Topic 6
Comparing Two Standard Deviations
Activity 6-1: Memorization Methods (cont’d.)
a)
Open the SPSS data file MEMORY.SAV. Use SPSS to draw side-by-side box plots for the memorization
score separated according to the memorization method. Do the plots indicate that there appears to be a
difference in the standard deviations of the amount of change?
12
10
Score
8
6
4
2
0
Repetition
Imagery
Method
The two boxes are roughly the same width. So, it appears reasonable that the two standard deviations are
the same.
b)
c)
Use SPSS to compute the standard deviation by gender, and complete the following table.
Method
Sample Size
Sample
Standard Deviation
Repetition
27
11.499
Imagery
27
11.892
Do the statistics computed seem to lead to a conclusion that is consistent with the one you arrived at using
the box plots? If not, what do you think is going on here?
Yes, the two sample standard deviations are virtually identical.
Activity 6-2: Computing the Test Statistic
a)
Compute the test statistic for the memorization method data.
f 
11.8922
 1.0695
11.4992
b)
If the null hypothesis is not true, and the population standard deviations differ, would you expect the value
of the test statistic to be greater than one, less than one, or is it impossible to tell? Explain why you think
so.
Since the test statistic is rigged to be greater than or equal to 1, evidence against the null hypothesis will be
an F value significantly greater than 1.
c)
Does the value of your test statistic appear to give evidence that the two standard deviations are not equal?
The value is just slightly greater than 1. This does not give evidence that the two standard deviations are
not equal.
Activity 6-3: Exploring the F Distribution
b)

Use the F table to find Pr  F
c)
Use the F table to find, as closely as you can,.01 < Pr F12,6  7.02 < .025.
d)
Use the F table to find, as closely as you can, Pr F12,9  10.03 <.001.
e)
Use the F table to find, as closely as you can,.1 < Pr F3,16  2.39 .
f)
Use the F table to find, as closely as you can, .025 < Pr F33,17   2.47 < .05.
g)
Use the F table to find f such that Pr F12,20  f  0.025 . f = 2.68
a)

  1.65  = .1.
Use the F table to find Pr F3,6  9.78 = .01.
25,28
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Activity 6-4: Computing the p-value
a)
Refer to Activity 6-2. What are the degrees of freedom for your F ratio?
dfn = dfd = 26
b) What is the value of f that should be used in the p-value calculation?
f = 1.0695
c) Use Table IV to calculate Pr  F  f  .
Pr( F(26,26)  1.0695)  .1
d)
e)
Calculate the p-value for the test.
p-value > .2
What conclusion do you draw from this test? Is it consistent with what you thought in Activity 6-1? If
not, what do you think is going on here?
There is no evidence that the two population standard deviations are not equal.
Activity 6-5: Hypothetical e-mail Times
You are going to test whether the standard deviation in e-mail time is the same for men as it is for women using the
hypothetical data given in Activity 3-1. The data are stored in the SPSS data file HYPOMAIL.SAV.
a)
Have SPSS compute the sample standard deviations for men and for women for the e-mail data given in
Activity 3-1.
sMen  3.226, sWomen  2.762
nMen  25, nWomen  20
b)
Compute and record the value of the F ratio.
3.2262
f 
 1.364
2.7622
c)
d)
e)
Record the degrees of freedom for your F ratio.
dfn = 24, dfd = 19
Compute the p-value for testing the equality of the population standard deviations.
p-value  2 Pr  F(24,19)  1.364   .2
Interpret your results. Does this test indicate that the use of the pooled t test is valid for testing the
equality of the mean e-mail times for men versus women? Explain.
There is no evidence against the null hypothesis. So, the pooled t test is appropriate.