Psych 318 Assignment 2: Tests of variances
Here is a key for questions 2 – 16 in the assignment you just completed. As you compare your
answers to the key, please keep these things in mind:
1. The computational effort for these tests is not the most important aspect of understanding
these tests. This includes looking up the critical values, etc. If you got a wrong critical
value, be sure you know what you did wrong, but don’t despair!
2. I expect you to be able to find the correct formula to use and to plug values into that
formula. Problems in this area are not statistical problems! They are simply problems of
using mathematical formulas! You’ve got to pay attention to subscripts!
KEY:
2. The appropriate test is the F test for two sample variances.
3. One critical assumption in this F test is that the populations that are sampled from are
normally distributed. A quick check on the tenability of this assumption can be made by
plotting out the sample data. When the samples look wildly non-normal, a researcher
might decide to use an alternate methodology.
4. H0: 2 (simple) = σ2 (disjunctive); H1: σ2 (simple) ≠ σ2 (disjunctive)
^ 2
5. Fobt 
1
^ 2
2
(28.12)2 790.7


 3.92
(14.21)2 201.9
6. Using Excel FDIST function (3.92, 11, 10), I get a one-tailed p-value of .02. I multiply
this by 2 to get the two-tailed p-value of .04.
7. I conclude that there is sufficient evidence to support a claim that the two population
variances are not equal because .04 (my two-tailed p-value) < .05 (my chosen alpha
level). Looking back at the data, I can further conclude that it appears that reaction times
are more variable (have a greater variance) for more complex tasks than for simple tasks.
8. The 95% confidence interval is:
^ 2
^ 2
2
 disjunctive
1
(
)
 ^ 2 (F /2;df ,df )
2
^ 2
F
 simple
 2  /2;df ,df
2
2
 disjunctive
1
3.92(
)
 3.92(3.526)
2
3.665
 simple
1
1
2
1
1
2
2
 disjunctive
 13.80
2
 simple
9. These limits are consistent with the results of our NHST since they indicate that the value
of the variance ratio under the null (which is 1) is not a plausible value.
10. The appropriate test for this situation is the chi-square test for a single sample variance.
11. H0: σ2 = 287.5 ; H1: σ2 ≠ 287.5
12. The best estimate of the population variance is unbiased variance from the sample: 42.2 2
= 1780.84
1.07 
^
(n  1)  2
(20  1)(1780.84)
 117.69

287.5
14. I would conclude that because 117.69 exceeds the upper critical value of 32.852 there is
sufficient evidence to reject the null hypothesis that the population variance is equal to
287.5. Looking at the data, I can further conclude that there is sufficient evidence to
support a claim that the population variance is greater than 287.5.
15. The 95% confidence interval:
13.  2 
2
0

^
(n  1)  2
2 /2,df
^
 
2
(n  1)  2
12  /2,df
(20  1)(1780.84)
(20  1)(1780.84)
 2 
33.852
8.907
2
1029.95    3798.81
16. The confidence interval is consistent with the NHST because 287.5 is not contained in
the interval.