fdist

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fdist 1 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

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Page 1: fdist

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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

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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 = s1

2/s12.

The distribution formed in this manner approximates an F distribution with the following degrees of freedom:v1 = n1 - 1 and v1 = n1 - 1

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Assumptions

Random, independent samples from 2 normal populations

Variability

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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.

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Example 1 Assume that we have two samples with:

n2 = 7 and n1 =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.

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Example 2: Hypothesis Testto Compare Two Variances1. Formulate the null and alternate hypotheses.

H0: 12= 1

2

Ha: 12> 2

2

[Note that we might also use 12 < 2

2 or 12 =/ 2

2]

2. Calculate the F ratio.F = s1

2/s12

[where s1 is the largest or the two variances]

3. Reject the null hypothesis of equal population variances if F(v1-1, v2-1) > F

[or F/2 in the case of a two tailed test]

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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 1 Sample 2

n 25 25s2 1.04 0.51

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Example 2 continued

H0: 12 = 22

Ha: 12 > 22

F(24,24) = s12/s22 = 1.04/.51 = 2.04

Assuming = 0.05

cv = 1.98 < 2.04

Thus, reject H0 and conclude that the variability in the new process (Sample 2) is less than the variability in the original process.

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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 H0 since 3.14 > 2.82]