F- Distribution Theoretically, we might define the F distribution to be the ratio of two independent chi-square distributions, each divided by their degrees of freedom. Hence, if f is a value of the random variable F, we have: F= = =Where X12 is a value of a chi-square distribution with v1= n1-1 degreesof freedom and X22 is a value of a chi-square distribution with v2= n2-1degrees of freedom. We say that f is a value of the F- distribution withv1 and v2 degrees of freedom.
To obtain an f value, first select a random sample of size n1 from a normalPopulation having a variance and compute .An independent sample of size n, is then selected from a secondnormal population with variance and is computed.The ratio of the two quantitiesand is the denominator is called the F- distribution, with v1 and v2 degrees of freedom.
• The number of degrees of freedom associated with the sample variance in the numerator is always stated first, followed by the number of degrees of freedom associated with the sample variance in the denominator. Thus the curve of the F distribution depends not only on the two parameters v1 and v2 but also on the order in which we state them. Once these two values are given, we can identify the curve.
6 and 24 degrees of freedom 6 and 10 degrees of freedom TYPICALF DISTRIBUTION
• Let be the f value above which we find an area equal to . This is illustrated by the shaded region.• Example: The f value with 6 and 10 degrees of freedom, leaving an area of 0.05 to the right, is = 3.22. By means of the following theorem, Table A.7 can be used to find values of and .
• The F distribution is applied primarily in the analysis of variance where we wish to test the equality of several means simultaneously. We shall use the F distribution to make inferences concerning the variance of two normal populations.
TEST ABOUT TWO VARIANCES OR STANDARD DEVIATION
FORMULA:Where : is the greater variance is the smaller variance
• Example 9.1.31. A group of Mass Communication students study telephone calls at the registrar’s office. The students monitor the time of the incoming and outgoing calls for one day. They that 17 incoming calls last an average of 4.13 minutes with a variance of 1.36 test the hypothesis that the variances of the samples are equal. Use the 0.05 level of significance.
• Solution: We follow the steps in hypothesis testing.a. H0 : The variances of the samples are equal.b. Ha : The variances of the samples are not equal
c. Setd. Tabular value of F. Use F-test (test about two variances or standard deviations) v = (17-1, 12-1)e. Computed value of F
f. Conclusion: Since , accept H0. Hence, the variances of the samples are equal.
Here’s the Solution:a. H0 : The two samples come from the population with equal variances.b. Ha : The two samples come from the population with smaller variances
SEATWORK: From sample of 25 observations, theestimate of the standards, the estimate of thevariance of the population was found to 15.0From another sample of 14 observations, theestimate variance was found to be 9.7 Can weaccept the hypothesis that the two samplescome from populations with equal variance,or must we conclude that the variance of thesecond population is smaller? Use the 0.01level of significance.
c. Setd. Tabular value of F. Use F-test. v = (25-1, 14-1)e. Computed value of Ff. Conclusion: Since , accept H0. Hence, the two samples come from the population with equal variances.
QUIZ Samples from two makers of ball bearings are collected,and their diameters (in inches) are measured, with thefollowing results: •Acme: n1=80, s1=0.0395 •Bigelow: n2=120, s2=0.0428 Assuming that the diameters of the bearings from bothcompanies are normally distributed, test the claim that there isno difference in the variation of the diameters between the twocompanies.
• The hypotheses are: H0: Ha :• α=0.05.• The test statistic is F test• Use formula: =• Since the first sample had the smaller standard deviation, this is a left-tailed test. The p-value is• Since > , we fail to reject H0 .• There is insufficient evidence to conclude that the diameters of the ball bearings in the two companies have different standard deviations.