# Thread: ANOVA chapter..... 2 problems , one anwered - did I do it correctly? one no clue

1. ## ANOVA chapter..... 2 problems , one anwered - did I do it correctly? one no clue

I actually have two problems - the first I have attempted to figure out, but not sure I have it correcct. Still don't understand how to interpret the answer I get.

A Stockbrocker at Critical Securities reported that the mean rate of return on a sample of 10 oil stocks was 12.6 percent with a satnadard deviation of 3.9 percent. The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can we conclude that there is more variation in the oil stocks?

I have F=3.9^/3.5^ =1.2416
n1 - 1, n2 - 1 = 9,7
Critical F value =3.667
Reject if F 3.667 ???
1.2416 < 3.667 so fail to reject?

Did I calculate correctly - and how does that result translate to answer is there more variation?

This next problem - need help on set up and interpretation.

The following is sample information. Test the hypothesis at the .05 significance level that the treatment means are equal.

Treatment 1
9,7,11,9,12,10

Treatment 2
13,20,14,13,

Treatment 3
10,9,15,14,15,

a.State the null hypothesis and the alternate hypotheses.
b.What is the decision rule?
c.Compute SST, SSE, and SS total.
d.Complete an ANOVA table.
e.State your decision regarding the null hypothesis.

2. ## Re: ANOVA chapter..... 2 problems , one anwered - did I do it correctly? one no clue

Originally Posted by cama62
I have F=3.9^/3.5^ =1.2416
n1 - 1, n2 - 1 = 9,7
Critical F value =3.667
Reject if F 3.667 ???
1.2416 < 3.667 so fail to reject?
Yes, if your calculated F-value is less than your critical F-value then fail to reject the null hypothesis.

3. ## Re: ANOVA chapter..... 2 problems , one anwered - did I do it correctly? one no clue

T1<>T2 (p<0.05)
T1=T3 (p>0.05)
T2=T3 (p>0.05)

4. ## Re: ANOVA chapter..... 2 problems , one anwered - did I do it correctly? one no clue

$P_f\left[\frac{3.9^2}{3.5^2}|\nu _1=9,\nu _2=7\right]=0.60356>.05$ Can't reject equal variance hypothesis.