GERTRUDE COX: SPREADING THE GOSPEL ACCORDING TO ST. GERTRUDE
PROCEDURE 11.3 Two-Standard-Deviations F -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 Ha:σ1< σ2 Ha:σ1> σ2
or or
(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 valueF0.
CRITICAL-VALUE APPROACH OR P-VALUE APPROACH Step 4 The critical value(s) are
F1−α/2andFα/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
Do not reject H0 Reject
H0
Do not reject H0 Reject H0
␣/2 ␣
F␣/2
Step 5 If the value of the test statistic falls in the rejection region, rejectH0; otherwise, do not reject H0.
Step 4 TheF-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 IfP ≤α, reject H0; otherwise, do not rejectH0.
Step 6 Interpret the results of the hypothesis test.
†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.
For the P-value approach, we could use Table VIII to estimate theP-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 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. The plots suggest that we can reasonably 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.
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.
11.2 Inferences for Two Population Standard Deviations, Using Independent Samples∗ 555
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/2andFα/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 550), 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 appar- ent.) ThusF0.975= 4.103=0.25.Figure 11.12A summa- rizes our results.
FIGURE 11.12A
F 0.025 0.025
4.03 0.25
Do not reject H0 Reject H0 Reject H0
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 reject H0. The test results are not statistically significant at the 5% level.
Step 4 TheF-statistic has df=(n1−1,n2−1).
Obtain the P-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
Step 5 If P≤α, reject H0; otherwise, do not rejectH0.
From Step 4,P=0.204. Because the P-value exceeds the specified significance level of 0.05, we do not re- ject H0. The test results are not statistically significant at the 5% level and (see Table 9.8 on page 408) provide at most weak evidence against the null hypothesis.
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 11.63 on page 560
Confidence Intervals for Two Population Standard Deviations
Using Key Fact 11.5 on page 552, we can also obtain a confidence-interval procedure, Procedure 11.4, for the ratio of two population standard deviations. We call it thetwo- standard-deviationsF-interval procedure.†Like the two-standard-deviationsF-test, this procedure is not at all robust to violations of the normality assumption.
†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.