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1. Elementary Statistics
Chapter 9: Inferences
from Two Samples
9.4 Two Variances or
Standard Deviations
1

2. Chapter 9: Inferences from Two Samples
9. 1 Two Proportions
9.2 Two Means: Independent Samples
9.3 Two Dependent Samples (Matched Pairs)
9.4 Two Variances or Standard Deviations
2
Objectives:
• Test the difference between two proportions.
• Test the difference between sample means, using the z test.
• Test the difference between two means for independent samples, using the t test.
• Test the difference between two means for dependent samples.
• Test the difference between two variances or standard deviations.

3. For the comparison of two variances or standard deviations, an F test is used.
• The F test should not be confused with the chi-square test, which compares a single sample
variance to a specific population variance.
Characteristics:
1. The values of F cannot be negative, because variances are always positive or zero.
2. The distribution is positively skewed.
3. The mean value of F is approximately equal to 1.
4. The F distribution is a family of curves based on the degrees of freedom of the variance of
the numerator and the degrees of freedom of the variance of the denominator.
• In some Texts the larger of the two variances is placed in the numerator regardless of the
subscripts. Therefore, 𝐹 =
𝑠𝑙𝑎𝑟𝑔𝑒𝑟
2
𝑠𝑠𝑚𝑎𝑙𝑙𝑒𝑟
2 in most cases unless otherwise mentioned. In others, this
is not the case! Therefore, follow what the text or author wants you to do for full credit.
• The F test has two terms for the degrees of freedom: that of the numerator, n1 – 1, and that of the
denominator, n2 – 1.
9.4 Two Variances or Standard Deviations
𝐹 =
𝑠1
2
𝑠2
2
3

4. 4
Find the critical value for a two-tailed F test with α = 0.05 when the sample size from
which the variance for the numerator was obtained was 21 and the sample size from
which the variance for the denominator was obtained was 12.
Example 1
2TT: α is split; and NORMALLY the right tail is used since F  1.
2TT & α = 0.05 ⇾ Use 0.05/2 = 0.025 table: d.f.N. = 21 – 1 = 20 & d.f.D. = 12 – 1 = 11.
F = 3.2261

5. 5
Find the critical value for a right-tailed F test when α = 0.05, the degrees of freedom
for the numerator (abbreviated d.f.N.) are 15, and the degrees of freedom for the
denominator (d.f.D.) are 21.
Example 2
F = 2.1757
Since this test is right-tailed with a 0.05, use the 0.05 table. The d.f.N. is listed across
the top, and the d.f.D. is listed in the left column. The critical value is found where
the row and column intersect in the table.
0.05
← 2.18
F-Distribution Calculator:
https://stattrek.com/online-calculator/f-
distribution.aspx
Enter values for degrees of freedom.
Enter a value for one, and only one, of the
remaining text boxes.
Click the Calculate button to compute a value
for the blank text box.

6. Key Concept: Use the F test for testing claims made about two population variances (or standard deviations). The F test
(named for statistician Sir Ronald Fisher) uses the F distribution introduced in this section. The F test requires that both
populations have normal distributions. Instead of being robust, this test is very sensitive to departures from normal distributions,
so the normality requirement is quite strict.
Conduct a hypothesis test of a claim about two population variances or standard deviations. (Any claim made about two
population standard deviations can be restated with an equivalent claim about two population variances, so the same procedure is
used for two population standard deviations or two population variances.)
1. The two populations are independent.
2. The two samples are simple random samples.
3. Each of the two populations must be normally distributed, regardless of their sample sizes. This F test is not robust
against departures from normality, so it performs poorly if one or both of the populations have a distribution that is not
normal. The requirement of normal distributions is quite strict for this F test.
Critical Values: Critical F values are determined by the following:
1. The significance level α (F distribution Tables includes critical values for α’s such as α = 0.025 & α = 0.05)
2. Numerator degrees of freedom n1 − 1 (determines column of the Table)
3. Denominator degrees of freedom n2 − 1 (determines row of the Table)
For significance level α = 0.05, refer to the Table and use the right-tail area of 0.025 or 0.05, depending on the type of test, as
shown below:
Two-tailed test: Use the Table with 0.025 in the right tail. (The significance level of 0.05 is divided between the two tails, so the area in
the right tail is 0.025.)
One-tailed test: Use the Table with α = 0.05 in the right tail.
9.4 Two Variances or Standard Deviations
6

7. 7
The standard deviation of the average waiting time to see a doctor for non-
life threatening problems in the emergency room at an urban hospital is 32
minutes. At a suburban hospital, the standard deviation is 28 minutes. If a
sample of 16 patients was used in the first case and 18 in the second case, is
there enough evidence to conclude that the standard deviation of the waiting
times in the first hospital is greater than the standard deviation of the waiting
times in the second hospital? α = 0.05.
Example 3
CV: α = 0.05 & RTT:
d.f.N. = 15, d.f.D. = 17 →F = 2.3077
Decision:
a. Do not Reject H0
b. The claim is False
c. There is not enough evidence to support the claim that
the standard deviation of the waiting times of the
Urban hospital is greater than the standard deviation of
the waiting times of the Suburban hospital.
Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is
/ is not sufficient evidence to
support the claim that…
𝑇𝑆: 𝐹 =
𝑠1
2
𝑠2
2
𝐻0: 𝜎1
2
= 𝜎2
2
or
𝜎1
2
𝜎2
2 =1 or 𝜎1 = 𝜎2 & 𝐻1: 𝜎1
2
> 𝜎2
2
(claim),RTT or
𝜎1
2
𝜎2
2 >1 or 𝜎1 > 𝜎2
=
322
282
= 1.3061
2.3077
0.05
Given: SRS:
𝑼𝒓𝒃𝒂𝒏: 𝑛1 = 16, 𝑠1 = 32𝑚𝑖𝑛
Suburban 𝑛2= 18, 𝑠2 = 28min
.
1.3061
𝐹 =
𝑠1
2
𝑠2
2
TI Calculator:
2 - Sample F - test
1. Stat
2. Tests
3. 2 ‒ Samp F Test
4. Enter Data or Stats
𝒔𝟏 , 𝒏𝟏, 𝒔𝟐 , 𝒏𝟐,
5. Choose RTT, LTT,
or 2TT
6. Calculate

8. 8
A medical researcher wishes to see whether the variance of the heart rates (in beats per minute) of
smokers is different from the variance of heart rates of people who do not smoke. Two samples are
selected, and the data are as shown. Using α = 0.05, is there enough evidence to support the claim?
Example 4
CV: α = 0.05 & 2TT: Use the 0.025 table
d.f.N. = 25 & d.f.D. = 17 ⇾ F = 2.5598
(d.f.N. 24 was used).
Decision:
a. Reject H0
b. The claim is True
c. There is enough evidence to
support the claim that the variance
of the heart rates of smokers and
nonsmokers is different.
Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is / is not
sufficient evidence to support the claim
that…
𝑇𝑆: 𝐹 =
𝑠1
2
𝑠2
2
𝐻0: 𝜎1
2
= 𝜎2
2
and 𝐻1: 𝜎1
2
≠ 𝜎2
2
(claim),2TT
=
36
10
= 3.6
𝐹 =
𝑠1
2
𝑠2
2

9. Decision:
a. Do not Reject H0
b. The claim is True
c. There is sufficient evidence
to support the claim of
equal standard deviations. 9
Listed below are stats for student course evaluation scores for courses taught by
female professors and male professors.
𝐹𝑒𝑚𝑎𝑙𝑒𝑠: 𝑥1= 3.8667, 𝑛1 = 12, 𝑠1 = 0.5630, 𝑀𝑎𝑙𝑒𝑠: 𝑥2 = 3.9933, 𝑛2 = 15, 𝑠2 = 0.3955
Use a 0.05 significance level to test the claim that course evaluation scores of
female professors and male professors have the same variation.
Example 5
𝐻0: 𝜎1
2
= 𝜎2
2
,claim 𝐻1: 𝜎1
2
≠ 𝜎2
2
,2TT
CV: α = 0.05 & 2TT: Use the 0.025 table
d.f.N. = 11 & d.f.D. = 14 ⇾
F is between 3.1469 and 3.0502.
𝑇𝑆: 𝐹 =
𝑠1
2
𝑠2
2 =
0.56302
0.39552
= 2.0269
P-Value Method:
2TT, α = 0.05
TS: F = 2.0269 < CV
The area to the right of the
test statistic is greater than
0.025 ⇾ for two-tailed test,
P-value > 0.05.
𝐹 =
𝑠1
2
𝑠2
2
TI Calculator:
2 - Sample F - test
1. Stat
2. Tests
3. 2 ‒ Samp F Test
4. Enter Data or Stats
𝒔𝟏 , 𝒏𝟏, 𝒔𝟐 , 𝒏𝟐,
5. Choose RTT, LTT,
or 2TT
6. Calculate

10. 10
Listed below are stats for the weights (in pounds) of samples of regular Coke and regular
Pepsi. Use the 0.05 significance level to test the claim that the weights of regular Coke and the
weights of regular Pepsi have the same variance (standard deviation).
Example 6
Decision:
a. Do not Reject H0
b. The claim is True
c. There is sufficient evidence to support the
claim of equal standard deviations.
Regular Coke Regular Pepsi
n 36 36
x 0.81682 0.82410
s 0.007507 0.005701
𝛼 = 0.05 → 𝐶𝑉:
𝐹 𝛼/2=0.025,𝑑𝑓1=35,𝑑𝑓2=35 = 1.8752
Note: We choose the df’s of 40 as
the table does not have df’s of 35.
=
0.007507 2
0.0057012
= 1.7339
𝐻0: 𝜎1
2
= 𝜎2
2
,claim 𝐻1: 𝜎1
2
≠ 𝜎2
2
,2TT
𝐹 =
𝑠1
2
𝑠2
2
Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is / is not
sufficient evidence to support the claim
that…
𝑇𝑆: 𝐹 =
𝑠1
2
𝑠2
2

11. Alternative Methods: Count Five & Levene-Brown-Forsythe Test
The count five method is a relatively simple alternative to the F test, and it
does not require normally distributed populations. If the two sample sizes
are equal, and if one sample has at least five of the largest mean
absolute deviations (MAD), then we conclude that its population has a
larger variance.
The Levene-Brown-Forsythe test (or modified Levene’s test) is another
alternative to the F test. It begins with a transformation of each set of
sample values. Within the first sample, replace each x value with
|x − median|, and apply the same transformation to the second sample.
Using the transformed values, conduct a t test of equality of means for
independent samples. Because the transformed values are now deviations,
the t test for equality of means is actually a test comparing variation in the
two samples.
11

12. The sampling distribution of F =
𝒔𝟏
𝟐
𝒔𝟐
𝟐 is an F distribution and There is a different F distribution
for each different pair of degrees of freedom for the numerator and denominator.
• The F distribution is not symmetric.
• Values of the F distribution cannot be negative.
• The exact shape of the F distribution depends on the two different degrees of freedom.
If the two populations have equal variances, then
𝑠1
2
𝑠2
2 will be close to 1 because 𝑠1
2
and 𝑠2
2
are close in value. Consequently, a
value of F near 1 will be evidence in favor of the conclusion that: 𝜎1
2
= 𝜎2
2
, but a large value of F will be evidence against
the conclusion of equality of the population variances.
1. In some Texts the larger of the two variances is placed in the numerator regardless of the subscripts. In others, this
is not the case! Therefore, follow what the text or author wants you to do for full credit.
2. For a two-tailed test, the α value must be divided by 2 and the critical value placed on the right side of the F curve.
3. If the standard deviations are given in the problem, they must be squared for the formula for the F test.
4. When the degrees of freedom cannot be found in the Table, the closest value on the smaller side should be used.
5. The populations from which the samples were obtained must be normally distributed. (Note: The test should not be used
when the distributions depart from normality.)
6. The samples must be independent of each other.
12
9.4 Two Variances or Standard Deviations / F Distribution