00:01
So we are given that in an experiment designed to test the output levels of different treatments, sum square between treatments, sst is we have been given 400.
00:18
Likewise, sstr, this we have been given 150 and nt is equals to 19.
00:26
So we just had to set up the anova table and test for the significant difference between the mean output levels of the three testaments with significance level of alpha equals to 0 .05.
00:38
So first of all we said the null hypothesis thus is h0 which should be said as mu1 equals to mu2 is equals to mu3 because we have to do for the three populations are equal and and alternative hypothesis as at least one, at least one mean mu i different.
01:01
So same of the population means are different.
01:05
So all this you can also represent like h a as mu 1 not equals to mu 2 not equals to mu 3.
01:13
In any of the way you can represent it.
01:17
So according to question if we say the k.
01:20
K.
01:20
So k is what? the hash of the treatment and that is nothing but equals to 3.
01:27
So can we calculate the degree of rhythms that should be equals to k minus 1 is equals to 3 minus 1 is equals to 2.
01:34
Now n is the sample size which is given to us 19 for that the degree of rhythm as error.
01:44
Okay this this is what the degree of freedom for a t let's say.
01:48
Means treatment.
01:49
And for error this will comes to n minus of k is equals to 19 minus 3 will becomes out 16...