# Fdist1 TESTING THE EQUALITY OF TWO VARIANCES: THE F TEST Application test assumption of equal variances that was made in using the t-test interest in actually.

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fdist1 TESTING THE EQUALITY OF TWO VARIANCES: THE F TEST Application test assumption of equal variances that was made in using the t-test interest in actually comparing the variance of two populations

fdist2 The F-Distribution Assume we repeatedly select a random sample of size n from two normal populations. Consider the distribution of the ratio of two variances:F = s 1 2 /s 1 2. The distribution formed in this manner approximates an F distribution with the following degrees of freedom: v 1 = n 1 - 1 and v 1 = n 1 - 1

fdist3 Assumptions Random, independent samples from 2 normal populations Variability

fdist4 F-Table The F table can be found on the appendix of our text. It gives the critical values of the F-distribution which depend upon the degrees of freedom.

fdist5 Example 1 Assume that we have two samples with: n 2 = 7 and n 1 =10 df = 7-1= 6 and df = 10-1= 9 Let v = F(6,9) where 6 is the df from the numerator and 9 is the df of the denominator. Using the table with the appropriate df, we find : P(v < 3.37) = 0.95.

fdist6 Example 2: Hypothesis Test to Compare Two Variances 1.Formulate the null and alternate hypotheses. H 0 : 1 2 = 1 2 H a : 1 2 > 2 2 [Note that we might also use 1 2 < 2 2 or 1 2 =/ 2 2 ] 2.Calculate the F ratio. F = s 1 2 /s 1 2 [where s 1 is the largest or the two variances] 3.Reject the null hypothesis of equal population variances if F(v 1 -1, v 2 -1) > F [or F /2 in the case of a two tailed test]

fdist7 Example 2 The variability in the amount of impurities present in a batch of chemicals used for a particular process depends on the length of time that the process is in operation. Suppose a sample of size 25 is drawn from the normal process which is to be compared to a sample of a new process that has been developed to reduce the variability of impurities. Sample 1Sample 2 n 2525 s 2 1.040.51

fdist8 Example 2 continued H 0 : 1 2 = 2 2 H a : 1 2 > 2 2 F(24,24) = s 1 2 /s 2 2 = 1.04/.51 = 2.04 Assuming = 0.05 cv = 1.98 < 2.04 Thus, reject H 0 and conclude that the variability in the new process (Sample 2) is less than the variability in the original process.

fdist9 Try This A manufacturer wishes to determine whether there is less variability in the silver plating done by Company 1 than that done by Company 2. Independent random samples yield the following results. Do the populations have different variances? [solution: reject H 0 since 3.14 > 2.82]

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