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F Test

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When we want to test the equality of variances of two normal populations, we make use of F- test based on F-distribution. When we want to test the equality of variances of two normal populations, we make use of F- test based on F-distribution. In such a situation, the null hypothesis happens to be Ho : σ 2 p1 = σ 2 p2 In such a situation, the null hypothesis happens to be Ho : σ 2 p1 = σ 2 p2 Σ 2 p1 and σ 2 p2 representing the variances of two normal populations Σ 2 p1 and σ 2 p2 representing the variances of two normal populations

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F test Uses of F test 1)F-test for testing the equality of several population means i.e. testing µ 1 = µ 2 = ……µ k ( for normal population) µ 1 = µ 2 = ……µ k ( for normal population) 2) F-test is for testing the significance of an observed sample correlation ratio. 3) F-test is for test the Linearity of Regression. 4) For testing the significances of an observed sample multiple correlation. sample multiple correlation.

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F Test This test is also known as variance- Ratio Test This test is also known as variance- Ratio Test To test this other ratio of F is calculated as follows: To test this other ratio of F is calculated as follows: s 1 2 s 1 2 F = ------- where s 1 2 > s 2 2 F = ------- where s 1 2 > s 2 2 s 2 2 s 2 2 s 1 2 = is calculated by s 1 2 = is calculated by ∑ (x 1 – x 1 ) 2 ∑ (x 1 – x 1 ) 2 s 1 2 = -------------------- s 1 2 = -------------------- n 1 - 1 n 1 - 1

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S 2 2 is calculated by: ∑ ( x 2 – x 2 ) 2 ∑ ( x 2 – x 2 ) 2 s 2 2 = -------------------- s 2 2 = -------------------- n 2 - 1 n 2 - 1 n1 and n2 refer to the number of items in first and second samples respectively. Here also reference is made to the calculated value of F to its table value (generally at 5% level of significance). If calculated value of F is less than the table value of F at a certain degree f freedom, it is said that the ratio is not significant and so the samples could have come from two normal populations with the same variance and vice- versa.

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It may be recalled that in the ‘F’ ratio, the numerator is greater than the denominator i.e. It may be recalled that in the ‘F’ ratio, the numerator is greater than the denominator i.e. Larger estimate of population variance Larger estimate of population variance F = ----------------------------------------------------- Smaller estimate of population variance Smaller estimate of population variance Thus the larger of the two sample variances is to be taken in the numerator Thus the larger of the two sample variances is to be taken in the numerator Let us assume that S1 2 > S2 2 Let us assume that S1 2 > S2 2 However S2 2 > S1 2, we would defined the ‘F” ratio as S2 2 / S1 2

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Example From the normal population two random samples drawn provided the following- Sample A : 20 16 26 27 23 22 18 24 25 19 Sample B : 27 33 42 35 32 34 38 28 41 43 30 37 s 1 2 s 1 2 F = ------- F = ------- s 2 2 s 2 2 ∑ (x 1 – x 1 ) 2 ∑ (x 1 – x 1 ) 2 s 1 2 = ----------------- s 1 2 = ----------------- n 1 – 1 n 1 – 1 ∑ ( x 2 – x 2 ) 2 ∑ ( x 2 – x 2 ) 2 s 2 2 = -------------------- s 2 2 = -------------------- n 2 - 1 n 2 - 1

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The calculation of s 1 2 and s 2 2 is made as is given under Sample A X 1 (x 1 – x 1 ) (x 1 – x 1 ) 2 Sample B X 2 (x 2 – x 2 )(x 2 – x 2 ) 2 20-2427-864 16-63633-24 2641642749 275253500 231132-39 2200341 18-4163839 242428-749 253941636 19-3943864 30-525 3724 X 1 = 220X 1 =0 ∑(x 1 – x 1 ) 2 = 120 X 2 = 420X 2 = 0 (x 2 – x 2 ) 2 = 314

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220 420 220 420 X1 = ------ = 22 x2 =------ = 35 10 12 10 12 ∑ (x 1 – x 1 ) 2 120 ∑ (x 1 – x 1 ) 2 120 s 1 2 = ----------------- = -------- = 13.33 n 1 - 1 9 n 1 - 1 9 ∑ ( x 2 – x 2 ) 2 314 ∑ ( x 2 – x 2 ) 2 314 s 2 2 = ---------------- = ------- = 28.55 s 2 2 = ---------------- = ------- = 28.55 n 2 - 1 11 n 2 - 1 11

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s 1 2 13.33 s 1 2 13.33 F = ------- = -------- F = ------- = -------- s 2 2 28.55 s 2 2 28.55. But since numerator is always the greater variance.Therefore, 28.55 28.55 F = ---------- = 2.14 13.33 13.33 Degrees of freedom (12-1) =11 and (10-1) = 9 f = 3.11 Degrees of freedom (12-1) =11 and (10-1) = 9 f 0.5 = 3.11 The table value of F at 11 and 9 degrees of freedom at 5 % level of significance is 3.11 which is greater that the calculated value of F (2.14). The table value of F at 11 and 9 degrees of freedom at 5 % level of significance is 3.11 which is greater that the calculated value of F (2.14). Therefore, the hypothesis set is accepted and the sample may well be drawn from the same population as the two populations have the same variance. Therefore, the hypothesis set is accepted and the sample may well be drawn from the same population as the two populations have the same variance.

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Assumptions for F-test The following assumptions are required for the validity of the F test for comparing variances of two populations The following assumptions are required for the validity of the F test for comparing variances of two populations 1.: i.e. the values in each populations are normally distributed 1.Normality: i.e. the values in each populations are normally distributed 2.Homogeneity: i.e. the variance within each population are equal ( σ 1 2 = σ 2 2 = σ 2 ) The assumption is needed in order to combine or pool the variances within the populations into a single source of variation ‘within populations’. The assumption is needed in order to combine or pool the variances within the populations into a single source of variation ‘within populations’. 3. Independence of errors: It stipulates that the error i.e. variation of each value around its own population mean, should be independent of the value. 3. Independence of errors: It stipulates that the error i.e. variation of each value around its own population mean, should be independent of the value.

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Example To compare the variability of dividends declared for a year, 9 companies were randomly selected from the IT sector and 8 companies were randomly selected from the pharmaceutical sector. To compare the variability of dividends declared for a year, 9 companies were randomly selected from the IT sector and 8 companies were randomly selected from the pharmaceutical sector. The sum of squares of the dividends from their respective means were found to be 160 to 91 respectively. Can we conclude that the variability in dividends of the companies in the two sectors is the same? The sum of squares of the dividends from their respective means were found to be 160 to 91 respectively. Can we conclude that the variability in dividends of the companies in the two sectors is the same?

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Solution Let σ 1 2 and σ 2 2 represent the variances of dividends paid by IT sector and pharmaceutical companies. According to the hypotheses are: H O : σ 1 2 = σ 2 2 H 1 : σ 1 2 ≠ σ 2 2 ‘F’ test is known to be the appropriate test for test of such hypothesis

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Accordingly, the appropriate statistic is s1 2 F = ----------- s2 2 Where, ∑ (x 1 – x 1 ) 2 ∑ (x 1 – x 1 ) 2 s 1 2 = ----------------- n 1 – 1 n 1 – 1 ∑ (x 2 – x 2 ) 2 ∑ (x 2 – x 2 ) 2 s 2 2 = ---------------- n 2 – 1 n 2 – 1 n 1 and n 2 being sample sizes from the two populations

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Further, it is presumed that S 1 2 > S 2 2 Further, it is presumed that S 1 2 > S 2 2 the d.f. for ‘F’ are (n 1 -1), (n 2 -1) the d.f. for ‘F’ are (n 1 -1), (n 2 -1) However if S 1 2 < S 2 2 then the ‘F’ ratio is defined as However if S 1 2 < S 2 2 then the ‘F’ ratio is defined as s 2 2 F = --------- with d.f. as (n 2 – 1), (n 1 -1) s 1 2 In the example, 160 91 160 91 s 1 2 = -------- = 20 s 2 2 = ------- = 13 s 1 2 = -------- = 20 s 2 2 = ------- = 13 9 -1 8 – 1 9 -1 8 – 1 20 20 ‘F’ = ------- = 1.54 ‘F’ = ------- = 1.54 13 13 The tabulated value of ‘F’ for 8, 7 d.f. at 5 % level of significance is 3.73 The tabulated value of ‘F’ for 8, 7 d.f. at 5 % level of significance is 3.73 The calculated value of ‘F’ (=1.54) is less than the tabulated value of ‘F’ (=3.73) therefore, the null hypothesis is accepted. i.e. the variances of two p0pulations are equal. The calculated value of ‘F’ (=1.54) is less than the tabulated value of ‘F’ (=3.73) therefore, the null hypothesis is accepted. i.e. the variances of two p0pulations are equal.

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Example With the help of following data about daily prices of stocks, one could compare variation in two stocks listed on BSE With the help of following data about daily prices of stocks, one could compare variation in two stocks listed on BSE DateONGC BSE RIL BSE 5.10.061143.61155.05 6.10.061143.11163.05 9.10.061150.051154.1 10.10.061145.41150.5 11.10.061128.91143.2 12.10.061132.851169.5 13.10.061153.751190.15 16.10.061172.151213.4 17.10.061167.91216.05 18.10.061164.31208

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DateONGC (x 1 ) (x1 - x 1)(x1 - x ) 2 RIL (x 2 ) (x 2 - x 2 ) (x 2 - x 2 ) 2 5.10.061143.6- 6.846.241155.05- 21.25451.56 6.10.061143.1- 7.860.841163.05-13.25175.56 9.10.061150.05- 0.350.12251154.1- 22.2492.84 10.10.061145.4- 5.0251150.5- 25.8665.64 11.10.061128.9-21.5462.251143.2- 33.11095.61 12.10.061132.85- 17.553081169.5- 6.846.24 13.10.061153.753.3511.221190.1513.85191.82 16.10.061172.1521.75473.061213.437.11376.41 17.10.061167.917.5306.251216.0539.751580.06 18.10.061164.313.9193.21120831.71004.89 ∑115042.51886.19117637080.56 xx 1150.41176.3

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∑ (x 1 – x 1 ) 2 1886.19 ∑ (x 1 – x 1 ) 2 1886.19 s 1 2 = ---------------- = ------------ = 209.57 n 1 – 1 10 -1 n 1 – 1 10 -1 ∑ ( x 2 – x 2 ) 2 ∑ ( x 2 – x 2 ) 2 7080.56 s 2 2 = ---------------- = ------------ = 786.72 n 2 – 1 10 – 1 n 2 – 1 10 – 1 S 1 2 < S 2 2 then the ‘F’ ratio is defined as S 1 2 < S 2 2 then the ‘F’ ratio is defined as s 2 2 786.72 F = --------- = ------------ = 3.75 s 1 2 209.57

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The tabulated value of ‘F’ for 9, 9 d.f. at 5 % level of significance is 3.18 The tabulated value of ‘F’ for 9, 9 d.f. at 5 % level of significance is 3.18 The calculated value of ‘F’ (= 3.75) is greater than the tabulated value of ‘F” 3.18 The calculated value of ‘F’ (= 3.75) is greater than the tabulated value of ‘F” 3.18 Therefore null hypothesis is rejected that is the variance of two populations are different. Therefore null hypothesis is rejected that is the variance of two populations are different.

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