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- 1. The F - Distribution
- 2. The F Distribution <ul><li>Named in honor of R.A. Fisher who studied it in 1924. </li></ul><ul><li>Used for comparing the variances of two populations. </li></ul><ul><li>It is defined in terms of ratio of the variances of two normally distributed populations. So, it sometimes also called variance ratio. </li></ul><ul><li>F–distribution: where </li></ul>
- 3. <ul><li>Degrees of freedom: v 1 = n 1 – 1 , v 2 = n 2 – 1 </li></ul><ul><li>For different values of v 1 and v 2 we will get different distributions, so v 1 and v 2 are parameters of F distribution. </li></ul><ul><li>If σ 1 2 = σ 2 2 , then , the statistic F = s 1 2 /s 2 2 follows F distribution with n 1 – 1 and n 2 – 1 degrees of freedom. </li></ul>
- 4. Probability density function <ul><li>Probability density function of F-distribution: </li></ul><ul><li>where Y 0 is a constant depending on the values of v 1 and v 2 such that the area under the curve is unity. </li></ul>
- 5. Properties of F-distribution <ul><li>It is positively skewed and its skewness decreases with increase in v 1 and v 2 . </li></ul><ul><li>Value of F must always be positive or zero, since variances are squares. So its value lies between 0 and ∞. </li></ul><ul><li>Mean and variance of F-distribution: </li></ul><ul><li>Mean = v 2 /(-v 2 -2), for v 2 > 2 </li></ul><ul><li>Variance = 2v 2 2 (v 1 +v 2 -2) , for v 2 > 4 </li></ul><ul><li>v 1 (v 2 -2) 2 (v 2 -4) </li></ul><ul><li>Shape of F-distribution depends upon the number of degrees of freedom. </li></ul>
- 6. Testing of hypothesis for equality of two variances <ul><li>It is based on the variances in two independently selected random samples drawn from two normal populations. </li></ul><ul><li>Null hypothesis H o : σ 1 2 = σ 2 2 </li></ul><ul><li>F = s 1 2 / σ 1 2 , which reduces to F = s 1 2 </li></ul><ul><li>s 2 2 / σ 2 2 s 2 2 </li></ul><ul><li>Degrees of freedom v 1 and v 2 . </li></ul><ul><li>Find table value using v 1 and v 2 . </li></ul><ul><li>If calculated F value exceeds table F value, null hypothesis is rejected. </li></ul>
- 7. Problem 1 <ul><li>Two sources of raw materials are under consideration by a company. Both sources seem to have similar characteristics but the company is not sure about their respective uniformity. </li></ul><ul><li>A sample of 10 lots from source A yields a variance of 225 and a sample of 11 lots from source B yields a variance of 200. Is it likely that the variance of source A is significantly greater than the variance of source B at α = 0.01 ? </li></ul>
- 8. Solution <ul><li>Null hypothesis H o : σ 1 2 = σ 2 2 i.e. the variances of source A and that of source B are the same. The F statistic to be used here is </li></ul><ul><li>F = s 1 2 / s 2 2 </li></ul><ul><li>where s 1 2 = 225 and s 2 2 = 200 </li></ul><ul><li>F = 225/200 = 1.1 </li></ul><ul><li>Value for v 1 =9 and v 2 =10 at 1% level of significance is 4.49. Since computed value of F is smaller than the table value of F, the null hypothesis is accepted. </li></ul><ul><li>Hence the variances of two populations are same. </li></ul>
- 9. Problem 2 <ul><li>In order to test the belief that incomes earned by college graduates show much greater variability than the earnings of those who did not attend college. A sample of 21 college graduates is taken and the sample std. deviation for the sample is s 1 = Rs 17,000. For the second sample of 25 non-graduates and obtains a std dev. s 2 = Rs 7,500. </li></ul><ul><li>s 1 = Rs 17,000 n 1 = 21 </li></ul><ul><li>s 1 = Rs 7,500 n 2 = 25 </li></ul>
- 10. Solution <ul><li>Null hypothesis H o : σ 1 2 = σ 2 2 or ( σ 1 2 / σ 2 2 = 1) </li></ul><ul><li>Alternative hypothesis H 1 : σ 1 2 > σ 2 2 </li></ul><ul><li>Level of significance α = 0.01 </li></ul><ul><li>F = s 1 2 / s 2 2 </li></ul><ul><li>F = (17,000) 2 / (7,500) 2 </li></ul><ul><li>F = 5.14 </li></ul><ul><li>Value for v 1 =20 and v 2 =24 at 1% level of significance is 2.74. Since computed value of F is greater than the table value of F, the null hypothesis is rejected. </li></ul>

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