ of \ all \ observations}{total \ observations} = \frac{372}{12} =31

Means for each tratment

1: Valerian Root, 2: Melatonin and 3: 5- HTP

\bar x_1 = \frac{Sum \ of \ all \ observations \ of \ valrian \ root}{total \ observations} = \frac{118}{4} =29.5

\bar x_2 = \frac{Sum \ of \ all \ observations \ of \ Melatonin}{total \ observations} = \frac{141}{4} =35.25

\bar x_3 = \frac{Sum \ of \ all \ observations \ of \ 5-HTP}{total \ observations} = \frac{113}{4} =28.25

The formula to find the total sum of squares

SST = \sum \sum (x_{ij}-\bar \bar x)^2

Subtact overall mean 31 from each 12 observations then take squares and add all of them to get the total sum of squares.

SST = \sum \sum (x_{ij}-\bar \bar x)^2 = 174

The formula for within sum of squares,

SSW = \sum_{j}\sum_{i}(x_{ij}-\bar x_j)^2

Subtract each treatment mean from the 4 observations of that same treatment, then take squares. Repeat same for the rest two treatments and then add all the squares.

The first treatment is 29.5 so subtract it from the observations 29, 32, 27 and 30 then take squares. Similarly subtract the means 35.25 and 28.25 from their corresponding observations and take square of them.

After that add all the 12 squares like 4 for first treatment, 4 for second and 4 for last treatment.

SSW = \sum_{j}\sum_{i}(x_{ij}-\bar x_j)^2= 62.5

Between sum of squares

SSB = SST - SSE = 174 - 62.5 = 111.5

Degrees of freedom

Degrees of freedom for between = k - 1 = 3 - 1 = 2

Degrees of freedom for within = n - k = 12 - 3 = 9

Degrees of freedom for total = n - 1 = 12 - 1 = 11

Mean square sums

Between mean sum of square(MSB)

MSB = \frac{SSB}{DF \ for \ between} = \frac{111.5}{2} = 55.75

Within mean sum of square (MSW)

MSW = \frac{SSW}{DF \ for \ within} = \frac{62.5}{9} = 6.944

F test statistics

F = \frac{MSB}{MSW} = \frac{55.75}{6.944} = 8.028

F test value = 8.028

The ANOVA table is,

Source of variation Sum of squares DF Mean square F
Between groups 111.5 2 55.75 8.028
Within groups 62.5 9 6.944
Total 174 11

The degrees of freedm for between and within groups are correct. Also the critical value at 0.05 is correct.

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