A textbook of Computer Based Numerical and Statiscal Techniques part 56 potx

536 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Similarly, we will get b – E(b)= − ad bc N – = c – E(c); d – E(d) = ad bc N – Substituting in (2), we get χ 2 = ad bc N – bg 2 2 1111 Ea Eb Ec Ed af af af af +++ L N M O Q P = ad bc N – bg 2 11 11 abac abbd accd bdcd++ + ++ R S | T | U V | W | + ++ + ++ R S | T | U V | W | L N M M O Q P P bgbgbgbg bgbgbgbg = ad bc N – bg 2 bdac abacbd bdac accdbd +++ +++ + +++ +++ L N M M O Q P P bgbgbgbgbgbg =(ad – bc) 2 cdab abacbdcd +++ ++++ L N M M O Q P P bgbgbgbg = Nad bc abacbdcd – bg bgbgbgbg 2 ++++ Example 11. From the following table regarding the colour of eyes of father and son test if the colour of son’s eye is associated with that of the father. Eye colour of son Light Not light Eye colour of father Light Not light 471 51 148 230 Sol. Null Hypothesis H 0 : The colour of son’s eye is not associated with that of the father i.e., they are independent. Under H 0 , we calculate the expected frequency in each cell as = Product of column total and row total whole total TESTING OF HYPOTHESIS 537 Expected frequencies are: Eye colour of son Light Not light Total Eye colour of father Light 619 522× 900 = 359.02 289 522× 900 = 167.62 522 Not Light 619 900 × 378 = 259.98 289 900 × 378 = 121.38 378 Total 619 289 900 χ 2 = 471 359 02 359 02 51 167 62 167 62 148 259 98 259 98 230 121 38 121 38 2222 –. . –. . –. . –. . bgbgbgbg ++ + = 261.498. Conclusion: At 5% level for 1 d.f., χ 2 is 3.841 (tabulated value) Since tabulated value of χ 2 < calculated value of χ 2 . Hence H 0 is rejected. Example 12. The following table gives the number of good and bad parts produced by each of the three shifts in a factory: Good parts Bad parts Total Day shift 960 40 1000 Evening shift 940 50 990 Night shift 950 45 995 Total 2850 135 2985 Test whether or not the production of bad parts is independent of the shift on which they were produced. Sol. Null Hypothesis H 0 : The production of bad parts is independent of the shift on which they were produced. i.e., the two attributes, production and shifts are independent. Under H 0 , χ 2 = i ij ij ij j AB AB AB == ∑∑ L N M M M M O Q P P P P 1 2 0 2 0 1 3 ejej ej – Calculation of expected frequencies Let A and B be the two attributes namely production and shifts. A is divided into two classes A 1 , A 2 and B is divided into three classes B 1 , B 2 , B 3 . 538 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (A 1 B 1 ) 0 = AB N 12 bgbg = 2850 1000 2985 afaf × = 954.77; (A 1 B 2 ) 0 = AB N 12 bgbg = 2850 990 2985 afaf × = 945.226 (A 1 B 3 ) 0 = AB N 13 bgbg = 2850 995 2985 afaf × = 950; (A 2 B 1 ) 0 = AB N 21 bgbg = 135 1000 2985 afa f × = 45.27 (A 2 B 2 ) 0 = AB N 22 bgbg = 135 990 2985 afaf × = 44.773; (A 2 B 3 ) 0 = AB N 23 bgbg = 135 995 2985 afaf = 45. To calculate the value of χ 2 . Class O i E i (O i – E i ) 2 (O i – E i ) 2 /E i (A 1 B 1 ) 960 954.77 27.3529 0.02864 (A 1 B 2 ) 940 954.226 27.3110 0.02889 (A 1 B 3 ) 950 950 0 0 (A 2 B 1 ) 40 45.27 27.7729 0.61349 (A 2 B 2 ) 50 44.773 27.3215 0.61022 (A 2 B 3 )45 45 0 0 1.28126 Conclusion: The tabulated value of χ 2 at 5% level of significance for 2 degrees of freedom (r – 1)(s –1) is 5.991. Since the calculated value of χ 2 is less than the tabulated value, we accept H 0 . i.e., the production of bad parts is independent of the shift on which they were produced. 12.7.2 Student’s t -distribution The t-distribution is used when sample size is less than equal to 30 (≤ 30) and the population standard deviation is unknown. Let X i , i = 1, 2, , n be a random sample of size n from a normal population with mean µ and variance σ 2 . Then student’s t is defined by t = X Sn – / µ ~ t (n –1 d.f.) where X = 1 1 n X i i n = ∑ is the sample mean TESTING OF HYPOTHESIS 539 S 2 = 1 1n – i n = ∑ 1 XX i – di 2 is an unbiased estimate of the population variance σ 2 . The t-distribution has different values for each d.f. and when the d.f. are infinitely large, the t-distribution is equivalent to normal distribution. Example 13. The 9 items of a sample have the following values 45, 47, 50, 52, 48, 47, 49, 53, 51. Does the mean of these values differ significantly from the assumed mean 47.5 ? Sol. H 0 : µ = 47.5 i.e., there is no significant difference between the sample and population mean. H 1 : µ ≠ 47.5 (two tailed test): Given: n = 9, µ = 47.5 X 45 47 50 52 48 47 49 53 51 XX– – 4.1 – 2.1 0.9 2.9 –1.1 –2.1 –0.1 3.9 1.9 XX– di 2 16.81 4.41 0.81 8.41 1.21 4.41 0.01 15.21 3.61 X = Σx n = 442 9 = 49.11; Σ XX– di 2 = 54.89; s 2 = Σ XX n – – di bg 2 1 = 6.86 ∴ s = 2.619 Applying t-test t = X sn – / µ = 49 1 47 5 2 619 8 .– . . = 16 8 2 619 . . af = 1.7279 t 0.05 = 2.31 for γ = 8. Conclusion: Since t < t 0.05 , the hypothesis is accepted i.e., there is no significant difference between their mean. Example 14. A random sample of 10 boys had the following I. Q’. s: 70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Do these data support the assumption of a population mean I.Q. of 100 ? Find a reasonable range in which most of the mean I.Q. values of samples of 10 boys lie. Sol. Null hypothesis, H 0 : The data are consistent with the assumption of a mean I.Q. of 100 in the population, i.e., µ = 100. Alternative hypothesis: H 1 : µ≠ 100 Test Statistic. Under H 0 , the test statistic is: t = x Sn – / µ di 2 ∼ t (n –1) where x and S 2 are to be computed from the sample values of I.Q.’s. 540 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Calculation for Sample Mean and S.D. X Xx– di Xx– di 2 70 –27.2 739.84 120 22.8 519.84 110 12.8 163.84 101 3.8 14.44 88 –9.2 84.64 83 –14.2 201.64 95 –2.2 4.84 98 0.8 0.64 107 9.8 96.04 100 2.8 7.84 Total 972 1833.60 Hence n = 10, x = 972 10 = 97.2 and S 2 = 1833 60 9 . = 203.73 ∴ t = 97 2 100 203 73 10 .– ./ = 28 20 37 . . = 28 4514 . . = 0.62 Tabulated t 0.05 for (10 – 1) i.e., 9 d.f. for two-tailed test is 2.262. Conclusion: Since calculated t is less than tabulated t 0.05 for 9 d.f., H 0 may be accepted at 5% level of significance and we may conclude that the data are consistent with the assumption of mean I.Q. of 100 in the population. The 95% confidence limits within which the mean I.Q. values of samples of 10 boys will lie are given by x ± t 0.05 Sn/ = 97.2 ± 2.262 × 4.514 = 97.2 ± 10.21 = 107.41 and 86.99 Hence the required 95% confidence intervals is [86.99, 107.41] Example 15. The mean weekly sales of soap bars in departmental stores was 146.3 bars per store. After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2. Was the advertising campaign successful? Sol. We are given: n = 22, x = 153.7, s = 17.2. Null Hypothesis: The advertising campaign is not successful, i.e., H 0 : µ = 146.3 Alternative Hypothesis: H 1 : µ > 146.3. (Right-tail). Test Statistic: Under the null hypothesis, the test statistic is: t = x sn – /– µ 2 1 bg ~ t 22 – 1 = t 21 Now t = 153 7 146 3 17 2 21 2 . − af = 74 21 17 2 . . × = 9.03 TESTING OF HYPOTHESIS 541 Conclusion: Tabulated value of t for 21 d.f. at 5% level of significance for single-tailed test is 1.72. Since calculated value is much greater than the tabulated value, therefore it is highly significant. Hence we reject the null hypothesis. Example 16. A machinist is making engine parts with axle diameters of 0.700 inch. A random sample of 10 parts shows a mean diameter of 0.742 inch with a standard deviation of 0.040 inch. Compute the statistic you would use to test whether the work is meeting the specifications. Also state how you would proceed further. Sol. Here we are given: µ = 0.700 inches, x = 0.742 inches, s = 0.040 inches and n = 10 Null Hypothesis, H 0 : µ = 0.700, i.e., the product is conforming to specifications. Alternative Hypothesis, H 1 : µ ≠ 0.700 Test Statistic : Under H 0 , the test statistic is: t = x sn – / µ 2 = x sn – /– µ 2 1 bg ∼ t (n – 1) Now, t = 9 0 742 0 700 0 040 .–. . bg = 3.15 Here the test statistic ‘t’ follows Student’s t-distribution with 10 – 1 = 9 d.f. We will now compare this calculated value with the tabulated value of t for 9 d.f. and at certain level of significance, say 5%. Let this tabulated value be denoted by t 0 . (i) If calculated ‘t’ viz., 3.15 > t 0 , we say that the value of t is significant. This implies that x differs significantly from µ and H 0 is rejected at this level of significance and we conclude that the product is not meeting the specifications. (ii) If calculated t < t 0 , we say that the value of t is not significant, i.e., there is no significant difference between x and µ. In other words, the deviation x – µ di is just due to fluctuations of sampling and null hypothesis H 0 may be retained at 5% level of significance, i.e., we may take the product conforming to specifications. Example 17. A random sample of size 16 has 53 as mean. The sum of squares of the derivation from mean is 135. Can this sample be regarded as taken from the population having 56 as mean ? Obtain 95% and 99% confidence limits of the mean of the population. Sol. H 0 : There is no significant difference between the sample mean and hypothetical population mean. H 0 : µ = 56; H 1 : µ ≠ 56 (Two tailed test) t : X sn – / µ ∼ t(n – 1 d.f.) Given: X = 53, µ = 56, n = 16, Σ XX– di 2 = 135 s = Σ XX n – – di 2 1 = 135 15 = 3; t = 53 56 316 – / = – 4 t = 4, d.f. = 16 – 1 = 15. 542 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Conclusion: t 0.05 = 1.753. Since t = 4 > t 0.05 = 1.753 i.e., the calculated value of t is more than the table value. The hypothesis is rejected. Hence, the sample mean has not come from a population having 56 as mean. 95% confidence limits of the population mean = X ± s n t 0.05 = 53 ± 3 16 (1.725) = 51.706; 54.293 99% confidence limits of the population mean = X ± s n t 0.01, = 53 ± 3 16 (2.602) = 51.048; 54.951. (i ) t-Test of Significance for Mean of a Random Sample: To test whether the mean of a sample drawn from a normal population deviates significantly from a stated value when variance of the population is unknown. H 0 : There is no significant difference between the sample mean x and the population mean µ i.e., we use the statistic. t = X sn – / µ where X is mean of the sample. s 2 = 1 1n – XX i i n – di 2 1 = ∑ with degrees of freedom (n – 1). At given level of significance α 1 and degrees of freedom (n – 1). We refer to t-table t α (two tailed or one tailed). If calculated t value is such that t < t α the null hypothesis is accepted and for t > t α , H 0 is rejected. (ii ) t-Test For Difference of Means of Two Samples: This test is used to test whether the two samples x 1 , x 2 , x n 1 , y 1 , y 2 , , y n 2 of sizes n 1 , n 2 have been drawn from two normal populations with mean µ 1 and µ 2 respectively under the assumption that the population variance are equal. (σ 1 = σ 2 = σ). H 0 : The samples have been drawn from the normal population with means µ 1 and µ 2 i.e., H 0 : µ 1 ≠ µ 2 . Let X, Y be their means of the two samples. Under this H 0 the test of statistic t is given by t = XY s nn – di 11 12 + – t(n 1 + n 2 – 2 d.f.) Also, if the two sample’s standard deviations s 1 , s 2 are given then we have s 2 = ns ns nn 11 2 22 2 12 2 + + – . And, if n 1 = n 2 = n, t = XY ss n – – 1 2 2 2 1 + can be used as a test statistic. TESTING OF HYPOTHESIS 543 If the pairs of values are in some way associated (correlated) we can’t use the test statistic as given in Note 2. In this case, we find the differences of the associated pairs of values and apply for single mean i.e., t = X sn – / µ with degrees of freedom n – 1. The test statistic is t = d sn/ or t = d sn/–1 , where d is the mean of paired difference. i.e., d i = x i – y i d i = XY– , where (x i , y i ) are the paired data i = 1, 2, , n. Example 18. Samples of sizes 10 and 14 were taken from two normal populations with S.D. 3.5 and 5.2. The sample means were found to be 20.3 and 18.6. Test whether the means of the two populations are the same at 5% level. Sol. H 0 : µ 1 = µ 2 , i.e., the means of the two populations are the same. H 1 : µ 1 ≠µ 2 . Given X = 20.3, X 2 = 18.6; n 1 = 10, n 2 = 14, s 1 = 3.5, s 2 = 5.2 s 2 = ns ns nn 11 2 22 2 12 2 + + – = 10 3 5 14 5 2 10 14 2 22 – af af + + = 22.775. ∴ s = 4.772 t = XX s nn 12 12 11 − + = 20 3 18 6 1 10 1 14 4 772 .– . .+ F H G I K J = 0.8604 The value of t at 5% level for 22 d.f. is t 0.05 = 2.0739. Conclusion: Since t = 0.8604 < t 0.05 the hypothesis is accepted i.e., there is no significant difference between their means. Example 19. Two samples of sodium vapour bulbs were tested for length of life and the following results were got: Size mean Sample S.D. Type I 8 1234 hrs 36 hrs Type II 7 1036 hrs 40 hrs Sample Is the difference in the means significant to generalise that Type I is superior to Type II regarding length of life ? Sol. H 0 : µ 1 = µ 2 , i.e., two types of bulbs have same lifetime. H 1 : µ 1 > µ 2 i.e., type I is superior to type II. s 2 = ns ns nn 11 2 22 2 12 2 + + – 544 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES = 836 740 872 22 af af + + – = 1659.076. ∴ s = 40.7317 The t-statistic t = XX s nn 12 12 11 − + = 1234 1036 40 7317 1 8 1 7 – . + = 18.1480 ~ t(n 1 + n 2 –2d.f) t 0.05 at d.f. 13 is 1.77 (one tailed test). Conclusion: Since calculated t > t 0.05 , H 0 is rejected i.e., H 1 is accepted. ∴ Type I is definitely superior to Type II. where X = i n = ∑ 1 1 X n i i , Y = j n = ∑ 1 2 Y n j 2 ; s 2 = 1 2 12 nn+ – EX X Y Y ij –– di ej 2 2 + L N M O Q P is an unbiased estimate of the population variance σ 2 . t follows t-distribution with n 1 + n 2 – 2 degrees of freedom. Example 20. The following figures refer to observations in live independent samples: Sample I Sample II 25 30 28 34 24 20 13 32 22 38 40 34 22 20 31 40 30 23 36 17 Analyse whether the samples have been drawn from the populations of equal means. Sol. H 0 : The two samples have been drawn from the population of equal means. i.e., there is no significant difference between their means. i.e., µ 1 = µ 2 H 1 : µ 1 ≠µ 2 (Two tailed test) Given n 1 = Sample I size = 10; n 2 = Sample II size = 10 To calculate the two sample mean and sum of squares of deviation from mean. Let X 1 be the Sample I and X 2 be the Sample II. X 1 25 30 28 34 24 20 13 32 22 38 XX– 1 – 1.6 3.4 1.4 7.4 –2.6 –6.6 –13.6 5.4 4.6 11.4 XX 1 1 2 – di 2.56 11.56 1.96 54.76 6.76 43.56 184.96 29.16 21.16 129.96 X 2 40 34 22 20 31 40 30 23 36 17 XX 2 2 – 10.7 4.7 – 7.3 – 9.3 1.7 10.7 0.7 – 6.3 6.7 –12.3 XX 2 2 2 – di 114.49 22.09 53.29 86.49 2.89 114.49 0.49 39.67 44.89 151.29 TESTING OF HYPOTHESIS 545 X 1 = i= ∑ 1 10 X n 1 1 = 26.6 X 2 = i= ∑ 1 10 X n 2 2 = 293 10 = 29.3 Σ XX 1 1 2 – di = 486.4 Σ XX 2 2 2 – di = 630.08 s 2 = 1 2 12 nn+ – ΣΣXX XX 1 1 2 2 2 2 –– didi + L N M O Q P = 1 10 10 2+ – [486.4 + 630.08] = 62.026. ∴ S = 7.875 Under H 0 the test statistic is given by t = XX s nn 12 12 11 – + = 26 6 29 3 7 875 1 10 1 10 .– . . + = – 0.7666 –t(n 1 + n 2 – 2 d.f) t = 0.7666. Conclusion: The tabulated value of t at 5% level of significance for 18 d.f. is 2.1. Since the calculated value t = 0.7666 < t 0.05 . H 0 is accepted. i.e., there is no significant difference between their means. i.e., the two samples have been drawn from the populations of equal means. Applications of t-Distribution: The t-distribution has a wide number of applications in statistics, some of them are: 1. To test if the sample mean X di differs significantly from the hypothetical value µ of the population mean. 2. To test the significance between two sample means. 3. To test the significance of observed partial and mutiple correlation coefficients. 4. To test the significance of an observed sample correlation co-efficient and sample regression coefficient. Also, the critical value or significant value of t at level of significance α and degree of freedom ν for two tailed test are given by P[ J > t ν (α)] = α ⇒ P[ J ≤ t ν (α)] = 1– α The significant value of t at level of significance ‘α’ for a single tailed test can be obtained from those of two tailed test by considering the values at level of significance ‘2α’. 12.7.3 Snedecor’s Variance Ratio Test or F-test Suppose we want to test (i) whether two independent samples x i and y j For i = 1, 2 , n 1 and j = 1, 2, , n 2 have been drawn from the normal populations with the same variance σ 2 , (say) or (ii) whether two independent estimates of the population variance are homogenous or not. . (tabulated value) Since tabulated value of χ 2 < calculated value of χ 2 . Hence H 0 is rejected. Example 12. The following table gives the number of good and bad parts produced by each of. attributes namely production and shifts. A is divided into two classes A 1 , A 2 and B is divided into three classes B 1 , B 2 , B 3 . 538 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (A 1 B 1 ) 0 = AB N 12 bgbg . may take the product conforming to specifications. Example 17. A random sample of size 16 has 53 as mean. The sum of squares of the derivation from mean is 135. Can this sample be regarded as