100 STATISTICAL TESTS phần 8 potx

GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 167 — #31 THE TESTS 167 rejection number. If θ 0 = 0.04 and θ 1 = 0.08, α = 0.15, β = 0.25, then the graphical representation of the sequential plan is: As soon as (m, S m ) lies on or below the line for a m or on or above the line for r m , sampling is to be stopped; the new process is to be considered in the former case and rejected in the latter. GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 168 — #32 168 100 STATISTICAL TESTS Test 91 Sequential probability ratio test Object A sequential test for the ratio between the mean and the standard deviation of a normal population where both are unknown. Limitations This test is applicable if the observations are normally distributed with unknown mean and variance. Method Let X ∼ N(µ, σ 2 ), where both µ and σ 2 are unknown. We want to find a sequential test for testing H 0 : µ/σ = r 0 against H 1 : µ/σ = r 1 . The sequential probability ratio test procedure is as follows. 1. Continue sampling if b n < t n < a n , where t n = n  i=1 X i     n  i=1 X 2 i = n  i=1 y i     n  i=1 y 2 i where y i = X i /|X i |, i = 1, 2, , n, a n = log 1 − β α and b n = log β 1 − α . 2. Fail to reject H 0 if t n  b n and reject H 0 if t n  a n . Example A useful measure is the ratio of mean divided by standard deviation since it is inde- pendent of measured units. Here we set up a sequential test for the ratio equal to 0.2 versus the alternative that it is equal to 0.4. The rule is that if the test statistic is less than log(7/15) do not reject the null hypothesis and if the test statistic is greater than log(13/5) then reject the null hypothesis; otherwise continue to sample. Numerical calculation Consider a sample from N(µ, σ 2 ) where µ and σ are both unknown. Then we want a sequential test for testing H 0 : µ/σ = 0.2 against H 1 : µ/σ = 0.4. Let α = 0.25, β = 0.35, a n = log(0.65/0.25), b n = log(0.35/0.75). If log 7/15 < t n < log 13/5 then continue sampling. If t n  log 7/15 do not reject H 0 , and if t n  log 13/5 reject H 0 . GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 169 — #33 THE TESTS 169 Test 92 Durbin–Watson test Object To test whether the error terms in a regression model are autocorrelated. Limitations This test is applicable if the autocorrelation parameter and error terms are independently normally distributed with mean zero and variance s 2 . Method This test is based on the first-order autoregressive error model ε t = ϕε t−1 + u t where ϕ is the autocorrelation parameter and the u t , are independently normally distributed with zero mean and variance σ 2 . When one is concerned with positive autocorrelation the alternatives are given as follows: H 0 : ϕ  0 H 1 : ϕ>0. Here H 0 implies that error terms are uncorrelated or negatively correlated, while H 1 implies that they are positively autocorrelated. This test is based on the difference between adjacent residuals ε t − ε t−1 and is given by d = n  t=2 (e t − e t−1 ) 2 n  t=1 e 2 t where e t is the regression residual for period t, and n is the number of time periods used in fitting the regression model. When the error terms are positively autocorrelated the adjacent residuals will tend to be of similar magnitude and the numerator of the test statistic d will be small. If the error terms are either not correlated or negatively correlated e t and e t−1 will tend to differ and the numerator of the test statistic will be larger. The exact action limit for this test is difficult to calculate and the test is used with lower bound d L and the upper bound d U . When the statistic d is less than the lower bound d L , we conclude that positive autocorrelation is present. Similarly, when the test statistic exceeds the upper bound d U , we conclude that positive autocorrelation is not present. When d L < d < d U , the test is inconclusive. Example Data on the sales (£m) of a large company (Y) compared with its sector total (X) have been collected over a five-year period. In order to test the significance of the regression of Y on X, and to obtain confidence intervals on predictions, it is necessary to perform a test of serial correlation on the error term or residuals. The Durbin–Watson test statistic GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 170 — #34 170 100 STATISTICAL TESTS (d) is equal to 0.4765 and is less than the lower d value (d L ) of 1.20 [Table 33]. So the residuals are positively autocorrelated. An appropriate adjustment to the error sum of squares is necessary. Numerical calculation n = 20, α = 0.05 Quarter-t Company sale Industry sales Y i X i 1 77.044 746.512 2 78.613 762.345 3 80.124 778.179 4–– ––– ––– ––– ––– 20 102.481 1006.882 A computer run of a regression package provided us with the value of the test statistic d = 0.4765: d L = 1.20 and d U = 1.41 from Table 33. Since d = 0.4765 < 1.20, the error terms are positively autocorrelated. GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 171 — #35 THE TESTS 171 Test 93 Duckworth’s test for comparing the medians of two populations Object A quick and easy test for comparing the medians of two populations which could be used for a wide range of m and n observations. Limitations This is not a powerful test but it is easy to use and the table values can be easily obtained. It works only on the largest and the smallest values of the observations from different populations. Method Consider the smallest observation from the X population and the largest from the Y population. Then the test statistic, D, is the sum of the overlaps, the number of X observations that are smaller than the smallest Y , plus the number of Y observations that are larger than the largest X. If either 3 +4n/3  m  2n or vice versa we subtract 1 from D. Under these circumstances, the table of critical values consists of the three numbers; 7, 10 and 13. If D  7 we reject the hypothesis of equal medians at α = 0.05. Example Two groups of workers are compared in terms of their daily rates of pay. Are they significantly different? We use Duckworth’s test since the median is an appropriate measure of central tendency for an income variable. The test statistic is 5 which is less than 7 and so not significant. We have no reason to assume any difference in rates of pay between the two groups. Numerical calculation m = n = 12 123456789101112 66.3 68.3 68.5 69.2 70.0 70.1 70.4 70.9 71.1 71.2 72.1 72.1 XXXYYXXYXXYY 13 14 15 16 17 18 19 20 21 22 23 24 72.1 72.7 72.8 73.3 73.6 74.1 74.2 74.6 74.7 74.8 75.5 75.8 XXYYXXYYYXYY We note that there are three X observations below all the Y observations and two Y s above all the Xs. The total is D = 5, which is less than 7 and so not significant. GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 172 — #36 172 100 STATISTICAL TESTS Test 94 χ 2 -test for a suitable probabilistic model Object Many experiments yield a set of data, say X 1 , X 2 , , X n , and the experimenter is often interested in determining whether the data can be treated as the observed values of the random sample X 1 , X 2 , , X n from a given distribution. That is, would this proposed distribution be a reasonable probabilistic model for these sample items? Limitations This test is applicable if both distributions have the same interval classification and the same number of elements. The observed data are observed by random sampling. Method Let X 1 denote the number of heads that occur when coins are tossed at random, under the assumptions that the coins are independent and the probability of heads for each coin has a binomial distribution. An experiment resulted in certain observed values at Y i corresponding to 0, 1, 2, 3 and 4 heads. Let A 1 ={0}, A 2 ={1}, A 3 ={2}, A 4 ={3}, A 5 ={4} be the corresponding heads and if π i = P(X ∈ A i ) when X is B(4, 1 2 ), then we have π 1 = π 5 =  4 0  1 2  4 = 0.0625 π 2 = π 4 =  4 1  1 2  4 = 0.25 π 3 =  4 2  1 2  4 = 0.375. If α = 0.05, then the null hypothesis H 0 : p 1 = π 1 , p 2 = π 2 , p 3 = π 3 , p 4 = π 4 , p 5 = π 5 is rejected if the calculated value is greater than the tabulated value using  q k−1 = k  i=1 (y i − nπ i ) 2 nπ i  ∼ χ 2 k−1 . Example We wish to test whether a binomial distribution is a good model for an application area. A full group of prisoner trainees consists of up to five members. Each full group must have two escorts if five members turn up for training. If up to three turn up for training then only one escort is needed. Data is collected over 100 group turnouts. Is a binomial GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 173 — #37 THE TESTS 173 model a good fit? The calculated chi-squared statistic of 4.47 is less than the critical value of 9.49 [Table 5] so a binomial model is an acceptable model for this data. Numerical calculation In this case y 1 = 7, y 2 = 18, y 3 = 40, y 4 = 31 and y 5 = 4. The computed value is q 4 = (7 − 6.25) 2 6.25 + (18 − 25) 2 25 + (40 −37.5) 2 37.5 + (31 − 25) 2 25 + (4 − 6.25) 2 6.25 = 4.47. Critical value χ 2 4 (0.05) = 9.49 [Table 5]. Hence the hypothesis is not rejected. Thus the data support the hypothesis that B(4, 1 2 ) is a reasonable probabilistic model for X. GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 174 — #38 174 100 STATISTICAL TESTS Test 95 V -test (modified Rayleigh) Object To test whether the observed angles have a tendency to cluster around a given angle indicating a lack of randomness in the distribution. Limitations For grouped data the length of the mean vector must be adjusted, and for axial data all angles must be doubled. Method Given a random sample of n angular values  1 ,  2 , ,  n and a given theoretical direction determined by an angle θ 0 , then the test statistic for the test of randomness is: V = (2n) 1 2 ϑ where ϑ = r cos( ¯  − θ 0 ) and r is the length of the mean vector r = 1 n +    cos  i  2 +   sin  i  2  = (¯x 2 +¯y 2 ) 1 2 ¯  = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ arctan  ¯y ¯x  if ¯x > 0 180 ◦ + arctan  ¯y ¯x  if ¯x < 0. If V is greater than or equal to the critical V (α), the null hypothesis, that the parent population is uniformly distributed (randomness), is rejected. Example A radar screen produces a series of traces; angles from a centre are measured. Do these cluster around a value of 265 degrees? The calculated V statistic is 3.884, which is greater than the critical value of 2.302 [Table 34]. So the angles are not random and do cluster. Numerical calculation n = 15  1 = 250 ◦ ,  2 = 275 ◦ ,  3 = 285 ◦ ,  4 = 285 ◦ ,  5 = 290 ◦ ,  6 = 290 ◦ ,  7 = 295 ◦ ,  8 = 300 ◦ ,  9 = 305 ◦ ,  10 = 310 ◦ ,  11 = 315 ◦ ,  12 = 320 ◦ ,  13 = 330 ◦ ,  14 = 330 ◦ ,  15 = 5 ◦ , θ 0 = 265 ◦ ¯x = 1 n (cos  1 +···+cos  15 ) = 7.287 15 = 0.4858 GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 175 — #39 THE TESTS 175 ¯y = 1 n (sin  1 +···+sin  15 ) = −11.367 15 =−0.7578 r = (¯x 2 +¯y 2 ) 1 2 = (0.4858 2 + 0.7578 2 ) 1 2 = 0.9001 ¯  = arctan  −0.7578 0.4858  =−57.3 ◦ , which is equivalent to 303 ◦ . ϑ = r cos( ¯  − θ 0 ) = 0.9001 × cos(303 ◦ − 265 ◦ ) = 0.9001 × 0.7880 = 0.7093 V = (2 × 15) 1 2 × 0.7093 = 5.4772 ×0.7092 = 3.885 Critical value V 15; 0.01 = 2.302 [Table 34]. Hence reject the null hypothesis of randomness. GOKA: “CHAP05D” — 2006/6/10 — 17:23 — PAGE 176 — #40 176 100 STATISTICAL TESTS Test 96 Watson’s U 2 n -test Object To test whether the given distribution fits a random sample of angular values. Limitations This test is suitable for both unimodal and the multimodal cases. The test is very practical if a computer program is available. It can be used as a test for randomness. Method Given a random sample of n angular values  1 ,  2 , ,  n rearranged in ascending order:  1   2  ···  n . Suppose F() is the distribution function of the given theoretical distribution and let V i = F( i ), i = 1, 2, , n ¯ V = 1 n  V i and C i = 2i − 1. Then the test statistic is: U 2 n = n  i=1 V 2 i − n  i=1  C i V i n  + n  1 3 −  ¯ V − 1 2  2  . If the sample value of U 2 n exceeds the critical value the null hypothesis is rejected. Otherwise the fit is satisfactory. Example A particle atomizer produces traces on a filter paper, which is calibrated on an angular scale. Are the particles equally spaced on an angular scale? The distribution is one of equal angles and the calculated U squared statistic is 0.1361. This is smaller than the critical value of 0.184 [Table 35] so the null hypothesis of no difference from the theoretical distribution is accepted. Numerical calculation n = 13, F() = /360 ◦ ,  1 = 20 ◦ ,  2 = 135 ◦ ,  3 = 145 ◦ ,  4 = 165 ◦ ,  5 = 170 ◦ ,  6 = 200 ◦ ,  7 = 300 ◦ ,  8 = 325 ◦ ,  9 = 335 ◦ ,  10 = 350 ◦ ,  11 = 350 ◦ ,  12 = 350 ◦ ,  13 = 355 ◦ V i =  i /13, i = 1, ,13  V i = 8.8889,  V 2 i = 7.2310,  C i V i /n = 10.9893, ¯ V = 0.68376 U 2 n = 7.2310 − 10.9893 + 13  1 3 − (0.18376) 2  = 0.1361 Critical value U 2 13; 0.05 = 0.184 [Table 35]. Do not reject the null hypothesis. The sample comes from the given theoretical distribution. [...]... 78, 90, 72, 91, 73, 79, 82 , 87 , 78, 83 , 74, 82 , 85 , 75, 67, 72, 78, 88 , 89 , 71, 73, 77, 90, 82 , 80 , 81 , 89 , 87 , 78, 73, 78, 86 , 73, 84 , 68, 75, 70, 89 , 54, 80 , 90, 88 , 81 , 82 , 88 , 82 , 75, 79, 83 , 82 88 , 69, 64, 78, 71, 68, 54, 80 , 73, 72, 65, 73, 93, 84 , 80 , 49, 78, 82 , 95, 69, 87 , 83 , 52, 79, 85 , 67, 82 , 84 , 87 , 83 , 88 , 79, 83 , 77, 78, 89 , 75, 72, 88 , 78, 62, 86 , 89 , 74, 71, 73, 84 , 56, 77, 71 q = 5,... 74, 71, 63, 83 , 74, 82 , 78, 87 , 87 , 82 , 71, 60, 66, 63, 85 , 81 , 78, 80 , 89 , 82 , 82 , 92, 80 , 81 , 74, 90, 78, 73, 72, 80 , 59, 64, 78, 73, 70, 79, 79, 77, 81 , 72, 76, 69, 73, 75, 84 , 81 , 51, 76, 88 70, 76, 79, 86 , 77, 86 , 77, 90, 88 , 82 , 84 , 70, 87 , 61, 71, 89 ,72, 90, 74, 88 , 82 , 68, 83 , 75, 90, 79, 89 , 78, 74, 73, 71, 80 , 83 , 89 , 68, 81 , 47, 88 , 69, 76, 71, 67, 76, 90, 84 , 70, 80 , 77, 93, 89 78, 90, 72,... 0.7967 0 .82 38 0.7054 0.7 389 0.7704 0.7995 0 .82 64 0.7 088 0.7422 0.7734 0 .80 23 0 .82 89 0.7123 0.7454 0.7764 0 .80 51 0 .83 15 0.7157 0.7 486 0.7794 0 .80 78 0 .83 40 0.7190 0.7517 0. 782 3 0 .81 06 0 .83 65 0.7224 0.7549 0. 785 2 0 .81 33 0 .83 89 1.0 1.1 1.2 1.3 1.4 0 .84 13 0 .86 43 0 .88 49 0.9032 0.9192 0 .84 38 0 .86 65 0 .88 69 0.9049 0.9207 0 .84 61 0 .86 86 0 .88 88 0.9066 0.9222 0 .84 85 0 .87 08 0 .89 07 0.9 082 0.9236 0 .85 08 0 .87 29 0 .89 25... 0.96 08 0.9 686 0.9750 0.94 18 0.9525 0.9616 0.9693 0.9756 0.9429 0.9535 0.9625 0.9699 0.9761 0.9441 0.9545 0.9633 0.9706 0.9767 2.0 2.1 2.2 2.3 2.4 0.9772 0. 982 1 0. 986 1 0. 989 3 0.99 18 0.97 78 0. 982 6 0. 986 4 0. 989 6 0.9920 0.9 783 0. 983 0 0. 986 8 0. 989 8 0.9922 0.9 788 0. 983 4 0. 987 1 0.9901 0.9925 0.9793 0. 983 8 0. 987 5 0.9904 0.9927 0.97 98 0. 984 2 0. 987 8 0.9906 0.9929 0. 980 3 0. 984 6 0. 988 1 0.9909 0.9931 0. 980 8 0. 985 0... 6.26 234.0 19.33 8. 94 6.16 236 .8 19.35 8. 89 6.09 2 38. 9 19.37 8. 85 6.04 240.5 19. 38 8 .81 6.00 5 6 7 8 9 6.61 5.99 5.59 5.32 5.12 5.79 5.14 4.74 4.46 4.26 5.41 4.76 4.35 4.07 3 .86 5.19 4.53 4.12 3 .84 3.63 5.05 4.39 3.97 3.69 3. 48 4.95 4. 28 3 .87 3. 58 3.37 4 .88 4.21 3.79 3.50 3.29 4 .82 4.15 3.73 3.44 3.23 4.77 4.10 3. 68 3.39 3. 18 10 11 12 13 14 4.96 4 .84 4.75 4.67 4.60 4.10 3. 98 3 .89 3 .81 3.74 3.71 3.59... 1.6 28 This is less than the tabulated value of 2.37 [Table 3] So the researcher concludes that all five magnification systems are equally effective Numerical calculation Magnification 100 200 400 1200 400 × 1.3 82 , 71, 85 , 89 , 78, 77, 74, 71, 68, 83 , 72, 73, 81 , 65, 62, 90, 92, 80 , 77, 93, 75, 80 , 69, 74, 77, 75, 71, 82 , 84 , 79, 78, 81 , 89 , 79, 82 , 81 , 85 , 76, 71, 80 , 94, 68, 72, 70, 59, 80 , 86 , 98, 82 ,... 2. 28 2.25 2.23 2.20 2. 18 2.20 2. 18 2.15 2.13 2.11 2.12 2.10 2.07 2.05 2.03 2. 08 2.05 2.03 2.01 1. 98 2.04 2.01 1. 98 1.96 1.94 1.99 1.96 1.94 1.91 1 .89 1.95 1.92 1 .89 1 .86 1 .84 1.90 1 .87 1 .84 1 .81 1.79 1 .84 1 .81 1. 78 1.76 1.73 25 26 27 28 29 2.24 2.22 2.20 2.19 2. 18 2.16 2.15 2.13 2.12 2.10 2.09 2.07 2.06 2.04 2.03 2.01 1.99 1.97 1.96 1.94 1.96 1.95 1.93 1.91 1.90 1.92 1.90 1 .88 1 .87 1 .85 1 .87 1 .85 1 .84 ... 0.9251 0 .85 31 0 .87 49 0 .89 44 0.9115 0.9265 0 .85 54 0 .87 70 0 .89 62 0.9131 0.92 78 0 .85 77 0 .87 90 0 .89 80 0.9147 0.9292 0 .85 99 0 .88 10 0 .89 97 0.9162 0.9306 0 .86 21 0 .88 30 0.9015 0.9177 0.9319 1.5 1.6 1.7 1 .8 1.9 0.9332 0.9452 0.9554 0.9641 0.9713 0.9345 0.9463 0.9564 0.9649 0.9719 0.9357 0.9474 0.9573 0.9656 0.9726 0.9370 0.9 484 0.9 582 0.9664 0.9732 0.9 382 0.9495 0.9591 0.9671 0.97 38 0.9394 0.9505 0.9599 0.96 78 0.9744... 2.7 18 2. 681 2.650 2.624 2.602 3.106 3.055 3.012 2.977 2.947 16 17 18 19 20 1.337 1.333 1.330 1.3 28 1.325 1.746 1.740 1.734 1.729 1.725 2.120 2.110 2.101 2.093 2. 086 2. 583 2.567 2.552 2.539 2.5 28 2.921 2 .89 8 2 .87 8 2 .86 1 2 .84 5 21 22 23 24 25 1.323 1.321 1.319 1.3 18 1.316 1.721 1.717 1.714 1.711 1.7 08 2. 080 2.074 2.069 2.064 2.060 2.5 18 2.5 08 2.500 2.492 2. 485 2 .83 1 2 .81 9 2 .80 7 2.797 2. 787 26 27 28 29... 0. 985 0 0. 988 4 0.9911 0.9932 0. 981 2 0. 985 4 0. 988 7 0.9913 0.9934 0. 981 7 0. 985 7 0. 989 0 0.9916 0.9936 2.5 2.6 2.7 2 .8 2.9 0.99 38 0.9953 0.9965 0.9974 0.9 981 0.9940 0.9955 0.9966 0.9975 0.9 982 0.9941 0.9956 0.9967 0.9976 0.9 982 0.9943 0.9957 0.99 68 0.9977 0.9 983 0.9945 0.9959 0.9969 0.9977 0.9 984 0.9946 0.9960 0.9970 0.99 78 0.9 984 0.99 48 0.9961 0.9971 0.9979 0.9 985 0.9949 0.9962 0.9972 0.9979 0.9 985 0.9951 . 89 1200 78, 90, 72, 91, 73, 79, 82 , 87 , 78, 83 , 74, 82 , 85 , 75, 67, 72, 78, 88 , 89 , 71, 73, 77, 90, 82 , 80 , 81 , 89 , 87 , 78, 73, 78, 86 , 73, 84 , 68, 75, 70, 89 , 54, 80 , 90, 88 , 81 , 82 , 88 , 82 , 75,. 79, 83 , 82 400 ×1.3 88 , 69, 64, 78, 71, 68, 54, 80 , 73, 72, 65, 73, 93, 84 , 80 , 49, 78, 82 , 95, 69, 87 , 83 , 52, 79, 85 , 67, 82 , 84 , 87 , 83 , 88 , 79, 83 , 77, 78, 89 , 75, 72, 88 , 78, 62, 86 , 89 ,. 79, 86 , 77, 86 , 77, 90, 88 , 82 , 84 , 70, 87 , 61, 71, 89 ,72, 90, 74, 88 , 82 , 68, 83 , 75, 90, 79, 89 , 78, 74, 73, 71, 80 , 83 , 89 , 68, 81 , 47, 88 , 69, 76, 71, 67, 76, 90, 84 , 70, 80 , 77, 93, 89 1200