The Following Is a Partial ANOVA Table

The following is a partial ANOVA table.

Source Sum of Squares df Mean Square F

Treatment 2

Error 20

Total 500 11

Complete the table and answer the following questions. Use the .05 significance level.

a. How many treatments are there?

b. What is the total sample size?

c. What is the critical value of F?

d. Write out the null and alternate hypothesis?

e. What is your conclusion regarding the null hypothesis?

The ANOVA Table

ANOVA Table
Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / SST / k  1 / SST/ ( k 1 ) = MST / MST / MSE
Error / SSE / n  k / SSE/ ( n  k ) = MSE
Total / SS Total / n 1

There are (k 1) degrees of freedom associated with the numerator of the formula for F, and (n k) degrees of freedom associated with the denominator, where k is the number of populations and n is the total number of sample observations.

From the given ANOVA table we have,

SS Total = 500,

k  1 = 2,

n 1 = 11

MSE = SSE/ ( n  k ) = 20

Therefore,

k = 2+1 =3, and n = 11 +1 = 12

Also, SSE = MSE*( n  k ) = 20*(12-3) =20*9 = 180

Now, SST =SS Total – SSE = 500 – 180 = 320

Thus, MST = SST/( k 1 ) = 320/2 = 160

Also, F = MST/MSE = 160/20 = 8

Thus the completed ANOVA table is

ANOVA Table
Source of Variation / Sum of Squares / Degrees of Freedom / Mean Square / F
Treatments / 320 / 2 / 160 / 8
Error / 180 / 9 / 20
Total / 500 / 11

a. How many treatments are there?

There are three treatments.

b. What is the total sample size?

The total sample size is 12

c. What is the critical value of F?

For a = 0.05, from the F distribution with (k-1, n-k) = (2,9) degrees of freedom, the critical value of F is 4.26.

d. Write out the null and alternate hypotheses.

Ho: There is no significant difference between the treatment means.

Ha: At least one treatment mean is different from the others.

e. What is your conclusion regarding the null hypothesis?

We reject the null hypothesis if F > 4.26.

Here F = 8 > 4.26.

Since the calculated value is greater than critical value, we reject Ho and conclude that at least one treatment mean is different from the others.