Question

To test whether or not there is a difference between treatments A, B, and C, a sample of 12 observations has been randomly assigned to the 3 treatments. You are given the results below.

Treatment | Observation
A | 20 | 30 | 25 | 33
B | 22 | 26 | 20 | 28
C | 40 | 30 | 28 | 22

1.The null hypothesis for this ANOVA problem is?

2.The mean square between treatments (MSTR) equals:
A. 1.872
B. 5.86
C.34
D.36

3.The mean square within treatments (MSE) equals:
A.1.872
B. 5.86
C. 34
D.36

4. The test statistic to test the null hypothesis equals:
A. .944
B.1.059
C. 3.13
D. 19.231

5. The null hypothesis is to be tested at the 1% level of significance. The critical value from the table is
A.4.26
B.8.02
C. 16.69
D. 99.39

Answer 1

1. Null hypothesis:
`\mu_(A)=\mu_(B)=\mu_(C)`

Alternative hypothesis: Not all the means are equal
`\mu_(i)\\eq \mu_(j), i,j=A,B,C`

2. D. 36

3. C. 34

4. B. 1.059

5. B. 8.02

Explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part 1

The hypothesis for this case are:

Null hypothesis:
`\mu_(A)=\mu_(B)=\mu_(C)`

Alternative hypothesis: Not all the means are equal
`\mu_(i)\\eq \mu_(j), i,j=A,B,C`

Part 2

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have
`p` groups and on each group from
`j=1,\dots,p` we have
`n_j` individuals on each group we can define the following formulas of variation:

`SS_(total)=\sum_(j=1)^p \sum_(i=1)^(n_j) (x_(ij)-\bar x)^2`

`SS_(between)=SS_(model)=\sum_(j=1)^p n_j (\bar x_(j)-\bar x)^2`

`SS_(within)=SS_(error)=\sum_(j=1)^p \sum_(i=1)^(n_j) (x_(ij)-\bar x_j)^2`

And we have this property

`SST=SS_(between)+SS_(within)`

We need to find the mean for each group first and the grand mean.

`\bar X =(\sum_(i=1)^n x_i)/(n)`

If we apply the before formula we can find the mean for each group

`\bar X_A = 27`,
`\bar X_B = 24`,
`\bar X_C = 30`. And the grand mean
`\bar X = 27`

Now we can find the sum of squares between:

`SS_(between)=SS_(model)=\sum_(j=1)^p n_j (\bar x_(j)-\bar x)^2`

Each group have a sample size of 4 so then
`n_j =4`

`SS_(between)=SS_(model)=4(27-27)^2 +4(24-27)^2 +4(30-27)^2=72`

The degrees of freedom for the variation Between is given by
`df_(between)=k-1=3-1=2`, Where k the number of groups k=3.

Now we can find the mean square between treatments (MSTR) we just need to use this formula:

`MSTR=(SS_(between))/(k-1)=(72)/(2)=36`

D. 36

Part 3

For the mean square within treatments value first we need to find the sum of squares within and the degrees of freedom.

`SS_(within)=SS_(error)=\sum_(j=1)^p \sum_(i=1)^(n_j) (x_(ij)-\bar x_j)^2`

`SS_(error)=(20-27)^2 +(30-27)^2 +(25-27)^2 +(33-27)^2 +(22-24)^2 +(26-24)^2 +(20-24)^2 +(28-24)^2 +(40-30)^2 +(30-30)^2 +(28-30)^2 +(22-30)^2 =306`

And the degrees of freedom are given by:

`df_(within)=N-k =3*4 -3 = 12-3=9`. N represent the total number of individuals we have 3 groups each one with a size of 4 individuals. And k the number of groups k=3.

And now we can find the mean square within treatments:

`MSE=(SS_(within))/(N-k)=(306)/(9)=34`

C. 34

Part 4

The test statistic F is given by this formula:

`F=(MSTR)/(MSE)=(36)/(34)=1.059`

B. 1.059

Part 5

The critical value is from a F distribution with degrees of freedom in the numerator of 2 and on the denominator of 9 such that we have 0.01 of the area in the distribution on the right.

And we can use excel to find this critical value with this function:

"=F.INV(1-0.01,2,9)"

And we will see that the critical value is
`F_(crit)=8.02`

B. 8.02