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Chapter 19 Nonparametric Methods Learning Objectives Learn the difference between parametric and nonparametric methods Know the advantages of nonparametric methods and when they are applicable Be able to use the sign test to conduct hypothesis tests about a median Learn how to use the sign test to test a hypothesis with matched samples Be able to use the Wilcoxon signed-rank test to test a hypothesis with matched samples and to test a hypothesis about the median of a symmetric population Be able to use the Mann-Whitney-Wilcoxon test for the comparison of two populations when using independent random samples from each population Be able to use the Kruskal-Wallis test for the comparison of k populations when using independent random samples from each population Be able to compute the Spearman rank-correlation coefficient and test for a significant rank correlation for two variables that use ordinal or rank-ordered data Solutions: 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 H0: Median < 150 Ha: Median > 150 There are 22 + = 27 observations where the value is different from 150 Use the normal approximation with µ = 5n = 5(27) = 13.5 and σ = 25n = 25(27) = 2.5981 With the number of plus signs = 22 in the upper tail, we use the continuity correction factor and the normal distribution approximation as follows 21.5 − 13.5 P ( x ≥ 21.5) = P z ≥ = P ( z ≥ 3.08) 2.5981 ÷ The upper-tail p-value = (1.0000 - 9990) = 0010 p-value < 01; reject H0 Conclusion: The population median is greater than 150 Let p = the probability of a preference for brand A H0 : p = 50 Ha : p ≠ 50 Dropping the individual with no preference, the binomial probabilities for n = and p = 50 are as follows x Probability 0.0020 0.0176 0.0703 0.1641 0.2461 0.2461 0.1641 0.0703 0.0176 0.0020 Number of plus signs = P(x > 7) = P(7) + P(8) +P(9) = 0703 + 0176 + 0020 = 0899 Two-tailed p-value = 2(.0899) = 1798 p-value > 05, not reject H0 There is no indication that a difference in preference exists A larger sample size should be considered H0: Median = 18 Ha: Median ≠ 18 Dropping the restaurant with 18 part-time employees, the binomial probabilities for n = and p 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods = 50 are as follows x Probabilit y 0.0039 0.0313 0.1094 0.2188 0.2734 0.2188 0.1094 0.0313 0.0039 Number of plus signs = P(x > 7) = P(7) + P(8) = 0313 + 0039 = 0352 Since this is an upper-tailed test, p-value = 0352 p-value < 05, reject H0 Conclude that the population median is greater than 18 There has been an increase in the median number of part-time employees a H0: Median > 15 Ha: Median < 15 b Dropping Vanguard GNMA with net assets of $15 billion, the binomial probabilities for n = and p = 50 are as follows x Probability 0.0020 0.0176 0.0703 0.1641 0.2461 0.2461 0.1641 0.0703 0.0176 0.0020 Number of plus signs = American Funds with net assets of $22.4 billion P(x < 1) =P(0) + P(1) = 0020 + 0176 = 0196 p-value = 0196 p-value 05;do not reject H0 We are unable to conclude that the median annual income for Popular Photography & Imagining subscribers differs from $75,000 H0: Median < 56.2 Ha: Median > 56.2 There are n = 31 + 17 = 48 observations where the value is different from 56.2 Use the normal approximation with µ = 5n = 5(48) = 24 and σ = 25n = 25(48) = 3.4641 With the number of plus signs = 31in the upper tail, we use the continuity correction factor and the normal distribution approximation as follows 30.5 − 24 P( x ≥ 30.5) = P z ≥ = P( z ≥ 1.88) 3.4641 ÷ Upper-tail p-value = (1.0000 - 9699) = 0301 p-value < 05; reject H0 Conclude that the median annual income for families living in Chicago is greater that $56.2 thousand a Let p = probability the shares held will be worth more after the split H0 : p ≤ 50 Ha : p > 50 b Let the number of plus signs be the number of increases in value Use the binomial probability tables with n = 18 (there were ties in the 20 observations) With x = 14 plus signs, P(x ≥ 14) = P(14) + P(15) + P(16) + P(17) + P(18) = 0117 + 0031 + 0006 + 0001 + 0000 = 0155 Upper-tailed p-value = 0155 p-value ≤ 05, reject H0 The results support the conclusion that stock splits are beneficial for shareholders a H0 : p = 50 Ha : p ≠ 50 Dropping the individual with no preference, selected binomial probabilities for n = 15 and p = 50 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods are as follows x Probability 0000 0002 0018 0085 0278 Let the response faster pace be a plus sign Number of plus signs = Since this is in the lower tail of binomial distribution, we compute P(x < 4) = P(0) + P(1) + P(2) + P(3) + P4) = 0000 + 0002 + 0018 + 0085 + 0278 = 0384 Two-tailed p-value = 2(.0384) = 0768 p-value > 05, not reject H0 The sample does not allow the conclusion that a difference in preference exists for the fast pace or slower pace of life b Of the original 16 respondents, 4/16(100) = 25% favored a faster pace of life and 11/16(100) = 68.8% favored a slower pace of life There is an almost to in favor of the slower pace of life In addition, the p-value 0768 is low, but not low enough to detect a difference in preference While we are unable to reject H0 and conclude a difference in preference exists from this sample of 16 respondents, continuing the study and taking a larger sample would be recommended Let p = proportion of adults who feel children will have a better future H0 : p = 50 Ha : p ≠ 50 Eliminating the responses that said about the same, we have n = 242 + 310 = 552 Using the normal distribution, we have µ = n = 5(552) = 276 σ = 25n = 25(552) = 11.7473 With the number of plus signs = 242 in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows: 242.5 − 276 P( x ≤ 242.5) = P z ≤ = P ( z ≤ −2.85) 11.7473 ÷ The two-tailed p-value =2(.0022) = 0044 p-value ≤ 05, reject H0 Conclude that there is a significant difference between the adults projecting a better future and adults projects a worse future In 2008, more adults were projecting a worse future for their children 10 Let p = proportion who favor America Idol 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 H0 : p = 50 Ha : p ≠ 50 Eliminating the responses that said about the same, we have n = 330+ 270 = 600 Using the normal distribution, we have µ = n = 5(600) = 300 σ = 25n = 25(600) = 12.2474 With the number of plus signs = 330 in the upper tail, we will use the continuity correction factor and the normal distribution approximation as follows 329.5 − 300 P( x ≥ 330) = P z ≥ = P ( z ≥ 2.41) 12.2474 ÷ The two-tailed p-value = 2(1.0000 – 9920) = 0160 p-value ≤ 05, reject H0 Conclude that there is a significant difference between the preference for American Idol and Dancing with the Stars Based on the data, American Idol is most preferred 11 Let p = proportion who purchase brand A computers H0 : p = 50 Ha : p ≠ 50 Eliminating purchases of other computers, we have n = 202+ 175 = 377 Using the normal distribution, we have µ = n = 5(377) = 188.5 σ = 25n = 25(377) = 9.7082 With the number of plus signs = 202 in the upper tail, we will use the continuity correction factor and the normal distribution approximation as follows 201.5 − 188.5 P ( x ≥ 201.5) = P z ≥ = P ( z ≥ 1.34) 9.7082 ÷ The two-tailed p-value = 2(1.0000 – 9099) = 1802 p-value > 05, not reject H0 We are unable to conclude that there is a difference between the market shares for the two brands of computers 12 H0: Median for Additive - Median for Additive = 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods Ha: Median for Additive - Median for Additive ≠ Additive Car 10 11 12 20.12 23.56 22.03 19.15 21.23 24.77 16.16 18.55 21.87 24.23 23.21 25.02 18.05 21.77 22.57 17.06 21.22 23.80 17.20 14.98 20.03 21.15 22.78 23.70 Differenc e 2.07 1.79 -0.54 2.09 0.01 0.97 -1.04 3.57 1.84 3.08 0.43 1.32 Absolute Difference 2.07 1.79 0.54 2.09 0.01 0.97 1.04 3.57 1.84 3.08 0.43 1.32 Signed Ranks Rank 10 12 11 Negative -3 10 -5 Sum of Positive Signed Ranks µT = + σT = + Positive 12 11 T + = 70 n(n + 1) 12(13) = = 39 4 n(n + 1)(2n + 1) 12(13)(25) = = 12.7475 24 24 T + = 70 is in the upper tail of the sampling distribution Using the continuity correction factor, we have: 69.5 − 39 P (T + ≥ 70) = P z ≥ = P ( z ≥ 2.39) 12.7475 ÷ Using z = 2.39, the p-value = 2(1.0000 - 9916) = 0168 13 p-value < 05, reject H0 Conclude that there is a significant difference in the median miles per gallon for the two additives H0: Median time without Relaxant - Median time with Relaxant < Ha: Median time without Relaxant - Median time with Relaxant > Relaxant Absolute Signed Ranks Differenc Subject No Yes e Difference Rank Negative Positive 15 10 5 9 12 10 2 3 22 12 10 10 10 10 11 -3 6.5 -6.5 10 1 1 2 3 10 -2 -3 10 3 6.5 6.5 14 11 19 - 6.5 6.5 Learning All Rights Reserved © 2010 Cengage 10 May not be 3 6.5 6.5 scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Sum of Positive Signed Ranks T += 45.5 Chapter 19 µT = + σT = + n(n + 1) 10(11) = = 27.5 4 n(n + 1)(2n + 1) 10(11)(21) = = 9.8107 24 24 T + = 45.5 is in the upper tail of the sampling distribution Using the continuity correction factor, we have: 45 − 27.5 P (T + ≥ 45.5) = P z ≥ = P ( z ≥ 1.78) 12.7475 ÷ Using z = 1.78, p-value = (1.0000 - 9925) = 0375 p-value < 05, reject H0 Conclude that there is a significant difference in the median times to fall asleep Without the relaxant has a significantly greater median time and with the relaxant 14 H0: Median percent on-time in 2006 - Median percent on time in 2007 = Ha: Median percent on-time in 2006 - Median percent on time in 2007 ≠ Year Airport 10 11 2006 71.78% 68.23% 77.98% 78.71% 77.59% 77.67% 76.67% 76.29% 69.39% 79.91% 75.55% 2007 69.69% 65.88% 78.40% 75.78% 73.45% 78.68% 76.38% 70.98% 62.84% 76.49% 72.42% Differenc e 2.09% 2.35% -0.42% 2.93% 4.14% -1.01% 0.29% 5.31% 6.55% 3.42% 3.13% Absolute Difference 2.09% 2.35% 0.42% 2.93% 4.14% 1.01% 0.29% 5.31% 6.55% 3.42% 3.13% Signed Ranks Rank 10 11 Negative -2 -3 Sum of Positive Signed Ranks 19 - Positive 10 11 T + = 61 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods µT = + σT = + n(n + 1) 11(12) = = 33 4 n(n + 1)(2n + 1) 11(12)(23) = = 11.2472 24 24 T + = 61 is in the upper tail of the sampling distribution Using the continuity correction factor, we have: 60.5 − 33 P(T + ≥ 61) = P z ≥ = P ( z ≥ 2.45) 11.2472 ÷ Using z = 2.45, p-value = 2(1.0000 - 9929) = 0142 p-value < 05, reject H0 Conclude that there is a significance difference between the median percentage of on-time flights for the two years Median percentage of on-time flights was better in 2006 15 H0: Median time for Service - Median time for Service = Ha: Median time for Service - Median time for Service ≠ Service Delivery 10 11 24.5 26.0 28.0 21.0 18.0 36.0 25.0 21.0 24.0 26.0 31.0 Absolute 28.0 25.5 32.0 20.0 19.5 28.0 29.0 22.0 23.5 29.5 30.0 Differenc e -3.50 0.50 -4.00 1.00 -1.50 8.00 -4.00 -1.00 0.50 -3.50 1.00 Difference 3.50 0.50 4.00 1.00 1.50 8.00 4.00 1.00 0.50 3.50 1.00 Signed Ranks Rank 7.5 1.5 9.5 11 9.5 1.5 7.5 Negative -7.5 1.5 -9.5 -6 11 -9.5 -4 1.5 -7.5 Sum of Positive Signed Ranks µT = + Positive T + = 22 n(n + 1) 11(12) = = 33 4 19 - © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 σT = + n(n + 1)(2n + 1) 11(12)(23) = = 11.2472 24 24 T + = 22 is in the lower tail of the sampling distribution Using the continuity correction factor, we have: 22.5 − 33 P(T + ≤ 22) = P z ≤ = P ( z ≤ −.93) 11.2472 ÷ Using z = -.93, p-value = 1762 p-value > 05, not reject H0 There is no significant difference between the median delivery times for the two services 19 - 10 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 23 H0: The two populations of daily crimes are identical Ha: The two populations of daily crimes are not identical Winter 18 20 15 16 21 20 12 16 19 20 W = Rank 6.5 11 3.5 13 11 3.5 11 71.5 Summer 28 18 24 32 18 29 23 38 28 18 Rank 16.5 6.5 15 19 6.5 18 14 20 16.5 6.5 µW = n1 (n1 + n2 + 1) = 10(10 + 10 + 1) = 105 2 σW = n n (n + n + 1) = 12 2 10(10)(10 + 10 + 1) = 13.2288 12 With W = 71.5 in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows 72 − 105 P(W ≤ 71.5) = P z ≤ ÷ = P( z ≤ −2.49) 13.2288 Using z = -.2.49, the two-tailed p-value =2(.0064) = 0128 p-value < 05; reject H0 Conclusion: The two populations of daily crimes are not identical The population of daily crimes in winter months tends to be less than the population of daily crimes in the summer months 19 - 18 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods 24 H0: The two populations of microwave prices are identical Ha: The two populations of microwave prices are not identical Dallas 445 489 405 485 439 449 436 420 430 405 W = Rank 14.5 23 1.5 22 13 16 12 8.5 1.5 San Antonio 460 451 435 479 475 445 429 434 410 422 425 459 430 Rank 19 17 11 21 20 14.5 10 18 8.5 116 µW = n1 (n1 + n2 + 1) = 10(10 + 13 + 1) = 120 2 σ W = n1n2 (n1 + n2 + 1) = 12 10(13)(10 + 13 + 1) = 16.1245 12 With W = 116 in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows 116.5 − 120 P (W ≤ 116) = P z ≤ = P ( z ≤ −.22) 16.1245 ÷ Using z = -.22, the two-tailed p-value =2(.4129) = 8258 p-value > 05; not reject H0 Conclusion: We cannot reject the null hypothesis that the two populations of microware prices are identical There is no indication that there is a difference between the populations of microwave prices for the two cities 19 - 19 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 25 H0: The two populations of draft positions are identical Ha: The two populations of draft positions are not identical Southeastern Conference Player's College Georgia Alabama Vanderbilt Florida Mississippi Mississippi Auburn Projected Draft Position 14 18 20 24 27 W= Atlantic Coast Conference Player's College Georgia Tech Wake Forrest Virginia Wake Forrest Florida State Maryland Virginia Rank 10 13 47 Projected Draft Position 23 25 26 29 Rank 11 12 14 µW = n1 (n1 + n2 + 1) = 7(7 + + 1) = 52.5 2 σ W = n1n2 (n1 + n2 + 1) = 12 7(7)(7 + + 1) = 7.8262 12 With W = 47 in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows 47.5 − 52.5 P (W ≤ 47) = P z ≤ = P ( z ≤ −.64) 7.8262 ÷ Using z = -.64, the two-tailed p-value =2(.2611) = 5211 p-value > 05; not reject H0 Conclusion: We cannot reject the null hypothesis that the two populations of draft positions are identical There is no indication that there is a difference between the populations of draft position preferences for the players from the two conferences 26 H0: All populations of product ratings are identical Ha: Not all populations of product ratings are identical A 50 62 75 48 Rank 10 B 80 95 98 87 19 - 20 Rank 11 14 15 12 C 60 45 30 58 Rank © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods 65 Sum of Ranks 90 13 34 57 65 21 k 12 34 652 212 Ri2 12 H = − 3( n + 1) = + + ÷ − 3(16) = 10.22 ∑ T 5 nT ( nT + 1) i =1 ni 15(16) Using the χ2 table with df = 2, χ2 = 10.22 shows the p-value is between 005 and 01 Using Excel or Minitab, the p-value for χ2 = 10.22 is 0060 p-value < 01, reject H0 Conclude that the populations of product ratings are not identical 27 H0: All populations of test scores are identical Ha: Not all populations of test score are identical A 540 400 490 530 490 610 Rank 11.5 2.5 10 18 Sum of Ranks 58 B 450 540 400 410 480 370 550 Rank 11.5 2.5 13 C 600 630 580 490 590 620 570 43 k 12 582 432 1092 Ri2 12 H = + + − 3(nT + 1) = ∑ 7 20(21) nT ( nT + 1) i =1 ni Rank 17 20 15 16 19 14 109 ÷ − 3(21) = 9.06 Using the χ2 table with df = 2, χ2 = 9.06 shows the p-value is between 025 and 01 Using Excel or Minitab, the p-value for χ2 = 9.06 is 0108 p-value < 05, reject H0 Conclude that the populations of test scores are not identical In particular, program C appears to provide the higher test scores 28 H0: All populations of calories burned are identical Ha: Not all populations of calories burned are identical Swimming 408 380 425 400 427 Sum of Ranks Rank 11 12 Tennis 415 485 450 420 530 41 Rank 14 13 10 15 61 19 - 21 Cycling 385 250 295 402 268 Rank 18 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 k 12 412 612 182 Ri2 12 H = + + ÷ − 3(16) = 9.26 − 3(nT + 1) = ∑ 5 nT (nT + 1) i =1 ni 15(16) Using the χ2 table with df = 2, χ2 = 9.26 shows the p-value is between 005 and 01 Using Excel or Minitab, the p-value for χ2 = 9.26 is 0098 p-value < 05, reject H0 Conclude that the populations of calories burned by the three activities are not identical Cycling tends to have the lowest calories burned 29 H0: All populations of cruise ship ratings are identical Ha: Not all populations of cruise ship ratings are identical Holland America 84.5 81.4 84.0 78.5 80.9 Sum of Ranks Rank 11 9.5 Princess 85.1 79.0 83.9 81.1 83.7 Rank 13 29.5 Royal Caribbean 84.8 81.8 84.0 85.9 87.4 34 Rank 12 9.5 14 15 56.5 k 12 29.52 34 56.52 Ri2 12 H = + + − 3(nT + 1) = ∑ 5 nT (nT + 1) i =1 ni 15(16) ÷ − 3(16) = 4.19 Using the χ2 table with df = 2, χ2 = 4.19 shows the p-value is greater than 10 Using Excel or Minitab, the p-value for χ2 = 4.19 is 1231 p-value > 05, not reject H0 We cannot reject the null hypothesis that the three populations of cruise ship rating are identical There is no indication that there is a difference among the populations of rating for the three cruise ship lines 30 H0: All populations of training courses are identical Ha: Not all populations of training courses are identical Course Sum of Ranks A 14 10 12 13 52 B 11 26 19 - 22 C 19 16 18 17 79 D 20 15 53 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods k 12 52 26 79 532 Ri2 12 H = + + + ÷ − 3(21) = 8.03 − 3(nT + 1) = ∑ 5 nT ( nT + 1) i =1 ni 20(21) Using the χ2 table with df = 3, χ2 = 8.03 shows the p-value is between 025 and 05 Using Excel or Minitab, the p-value for χ2 = 8.03 is 0454 p-value < 05, reject H0 Conclude that the populations are not identical There is a significant difference in the quality of the courses offered at the four management development centers 31 H0: All populations of calories are identical Ha: Not all populations of calories are identical M&Ms 230 210 240 250 230 Sum of Ranks Rank 10 13 15 11 Kit Kat 225 205 245 235 220 56 Rank 14 12 Milky Way II 200 208 202 190 180 Rank 48 16 k 12 562 482 152 Ri2 12 H = − 3( n + 1) = + + ÷ − 3(16) = 8.96 ∑ T 5 nT (nT + 1) i =1 ni 15(16) Using the χ2 table with df = 2, χ2 = 8.96 shows the p-value is between 01 and 025 Using Excel or Minitab, the p-value for χ2 = 8.96 is 0113 p-value < 05, reject H0 Conclude that the populations of calories are not identical for the three candies Milky Way II appears to have the lowest number of calories 32 a b Σdi2 = 52 rs = − 6Σd i2 6(52) = 1− = 685 10(99) n(n − 1) σr = = n −1 s z= = 3333 rs − 685 = = 2.05 σ rs 3333 p-value = 2(1.0000 - 9798) = 0404 p-value ≤ 05, reject H0 Conclude that significant positive rank correlation exists 33 Case 1: Σdi2 = 19 - 23 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 rs = − 6Σd i2 6(0) = 1− =1 6(36 − 1) n(n − 1) Case 2: Σdi2 = 70 rs = − 6Σd i2 6(70) = 1− = −1 6(36 − 1) n(n − 1) With perfect agreement, rank correlation coefficient rs = With exact opposite ranking, rank correlation coefficient rs = -1 Σdi2 = 250 34 rs = − 6Σd i2 6(250) = 1− = −.136 11(120) n(n − 1) σr = 1 = = 3162 n −1 10 s z= rs − −.136 = = −.43 σ rs 3162 p-value = 2(.3336) = 6672 p-value > 05, not reject H0 We cannot conclude that there is a significant relationship between ranking based on the expenditure per student and the ranking based on the student-teacher ratio 35 a b Σdi2 = 54 r = 1− 6Σd i2 6(54) = 1− = 673 n(n − 1) 10(102 − 1) σr = 1 = = 3333 n −1 10 − s z= rs − µrs σr s = 673 − = 2.02 3333 p-value = 2(1.0000 - 9783) = 0434 c p-value ≤ 05, reject H0 Conclude that there is a significant positive rank correlation between a company’s reputation and having stock that is desirable to purchase 19 - 24 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods 36 Driving Distance Putting di 10 10 -4 -1 -6 -1 -6 -6 rs = − 6Σd i2 6(282) = 1− = −.709 10(100 − 1) n(n − 1) σr = = n −1 s z= rs − µrs σr s = d i2 16 36 49 49 36 36 49 Σdi2 = 282 = 3333 −.709 − = −2.13 3333 p-value = 2(.0166) = 0332 p-value ≤ 10, reject H0 There is a significant negative rank correlation between driving distance and putting Professional golfers who rank high in driving distance tend to rank low in putting 37 Σdi2 = 38 rs = − 6Σdi2 6(38) = 1− = 77 10(99) n(n − 1) σr = = n −1 s z= = 3333 rs − 77 = = 2.31 σ rs 3333 p-value = 2(1.0000 - 9896) = 0208 p-value ≤ 10, reject H0 Conclude that there is a significant positive rank correlation between current students and recent graduates in terms of professor teaching ability 38 Let p = proportion who favor the proposal 19 - 25 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 H0 : p = 50 Ha : p ≠ 50 Eliminating the 60 responses that offered no opinion, we have n = 905 + 1045 = 1950 Using the normal distribution, we have µ = n = 5(1950) = 975 σ = 25n = 25(1950) = 22.0794 With the number of plus signs = 905 in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows 905.5 − 975 P ( x ≤ 905.5) = P z ≤ = P ( z ≤ −3.15) 22.0794 ÷ Area in lower tail area is less than 0010 The two-tailed p-value is less than 2(.0010) = 0020 p-value ≤ 05, reject H0 Conclude that there is a significant difference between the preferences for the tax-funded vouchers or tax deductions for parents who send their children to private schools The greater percentage opposed the proposal 39 St Louis H0: Median > 180,000 Ha: Median < 180,000 There are n = 32 + 18 = 50 observations where the value is different from 180,000 Use the normal approximation with µ = 5n = 5(50) = 25 and σ = 25n = 25(50) = 3.5355 With the number of plus signs = 18 in the lower tail, we use the continuity correction factor and the normal distribution approximation as follows 18.5 − 25 P( x ≤ 18.5) = P z ≤ = P ( z ≤ −1.84) 3.5355 ÷ Lower-tail p-value = 0329 p-value < 05; reject H0 Conclude that the median sale price for single-family homes in St Louis is less than the national median price of $180,000 Denver H0: Median < 180,000 19 - 26 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods Ha: Median > 180,000 There are n = 13 + 27 = 40 observations where the value is different from 180,000 Use the normal approximation with µ = 5n = 5(40) = 20 and σ = 25n = 25(40) = 3.1623 With the number of plus signs = 27 in the upper tail, we use the continuity correction factor and the normal distribution approximation as follows 26.5 − 20 P ( x ≥ 26.5) = P z ≥ = P ( z ≥ 2.06) 3.1623 ÷ Upper-tail p-value = (1.0000 - 9803) = 0197 p-value < 05; reject H0 Conclude that the median sale price for single-family homes in Denver is greater than the national median price of $180,000 40 H0: Median price for Model - Median price for Model = Ha: Median price for Model - Median price for Model ≠ Homemaker 10 11 12 Model 850 960 940 900 790 820 900 890 1100 700 810 920 Model 1100 920 890 1050 1120 1000 1090 1120 1200 890 900 900 Difference -250 40 50 -150 -330 -180 -190 -230 -100 -190 -90 20 Absolute Difference 250 40 50 150 330 180 190 230 100 190 90 20 Rank 11 12 8.5 10 8.5 Signed Ranks Negative Positive -11 -6 -12 -7 -8.5 -10 -5 -8.5 -4 Sum of Positive Signed Ranks µT = + σT = + + T =6 n(n + 1) 12(13) = = 39 4 n(n + 1)(2n + 1) 12(13)(25) = = 12.7475 24 24 T + = is in the lower tail of the sampling distribution Using the continuity correction factor, we have: 6.5 − 39 P(T + ≤ 6) = P z ≤ ÷ = P ( z ≤ −2.55) 12.7475 19 - 27 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 Using z = -2.55, the p-value = 2(.0054) = 0108 p-value < 05, reject H0 Conclude that there is a significance difference between the median prices of the two models The housewives are estimating Model to have the higher median price 41 H0: Median weight After - Median weight Before < Ha: Median weight After - Median weight Before > Chicken 10 11 12 Difference 1.5 1.2 -0.2 0.5 0.7 0.8 0.6 0.2 -0.01 Absolute Difference 1.5 1.2 0.2 0.5 0.7 0.8 0.6 0.2 0.01 Rank 10 2.5 Signed Ranks Negative Positive 10 -2.5 8 2.5 2.5 -1 + T = 51.5 Sum of Positive Signed Ranks Chickens and had no weight change They are removed from the sample making n = 10 µT = + σT = + n(n + 1) 10(11) = = 27.5 4 n(n + 1)(2n + 1) 10(11)(21) = = 9.8107 24 24 T + = 51.5 is in the upper tail of the sampling distribution Using the continuity correction factor, we have: 51 − 27.5 P (T + ≥ 51.5) = P z ≥ = P ( z ≥ 2.40) 9.8107 ÷ Using z = 2.40, p-value = (1.0000 - 9918) = 0082 before p-value < 05, reject H0 Conclude that median weight after is greater than the median weight The feed provides a significant gain in the median weight 19 - 28 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Nonparametric Methods 42 H0: The two populations of product weights are identical Ha: The two populations of product weights are not identical Line 13.6 13.8 14.0 13.9 13.4 13.2 13.3 13.6 12.9 14.4 W = Rank 6.5 11.5 10 6.5 16.5 Line 13.7 14.1 14.2 14.0 14.6 13.5 14.4 14.8 14.5 14.3 15.0 14.9 Rank 13 14 11.5 19 16.5 20 18 15 22 21 70 µW = n1 (n1 + n2 + 1) = 10(10 + 12 + 1) = 115 2 σW = n n (n + n + 1) = 12 2 10(12)(10 + 12 + 1) = 15.1658 12 With W = 70 is in the lower tail, we will use the continuity correction factor and the normal distribution approximation as follows 70.5 − 115 P (W ≤ 70) = P z ≤ = P ( z ≤ −2.93) 15.1658 ÷ Using z = -2.93, the two-tailed p-value = 2(.0017) = 0034 p-value < 05; reject H0 Conclusion: The two populations of product weights are not identical The population of product weights from production line tends to be less that the population of product weights from production line 43 H0: All populations of times are identical Ha: Not all populations of times are identical Method 68 74 65 76 77 72 Sum of Ranks Rank 8.5 15 17 18 12.5 Method 62 73 75 68 72 70 77 Rank 4.5 14 16 8.5 12.5 11 66.5 19 - 29 Method 58 67 69 57 59 62 Rank 10 4.5 27.5 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 k 12 77 66.52 27.52 Ri2 12 H = + + − 3(nT + 1) = ∑ 6 nT (nT + 1) i =1 ni 18(19) ÷ − 3(19) = 7.96 Using the χ2 table with df = 2, χ2 = 7.96 shows the p-value is between 01 and 025 Using Excel or Minitab, the p-value for χ2 = 7.96 is 0187 p-value < 05, reject H0 Conclude that the three populations of times to complete a program evaluation are not identical Method tends to require the lowest times 44 H0: All populations of managerial potential ratings are identical Ha: Not all populations of managerial potential ratings are identical No Program 16 10 15 11 13 Sum of Ranks 74 Company 12 20 17 19 18 14 106 Off-Site 30 k 12 74 106 302 Ri2 12 H = + + − 3(nT + 1) = ∑ 7 20(21) nT (nT + 1) i =1 ni ÷ − 3(21) = 12.61 Using the χ2 table with df = 2, χ2 = 12.61 shows the p-value is less than 005 Using Excel or Minitab, the p-value for χ2 = 12.61 is 0018 p-value < 05, reject H0 Conclude that the three populations of managerial potential ratings are not identical Engineers who attended the Off-Site program tend to have the highest potential ratings 45 H0: All populations of teaching evaluations are identical Ha: Not all populations of teaching evaluations courses are identical Black 88 80 79 68 96 69 Rank 23.5 9.5 2.5 27 Sum of Ranks 74.5 Jennings 87 78 82 85 99 99 85 94 Rank 21.5 13 18.5 28.5 28.5 18.5 26 161.5 19 - 30 Swanson 88 76 68 82 85 82 84 83 81 Rank 23.5 2.5 13 18.5 13 16 15 11 Wilson 80 85 56 71 89 87 118.5 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Rank 9.5 18.5 25 21.5 80.5 Nonparametric Methods k 12 74.52 161.52 118.52 80.52 Ri2 12 H = + + + − 3(nT + 1) = ∑ nT ( nT + 1) i =1 ni 29(30) ÷ − 3(30) = 4.15 Using the χ2 table with df = 3, χ2 = 4.15 shows the p-value is greater than 10 Using Excel or Minitab, the p-value for χ2 = 4.15 is 2457 p-value >.05, not reject H0 We cannot reject the null hypothesis that the four populations of teaching evaluations are identical There is no indication that there is a difference among the instructors based on these ratings 46 Σdi2 = 136 rs = − 6Σdi2 6(136) = 1− = 757 15(224) n(n − 1) σr = 1 = = 2673 n −1 14 s z= rs − µrs σr s = 757 = 2.83 2673 p-value = 2(1.0000 - 9977) = 0046 p-value ≤ 10, reject H0 Conclude that there is a significant positive rank correlation between the two exams Students who rank high on the midterm exam tend to rank high on the final exam 47 Use the Kruskal-Wallis test for this exercise Due to the size of the data set, we recommend using a computer solution Here are some of the calculations The total sample size is 84 H0: All populations of show ratings are identical Ha: Not all populations of show ratings are identical Sample Size Sum of Ranks ABC 19 820.5 CBS 23 764.5 FOX 20 956 NBC 22 1029 k 12 820.52 764.52 956 1029 Ri2 12 H = − 3( n + 1) = + + + ÷ − 3(85) = 4.95 ∑ T 23 20 22 84(85) 19 nT ( nT + 1) i =1 ni Using the χ2 table with df = 3, χ2 = 4.95 shows the p-value is greater than 10 Using Excel or Minitab, the p-value for χ2 = 4.95 is 1755 p-value >.05, not reject H0 We cannot reject the null hypothesis that the four populations of network shows are identical There is no indication that there is a difference among the ratings for the four networks based on the sample ratings The printout from the Minitab Kruskal-Wallis test is as follows 19 - 31 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 Kruskal-Wallis Test on Ratings Network ABC CBS FOX NBC Overall H = 4.95 N 19 23 20 22 84 Median 42.00 31.00 55.50 48.00 DF = Ave Rank 43.2 33.2 47.8 46.8 42.5 Z 0.14 -2.14 1.11 0.96 P = 0.176 Although we not reject H0, the printout shows CBS with a median rating of 31 and an average rank of 33.2 appears to be doing the best of the four networks in terms of Nielsen ratings 19 - 32 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... per gallon 19 - 13 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 19 a H0: The... the two cities 19 - 19 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 19 25 H0: The... 1200 890 900 900 Difference -250 40 50 -150 -330 -180 -190 -230 -100 -190 -90 20 Absolute Difference 250 40 50 150 330 180 190 230 100 190 90 20 Rank 11 12 8.5 10 8.5 Signed Ranks Negative Positive
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