Question

Question 4:

Consider the following sample information for three treatment groups.

Treatment 1 Treatment 2 Treatment 3

$8 $3 $3

6 2 4

10 4 5

9 3 4

Using the 0.05 significance level, use the ANOVA method to test the null hypothesis that the three treatment means are equal.

Answer 1

Consider the given problem here there are three treatment and the observation corresponding to each treatment is given in the question, => the “mean value” corresponding to each treatment are given by, “m1=8.25”, “m2=3” and “m3=4” and the “grand mean” is given by “m=5.08” the overall mean of all observations. So, here the null and the alternative hypothesis is given below.

=> “H0: μ1 = μ2 = μ3”, and the corresponding alternative hypothesis is given by “H1: the means are not equal”.

So, here the “F statistic” is given below.

=> F = MSB/MSE, where “MSB = SSB/k-1” and “SSB = Summation[nj(mj-m)^2], where “mj” be the mean of the “jth treatment” and “m=the grand mean”.

Similarly, “MSE=SSE/N-k”, where “SSE = Summation Summation(X-mj)^2”.

Now, the “SSB” is given below.

=> SSB = Summation[nj(mj-m)^2] = 4*(8.25-5.08)^2 + 4*(3-5.08)^2 + 4*(4-5.08)^2 = 40.1956 + 17.3056 + 4.6656 = 62.1668, => SSB = 62.1668.

Now, the “SSE” is given by, => SSE = Summation Summation (X-mj)^2, where “X=oberservation”.

So, the “SSE” is given by, “SSE = 8.75 + 2 + 2 = 12.75”. Consider the following “ANOVA table”.

So, here the “F statistic” is “F=21.9407”. Now, the critical value of the hypothesis testing at “5% level of significance” is given by, => F0.05(2,9) = 19.38 < 21.94, => the value of “F” is on the critical region, => the mean value of the treatments are not same they are differ to each other.