Step 2 The confidence interval for σ is from
11.2 Inferences for Two Population Standard Deviations,
In Section 11.1, we discussed hypothesis tests and confidence intervals for one popula- tion standard deviation. We now introduce hypothesis tests and confidence intervals for two population standard deviations. More precisely, we examine inferences to compare the standard deviations of one variable of two different populations. Such inferences are based on a distribution called theF-distribution, named in honor of Sir Ronald Fisher.
The F -Distribution
A variable is said to have an F-distribution if its distribution has the shape of a special type of right-skewed curve, called an F-curve. Actually, there are infinitely many F-distributions, and we identify the F-distribution (and F-curve) in question by its number of degrees of freedom, just as we did fort-distributions and chi-square distributions.
An F-distribution, however, has two numbers of degrees of freedom instead of one. Figure 11.7 depicts two different F-curves; one has df=(10,2), and the other has df=(9,50).
FIGURE 11.7 Two differentF-curves df = (9, 50)
df = (10, 2)
0 1 2 3 F
The first number of degrees of freedom for an F-curve is called thedegrees of freedom for the numeratorand the second thedegrees of freedom for the denomi- nator.(This terminology will become clear shortly.) Thus, for theF-curve in Fig. 11.7 with df=(10,2), we have
df = (10,2).
Degrees of freedom Degrees of freedom for the numerator for the denominator
KEY FACT 11.3 Basic Properties ofF-Curves
Property 1: The total area under anF-curve equals 1.
Property 2: AnF-curve starts at 0 on the horizontal axis and extends indef- initely to the right, approaching, but never touching, the horizontal axis as it does so.
Property 3: AnF-curve is right skewed.
Percentages (and probabilities) for a variable having anF-distribution equal areas under its associated F-curve. To perform a hypothesis test or construct a confidence
interval for comparing two population standard deviations, we need to know how to find the F-value that corresponds to a specified area under an F-curve. The symbol
Fαdenotes theF-value having areaαto its right.
Table VIII in Appendix A gives F-values with areas 0.005, 0.01, 0.025, 0.05, and 0.10 to their right for various degrees of freedom. The degrees of freedom for the denominator (dfd) are displayed in the outside columns of the table; the values of α are in the next columns; and the degrees of freedom for the numerator (dfn) are along the top.
EXAMPLE 11.9 Finding the F -Value Having a Specified Area to Its Right
For an F-curve with df = (4, 12), find F0.05; that is, find the F-value having area 0.05 to its right, as shown in Fig. 11.8(a).
FIGURE 11.8 Finding theF-value having area 0.05 to its right
F -curve df = (4, 12)
0 F
F0.05= ? F0.05= 3.26
(a) (b)
F -curve df = (4, 12)
0 F
Area = 0.05 Area = 0.05
Solution To obtain the F-value, we use Table VIII. In this case,α=0.05, the degrees of freedom for the numerator is 4, and the degrees of freedom for the denominator is 12.
We first go down the dfd column to “12.” Next, we concentrate on the row forα labeled 0.05. Then, going across that row to the column labeled “4,” we reach 3.26.
This number is the F-value having area 0.05 to its right, as shown in Fig. 11.8(b).
In other words, for anF-curve with df=(4,12),F0.05=3.26.
Exercise 11.53 on page 538
In many statistical analyses that involve the F-distribution, we also need to deter- mine F-values having areas 0.005, 0.01, 0.025, 0.05, and 0.10 to their left. Although suchF-values aren’t available directly from Table VIII, we can obtain them indirectly from the table by using Key Fact 11.4.
KEY FACT 11.4 Reciprocal Property ofF-Curves
For an F-curve with df=(ν1, ν2), theF-value having areaαto its left equals the reciprocal of the F-value having areaα to its right for an F-curve with df=(ν2, ν1).
EXAMPLE 11.10 Finding the F -Value Having a Specified Area to Its Left For anF-curve with df=(60,8), find the F-value having area 0.05 to its left.
Solution We apply Key Fact 11.4. Accordingly, the requiredF-value is the recip- rocal of the F-value having area 0.05 to its right for anF-curve with df=(8,60). From Table VIII, this latter F-value equals 2.10. Consequently, the required F-value is 2.110, or 0.48, as shown in Fig. 11.9 on the next page.
528 CHAPTER 11 Inferences for Population Standard Deviations∗ FIGURE 11.9
Finding theF-value having area 0.05 to its left
F -curve df = (60, 8) F -curve
df = (8, 60)
0.05 0.05
2.10
F F
1 2.10= 0.48 Exercise 11.57
on page 538
EXAMPLE 11.11 Finding the F -Values for a Specified Area
For an F-curve with df=(9,8), determine the two F-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas, as shown in Fig. 11.10(a).
FIGURE 11.10 Finding the twoF-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas
F -curve df = (9, 8)
0 F
F = ? 4.36
(a) (b)
F = ?
F -curve df = (9, 8)
0 F 0.24 0.025
0.025 0.95 0.025 0.95 0.025
Solution First, we find the F-value on the right in Fig. 11.10(a). Because the shaded area on the right is 0.025, theF-value on the right isF0.025. From Table VIII with df=(9,8),F0.025=4.36.
Next, we find the F-value on the left in Fig. 11.10(a). By Key Fact 11.4, that F-value is the reciprocal of theF-value having area 0.025 to its right for anF-curve with df=(8,9). From Table VIII, we find that this latterF-value equals 4.10. Thus theF-value on the left in Fig. 11.10(a) is 4.110, or 0.24.
Consequently, for anF-curve with df=(9,8), the twoF-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas are 0.24 and 4.36, as shown in Fig. 11.10(b).
Exercise 11.59 on page 538
The Logic Behind Hypothesis Tests for Comparing Two Population Standard Deviations
We illustrate the logic behind hypothesis tests for comparing two population standard deviations in the next example.
EXAMPLE 11.12 Hypothesis Tests for Two Population Standard Deviations Elmendorf Tear Strength Variation within a method used for testing a product is an essential factor in deciding whether the method should be employed. Indeed, when the variation of such a test is high, ascertaining the true quality of a product is difficult.
Manufacturers use the Elmendorf tear test to evaluate material strength for vari- ous manufactured products. In the article “Using Repeatability and Reproducibility
Studies to Evaluate a Destructive Test Method” (Quality Engineering, Vol. 10(2), pp. 283–290), A. Phillips et al. investigated the variation of that test. In one aspect of the study, the researchers randomly and independently obtained the data shown in Table 11.3 on Elmendorf tear strength, in grams, of two different vinyl floor coverings.
TABLE 11.3 Results of Elmendorf tear test on two different vinyl floor coverings (data in grams)
Brand A Brand B 2288 2384 2592 2384 2368 2304 2512 2432 2528 2240 2576 2112 2144 2208 2176 2288 2160 2112 2304 2752
Suppose that we want to decide whether the standard deviations of tear strength differ between the two vinyl floor coverings.
a. Formulate the problem statistically by posing it as a hypothesis test.
b. Explain the basic idea for carrying out the hypothesis test.
c. Discuss the use of the data in Table 11.3 to make a decision concerning the hypothesis test.
Solution
a. We want to perform the hypothesis test
H0:σ1=σ2(standard deviations of tear strength are the same) Ha:σ1 =σ2(standard deviations of tear strength are different), whereσ1andσ2denote the population standard deviations of tear strength for Brand A and Brand B, respectively.
b. We carry out the hypothesis test by comparing the sample standard devia- tions,s1ands2, of the two sets of sample data presented in Table 11.3. Specifi- cally, we compute the square of the ratio ofs1tos2, or, equivalently, the quotient of the sample variances. That statistic is called theF-statistic.
If the population standard deviations,σ1andσ2, are equal, the sample stan- dard deviations,s1ands2, should be roughly the same, which means that the value of theF-statistic should be close to 1. When the value of the F-statistic differs from 1 by too much, it provides evidence against the null hypothesis of equal population standard deviations.
c. For the data in Table 11.3,s1=128.3 g ands2=199.7 g. Thus the value of theF-statistic is
F= s12
s22 = 128.32
199.72 =0.413.
Does this value ofFdiffer from 1 by enough to conclude that the null hypoth- esis of equal population standard deviations is false? To answer that question, we need to know the distribution of theF-statistic. We discuss that distribution and then return to complete the hypothesis test.
The Distribution of the F -Statistic
To perform hypothesis tests and obtain confidence intervals for two population stan- dard deviations, we need Key Fact 11.5.
KEY FACT 11.5 Distribution of theF-Statistic for Comparing Two Population Standard Deviations
Suppose that the variable under consideration is normally distributed on each of two populations. Then, for independent samples of sizesn1andn2from the two populations, the variable
F =s12/σ12 s22/σ22 has theF-distribution with df=(n1−1,n2−1).
530 CHAPTER 11 Inferences for Population Standard Deviations∗
EXAMPLE 11.13 The Distribution of the F -Statistic
Elmendorf Tear Strength In Example 11.12, suppose that the Elmendorf tear strengths for Brands A and B vinyl floor coverings are normally distributed with means 2275 g and 2405 g, respectively, and equal standard deviations 168 g. Then, according to Key Fact 11.5, for independent random samples, each of size 10, from Brands A and B, the variable F =s21/s22 has the F-distribution with df=(9,9). Use simulation to make that fact plausible.
Solution We first simulated 1000 samples of 10 tear strengths each for Brand A vinyl floor covering, that is, 1000 samples of 10 observations each of a normally distributed variable with mean 2275 and standard deviation 168. Next we simu- lated 1000 samples of 10 tear strengths each for Brand B vinyl floor covering, that is, 1000 samples of 10 observations each of a normally distributed variable with mean 2405 and standard deviation 168. Then, for each of the 1000 pairs of sam- ples from the two brands, we determined the sample standard deviations,s1ands2, and obtained the value of the variable F=s12/s22. Output 11.3 shows a histogram of those 1000 values of F, which is shaped like the superimposed F-curve with df=(9,9).
OUTPUT 11.3 Histogram ofFfor 1000 independent samples with superimposedF-curve
8 7 6 5 4 3 2 1 0
F
Hypothesis Tests for Two Population Standard Deviations
In light of Key Fact 11.5, for a hypothesis test with null hypothesisH0:σ1=σ2(pop- ulation standard deviations are equal), we can use the variable
F= s12 s22
as the test statistic and obtain the critical value(s) from theF-table, Table VIII. We call this hypothesis-testing procedure thetwo-standard-deviationsF-test.†
Procedure 11.3 gives a step-by-step method for performing a two-standard- deviationsF-test by using either the critical-value approach or the P-value approach.
For theP-value approach, we could use Table VIII to estimate the P-value, but to do so is awkward and tedious; thus, we recommend using statistical software.
Unlike thez-tests andt-tests for one and two population means, the two-standard- deviations F-test is not robust to moderate violations of the normality assumption.
In fact, it is so nonrobust that many statisticians advise against its use unless there is considerable evidence that the variable under consideration is normally distributed, or very nearly so, on each population.
Consequently, before applying Procedure 11.3, construct a normal probability plot of each sample. If either plot creates any doubt about the normality of the variable under consideration, do not use Procedure 11.3.
We note that nonparametric procedures, which do not require normality, have been developed to perform inferences for comparing two population standard deviations. If you have doubts about the normality of the variable on the two populations under consideration, you can often use one of those procedures to perform a hypothesis test or find a confidence interval for two population standard deviations.
EXAMPLE 11.14 The Two-Standard-Deviations F-Test
Elmendorf Tear Strength We can now complete the hypothesis test proposed in Example 11.12. Independent random samples of two vinyl floor coverings
†The two-standard-deviationsF-test is also known as theF-test for two population standard deviationsand thetwo-sampleF-test.This test is often formulated in terms of variances instead of standard deviations.
PROCEDURE 11.3 Two-Standard-DeviationsF-Test
Purpose To perform a hypothesis test to compare two population standard devi- ations,σ1andσ2
Assumptions
1. Simple random samples 2. Independent samples 3. Normal populations
Step 1 The null hypothesis is H0:σ1=σ2, and the alternative hypothesis is Ha:σ1=σ2 or Ha:σ1< σ2 or Ha:σ1 > σ2
(Two tailed) (Left tailed) (Right tailed) Step 2 Decide on the significance level,α.
Step 3 Compute the value of the test statistic
F = s12 s22 and denote that value F0.
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH
Step 4 The critical value(s) are
F1−α/2and Fα/2 F1−α Fα
or or
(Two tailed) (Left tailed) (Right tailed) with df=(n1−1,n2−1). Use Table VIII to find the critical value(s).
Two tailed Left tailed Right tailed
F F F
F1−/2 F1− F
/2
Do not reject H0 Reject
H0
Reject
H0 Reject Do not reject H0
H0 Do not reject H0 Reject
H0
/2
F/2
Step 5 If the value of the test statistic falls in the rejection region, reject H0; otherwise, do not rejectH0.
Step 4 The F-statistic has df=(n1−1,n2−1). Obtain theP-value by using technology.
Two tailed Left tailed Right tailed
F F F
F0 F0
F0 P-value
P-value P-value
Step 5 If P ≤α, reject H0; otherwise, do not rejectH0.
Step 6 Interpret the results of the hypothesis test.
yield the data on Elmendorf tear strength repeated here in Table 11.4. At the 5% significance level, do the data provide sufficient evidence to conclude that the population standard deviations of tear strength differ for the two vinyl floor coverings?
TABLE 11.4 Results of Elmendorf tear test on two different vinyl floor coverings (data in grams)
Brand A Brand B 2288 2384 2592 2384 2368 2304 2512 2432 2528 2240 2576 2112 2144 2208 2176 2288 2160 2112 2304 2752
Solution To begin, we construct normal probability plots for the two samples in Table 11.4, shown in Fig. 11.11 (next page). The plots suggest that we can rea- sonably presume that tear strength is normally distributed for each brand of vinyl flooring. Hence we can use Procedure 11.3 to perform the required hypothesis test.
532 CHAPTER 11 Inferences for Population Standard Deviations∗
FIGURE 11.11 Normal probability plots of the sample data for (a) Brand A and (b) Brand B
−3
−2
−1 0 1 2 3
Tear strength (g) (a) Brand A
Normal score
−3
−2
−1 0 1 2 3
Tear strength (g) (b) Brand B
Normal score
2200 2400 2600 2800
2100 2200 2300 2400 2500 2600
Step 1 State the null and alternative hypotheses.
Letσ1andσ2denote the population standard deviations of tear strength for Brand A and Brand B, respectively. Then the null and alternative hypotheses are, respectively,
H0:σ1 =σ2(standard deviations of tear strength are the same) Ha:σ1 =σ2(standard deviations of tear strength are different).
Note that the hypothesis test is two tailed.
Step 2 Decide on the significance level,α.
The test is to be performed at the 5% level of significance, orα=0.05.
Step 3 Compute the value of the test statisticF =s12/s22.
We computed the value of the test statistic at the end of Example 11.12, where we found thatF=0.413.
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH Step 4 The critical values for a two-tailed test
areF1−α/2and Fα/2with df=(n1−1,n2−1). Use Table VIII to find the critical values.
We have α=0.05. Also, n1=10 and n2=10, so df=(9, 9). Therefore the critical values are F1−α/2 = F1−0.05/2 = F0.975 and Fα/2= F0.05/2=F0.025. From Table VIII,F0.025=4.03.To obtainF0.975, we first note that it is theF-value having area 0.025 to its left. Apply- ing the reciprocal property ofF-curves (see page 527), we conclude that F0.975 equals the reciprocal of the F-value having area 0.025 to its right for an F-curve with df=(9, 9). (We switched the degrees of freedom, but because they are the same, the difference isn’t ap- parent.) ThusF0.975=4.103=0.25.Figure 11.12A sum- marizes our results.
FIGURE 11.12A
F 0.025 0.025
4.03 0.25
Do not reject H0 Reject H0 Reject H0
Step 4 The F-statistic has df=(n1−1,n2−1). Obtain theP-value by using technology.
We have n1=10 and n2=10, so df = (9, 9). Using technology, we find that the P-value for the hypothesis test isP=0.204, as depicted in Fig. 11.12B.
FIGURE 11.12B
F F = 0.413
P = 0.204
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH Step 5 If the value of the test statistic falls in
the rejection region, rejectH0; otherwise, do not rejectH0.
From Step 3, the value of the test statistic isF=0.413.
This value does not fall in the rejection region shown in Fig. 11.12A, so we do not rejectH0. The test results are not statistically significant at the 5% level.
Step 5 If P≤α, rejectH0; otherwise, do not rejectH0.
From Step 4, P=0.204. Because the P-value ex- ceeds the specified significance level of 0.05, we do not reject H0. The test results are not statistically sig- nificant at the 5% level and (see Table 9.8 on page 378) provide at most weak evidence against the null hypoth- esis.
Step 6 Interpret the results of the hypothesis test.
Interpretation At the 5% significance level, the data do not provide sufficient evidence to conclude that the population standard deviations of tear strength differ for the two vinyl floor coverings.
Report 11.3
Exercise 11.69 on page 538
Confidence Intervals for Two Population Standard Deviations
Using Key Fact 11.5 on page 529, we can also obtain a confidence-interval procedure, Procedure 11.4, for the ratio of two population standard deviations. We call this pro- cedure thetwo-standard-deviationsF-interval procedure.† Like the two-standard- deviations F-test, this procedure is not at all robust to violations of the normality assumption.
PROCEDURE 11.4 Two-Standard-DeviationsF-Interval Procedure
Purpose To find a confidence interval for the ratio of two population standard deviations,σ1andσ2
Assumptions
1. Simple random samples 2. Independent samples.
3. Normal populations
Step 1 For a confidence level of 1−α, use Table VIII to find F1−α/2 and Fα/2with df=(n1−1,n2−1).
Step 2 The confidence interval forσ1/σ2is from 1
Fα/2 ã s1
s2 to 1
F1−α/2 ã s1 s2,
where F1−α/2 and Fα/2 are found in Step 1,n1 and n2 are the sample sizes, ands1ands2are computed from the sample data obtained.
Step 3 Interpret the confidence interval.
To interpret confidence intervals for the ratio,σ1/σ2, of two population standard deviations, considering three cases is helpful.
†The two-standard-deviationsF-interval procedure is also known as theF-interval procedure for two popula- tion standard deviationsand thetwo-sampleF-interval procedure.This confidence-interval procedure is often formulated in terms of variances instead of standard deviations.
534 CHAPTER 11 Inferences for Population Standard Deviations∗
Case 1: The endpoints of the confidence interval are both greater than 1.
To illustrate, suppose that a 95% confidence interval forσ1/σ2 is from 5 to 8. Then we can be 95% confident thatσ1/σ2lies somewhere between 5 and 8 or, equivalently, 5σ2< σ1<8σ2. Thus, we can be 95% confident thatσ1is somewhere between 5 and 8 times greater thanσ2.
Case 2: The endpoints of the confidence interval are both less than 1.
To illustrate, suppose that a 95% confidence interval for σ1/σ2 is from 0.5 to 0.8.
Then we can be 95% confident that σ1/σ2 lies somewhere between 0.5 and 0.8 or, equivalently, 0.5σ2< σ1<0.8σ2. Thus, noting that 1/0.5=2 and 1/0.8=1.25, we can be 95% confident that σ1 is somewhere between 1.25 and 2 times less thanσ2.
Case 3: One endpoint of the confidence interval is less than 1 and the other is greater than 1.
To illustrate, suppose that a 95% confidence interval forσ1/σ2is from 0.5 to 8. Then we can be 95% confident thatσ1/σ2 lies somewhere between 0.5 and 8 or, equiva- lently, 0.5σ2< σ1<8σ2. Thus, we can be 95% confident thatσ1 is somewhere be- tween 2 times less than and 8 times greater thanσ2.
EXAMPLE 11.15 The Two-Standard-Deviations F-Interval Procedure
Elmendorf Tear Strength Use the sample data in Table 11.4 on page 531 to de- termine a 95% confidence interval for the ratio,σ1/σ2, of the standard deviations of tear strength for Brand A and Brand B vinyl floor coverings.
Solution As found in Example 11.14, we can reasonably presume that tear strengths are normally distributed for both Brand A and Brand B vinyl floor cover- ings. Consequently, we can apply Procedure 11.4 to obtain the required confidence interval.
Step 1 For a confidence level of 1−α, use Table VIII to findF1−α/2andFα/2 with df=(n1−1,n2−1).
We want to obtain a 95% confidence interval; consequently,α=0.05. Hence we need to findF0.975andF0.025for df=(n1−1,n2−1)=(9,9). We did so earlier (Example 11.14, Step 4 of the critical-value approach), where we determined that F0.975=0.25 andF0.025=4.03.
Step 2 The confidence interval forσ1/σ2is from 1
Fα/2 ã s1
s2 to 1
F1−α/2 ã s1 s2.
For the data in Table 11.4,s1=128.3 g ands2=199.7 g. From Step 1, we know thatF0.975=0.25 and F0.025=4.03. Consequently, the required 95% confidence interval is from
√1
4.03ã 128.3
199.7 to 1
√0.25ã 128.3 199.7, or 0.32 to 1.28.
Step 3 Interpret the confidence interval.