Statistics for Management and Economics 11e Keller SM TestBank

Statistics for Management and Economics 11e Keller SM Statistics for Management and Economics 11e Keller SM331 Appendix 13 A13.1 Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances This Year 3 Years Ago Mean 8.29 10.36 Variance 8.13 8.43 Observations 100 100 Pooled Variance 8.28 Hypothesized Mean Difference 0 df 198 t Stat -5.09 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6526 P(T<=t) two-tail 0.0000 t Critical two-tail 1.9720 t = –5.09, p-value = 0. There is overwhelming evidence to conclude that there has been a decrease over the past three years. A13.2 a z-test of p1  p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 1 2 3 4 5 6 7 A B C D E z-Test of the Difference Between Two Proportions (Case 1) Sample 1 Sample 2 z Stat 2.83 Sample proportion 0.4336 0.2414 P(Z<=z) one-tail 0.0024 Sample size 113 87 z Critical one-tail 1.6449 Alpha 0.05 P(Z<=z) two-tail 0.0047 z Critical two-tail 1.9600 z = 2.83, p-value = .0024. There is enough evidence to infer that customers who see the ad are more likely to make a purchase than those who do not see the ad. b Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0332 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Ad No Ad Mean 97.38 92.01 Variance 621.97 283.26 Observations 49 21 Pooled Variance 522.35 Hypothesized Mean Difference 0 df 68 t Stat 0.90 P(T<=t) one-tail 0.1853 t Critical one-tail 1.6676 P(T<=t) two-tail 0.3705 t Critical two-tail 1.9955 t = .90, p-value = .1853. There is not enough evidence to infer that customers who see the ad and make a purchase spend more than those who do not see the ad and make a purchase. c z-estimate of p 1 2 3 4 5 6 A B C D E z-Estimate of a Proportion Sample proportion 0.4336 Confidence Interval Estimate Sample size 113 0.4336 0.0914 Confidence level 0.95 Lower confidence limit 0.3423 Upper confidence limit 0.5250 ± We estimate that between 34.23% and 52.50% of all customers who see the ad will make a purchase. d t-estimate of µ 1 2 3 4 5 6 7 A B C D t-Estimate: Mean Ad Mean 97.38 Standard Deviation 24.94 LCL 90.22 UCL 104.55 We estimate that the mean amount spent by customers who see the ad and make a purchase lies between $90.22 and $104.55. A13.3 t-test of µD H0: µD = 0 H1: µD > 0333 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means Before After Mean 381.00 373.12 Variance Observations 25 25 Pearson Correlation 0.96 Hypothesized Mean Difference 0 df 24 t Stat 0.70 P(T<=t) one-tail 0.2438 t Critical one-tail 1.7109 P(T<=t) two-tail 0.4876 t Critical two-tail 2.0639 t = .70, p-value = .2438. There is not enough evidence to conclude that the equipment is effective. A13.4 Frequency of accidents: z -test of p1 – p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 1 2 3 4 5 6 7 A B C D E z-Test of the Difference Between Two Proportions (Case 1) Sample 1 Sample 2 z Stat 0.47 Sample proportion 0.0840 0.0760 P(Z<=z) one-tail 0.3205 Sample size 500 500 z Critical one-tail 1.6449 Alpha 0.05 P(Z<=z) two-tail 0.6410 z Critical two-tail 1.9600 z = .47, p-value = .32053. There is not enough evidence to infer that ABS-equipped cars have fewer accidents than cars without ABS. Severity of accidents Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0334 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances No ABS ABS Mean Variance 450,343 390,409 Observations 42 38 Pooled Variance 421,913 Hypothesized Mean Difference 0 df 78 t Stat 2 P(T<=t) one-tail 0.0077 t Critical one-tail 1.6646 P(T<=t) two-tail 0.0153 t Critical two-tail 1.9908 Estimate of the difference between two means (equal-variances) 1 2 3 4 5 6 7 8 A B C D E F t-Estimate of the Difference Between Two Means (Equal-Variances) Sample 1 Sample 2 Confidence Interval Estimate Mean 360.48 ± 290 Variance 450,343 390,409 Lower confidence limit 71 Sample size 42 38 Upper confidence limit 650 Pooled Variance 421,913 Confidence level 0.95 We estimate that the mean repair cost for non-ABS-equipped cars will be between $71 and $650 more than the mean repair cost for ABS-equipped cars. A13.5 Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0335 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Discount No Discount Mean 13.06 18.22 Variance 30.26 38.13 Observations 50 50 Pooled Variance 34.20 Hypothesized Mean Difference 0 df 98 t Stat -4.41 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6606 P(T<=t) two-tail 0.0000 t Critical two-tail 1.9845 t = –4.41, p-value = 0. There is enough evidence to infer that the discount plan works. A13.6 Speeds: Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Speeds Before Speeds After Mean 31.74 31.42 Variance 4.50 4.41 Observations 100 100 Pooled Variance 4.45 Hypothesized Mean Difference 0 df 198 t Stat 1.07 P(T<=t) one-tail 0.1424 t Critical one-tail 1.6526 P(T<=t) two-tail 0.2849 t Critical two-tail 1.9720 t = 1.07, p-value = .1424. There is not enough evidence to infer that speed bumps reduce speeds. Proper stops: Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0336 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Stops Before Stops After Mean 7.82 7.98 Variance 1.83 1.84 Observations 100 100 Pooled Variance 1.83 Hypothesized Mean Difference 0 df 198 t Stat -0.84 P(T<=t) one-tail 0.2021 t Critical one-tail 1.6526 P(T<=t) two-tail 0.4042 t Critical two-tail 1.9720 t = –.84, p-value = .2021. There is not enough evidence to infer that speed bumps increase the number of proper stops. A13.7 t-estimate of µ 1 2 3 4 5 6 7 A B C D t-Estimate: Mean PSI Mean 4.20 Standard Deviation 1.93 LCL 3.93 UCL 4.46 LCL = 3.93, UCL = 4.46. We estimate that on average tires are between 3.93 and 4.46 pounds per square inch below the recommended amount. Tire life: LCL = 100(3.93) = 393, UCL = 100(4.46) = 446. We estimate that the average tire life is decreased by between 393 and 446 miles. Gasoline consumption: LCL = .1(3.93) = .393, UCL = .1(4.46) = .446. We estimate that average gasoline consumption increases by between .393 and .446 gallons per mile. A13.8 t-test of µD H0: µD = 0 H1: µD > 0337 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means Before After Mean 28.94 26.22 Variance 61.45 104.30 Observations 50 50 Pearson Correlation 0.87 Hypothesized Mean Difference 0 df 49 t Stat 3.73 P(T<=t) one-tail 0.000 t Critical one-tail 1.677 P(T<=t) two-tail 0.000 t Critical two-tail 2.010 t = 3.73, p-value = .0002. There is enough evidence to infer that the law discourages bicycle use. A13.9 z -test of p1 – p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Cardizem Placebo Sample Proportions 0.084 0.0797 Observations 607 301 Hypothesized Difference 0 z Stat 0.22 P(Z<=z) one tail 0.4126 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.8252 z Critical two-tail 1.9600 z = .22, p-value = .4126. There is not enough evidence to indicate that Cardizem users are more likely to suffer headache and dizziness side effects than non-users. A13.10 t-test of µ H0: µ = 0 H1: µ > 0338 1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean Pedestrians Mean 209.13 Standard Deviation 60.01 Hypothesized Mean 200 df 39 t Stat 0.96 P(T<=t) one-tail 0.1711 t Critical one-tail 1.6849 P(T<=t) two-tail 0.3422 t Critical two-tail 2.0227 t = .96, p-value = .1711. There is not enough evidence to infer that the franchiser should build on this site. A13.11 Unequal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C t-Test: Two-Sample Assuming Unequal Variances Quit Did not quit Mean 2.04 0.72 Variance 2.05 1.40 Observations 259 1626 Hypothesized Mean Difference 0 df 316 t Stat 14.06 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6497 P(T<=t) two-tail 0.0000 t Critical two-tail 1.9675 t = 14.06, p-value = 0. There is enough evidence to infer that quitting smoking results in weight gains. A13.12 F-test of 12 / 2 2 H : / 1 H : / 1 22 21 1 22 21 0      339 1 2 3 4 5 6 7 8 9 10 A B C F-Test Two-Sample for Variances Brand A Brand B Mean 145.95 144.78 Variance 16.45 4.25 Observations 100 100 df 99 99 F 3.87 P(F<=f) one-tail 0.0000 F Critical one-tail 1.3941 F = 3.87, p-value = 0. There is overwhelming evidence to infer that Brand B is superior to Brand A. A13.13 a z-test of p1 - p2 (case 1) (Success = 2) H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Male most Male less Sample Proportions 0.5962 0.8077 Observations 52 52 Hypothesized Difference 0 z Stat -2.36 P(Z<=z) one tail 0.0092 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0184 z Critical two-tail 1.96 z = –2.36, p-value = .0184. There is enough evidence that men’s choices are affected by the attractiveness of women’s pictures b z-test of p1 - p2 (case 1) (Success = 2) H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Female most Female less Sample Proportions 0.7885 0.8269 Observations 52 52 Hypothesized Difference 0 z Stat -0.50 P(Z<=z) one tail 0.3094 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.6188 z Critical two-tail 1.96340 z = –.50, p-value = .6188. There is not enough evidence to infer that women’s choices are affected by the attractiveness of men’s pictures. A13.14 z-test of p1 - p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Exercisers Watchers Sample Proportions 0.4250 0.3675 Observations 400 400 Hypothesized Difference 0 z Stat 1.66 P(Z<=z) one tail 0.0482 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0964 z Critical two-tail 1.9600 z = 1.66, p-value = .0482. There is evidence to infer that exercisers are more likely to remember the sponsor's brand name than those who only watch. A13.15 a z-test of p H : 0 p = 104,320/425,000 = .245 H1 :p > .245 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Proportion Deliver Sample Proportion 0.2825 Observations 400 Hypothesized Proportion 0.245 z Stat 1.74 P(Z<=z) one-tail 0.0406 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0812 z Critical two-tail 1.9600 z = 1.74, p-value = .0406. There is enough evidence to indicate that the campaign will increase home delivery sales. b z-test of p H : 0 p = 110,000/425,000 = .259 H1 :p > . 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Proportion Deliver Sample Proportion 0.2825 Observations 400 Hypothesized Proportion 0.259 z Stat 1.07 P(Z<=z) one-tail 0.1417 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.2834 z Critical two-tail 1.9600 z = 1.07, p-value = .1417. There is not enough evidence to conclude that the campaign will be successful. A13.16 t-tests of µD a H0: µD = 40 H1: µD > 40 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means SAT after SAT before Mean Variance Observations 40 40 Pearson Correlation 0.94 Hypothesized Mean Difference 40 df 39 t Stat 2.98 P(T<=t) one-tail 0.0024 t Critical one-tail 1.6849 P(T<=t) two-tail 0.0049 t Critical two-tail 2.0227 t = 2.98, p-value = .0024. There is enough evidence to conclude that the ETS claim is false. b. H0: µD = 110 H1: µD < 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means SAT after SAT before Mean Variance Observations 40 40 Pearson Correlation 0.94 Hypothesized Mean Difference 110 df 39 t Stat -3.39 P(T<=t) one-tail 0.0008 t Critical one-tail 1.6849 P(T<=t) two-tail 0.0016 t Critical two-tail 2.0227 t = –3.39, p-value = .0008. There is enough evidence to conclude that the Kaplan claim is also false. A13.17a t-test of µ H0: µ = 1,000 H1: µ < 1,000 1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean Potatoes Mean 985.3 Standard Deviation 49.2 Hypothesized Mean 1000 df 49 t Stat -2.11 P(T<=t) one-tail 0.0198 t Critical one-tail 1.6766 P(T<=t) two-tail 0.0396 t Critical two-tail 2.0096 t = −2.11; p - value = .0198. There is enough evidence to infer that the supplier is cheating him. b t-estimate of µ 1 2 3 4 5 6 7 A B C D t-Estimate: Mean Potatoes Mean 985.3 Standard Deviation 49.2 LCL 971.3 UCL 999.3 Total: LCL = 15,000 (971.3) = 14,569,500, UCL = 15,000(999.3) = 14,989, A13.18 z-Estimate of a proportion LCL = .0615, UCL = .0899 Total: LCL = 221,963,000(.0615) = 13,654,376, UCL = 221,963,000(.0899) = 19,956,213 A13.19 z-test of p1  p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z = -.58, p-value = .2826. A13.20 z-Estimate of a proportion LCL = .1287, UCL = .1760 Total: LCL = 106,744,000(.1287) = 13,737,953, UCL = 106,744,000(.1760) = 18,786,944 A13.21 t-Estimate of µ344 LCL = 47.41, UCL = 48.98 A13.22 z-Estimate of p LCL = .2073, UCL = .2525 Total: LCL = 221,963,000(.2073) = 46.012,347, UCL = 221,963,000(.2525) = 56,047,274 A13.23 t-Estimate of µ LCL = 2.93, UCL = 3.24 A13.24 z-test of p H0: p = .5 H1: p > .5 z = 4.55, p-value = 0.345 A13.25 z-test of p H0: p = .5 H1: p > .5 z = 16.89, p-value = 0. A13.26 z-Estimate of p LCL = .7788, UCL = .8280 A13.27 F = 1.05, p-value = .4140. Use equal-variances t-test H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0 t = -.285, p-value = .3880. A13.28 F = .611, p-value = 0. Use unequal-variances t-test346 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 t = 3.25, p-value = .0006. A13.29 F = 3.77, p-value = 0: Unequal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 t = 15.02, p-value = 0. A13.30 t-Estimate of µ LCL = 57.02, UCL = 58.86347 A13.31 z-Estimate of p LCL = .3469, UCL = .3702 Total: LCL = 220,958,853(.3469) = 76,641,046, UCL = 220,958,853(.3702) = 81,800,292 A13.32 t-Estimate of µ LCL = 2809, UCL = 3555 A13.33 t-test of µ H0: µ = 3624 H1: µ > 3624 t = 23.73, p-value = 0. A13.34 z-Estimate of p348 LCL = .0913, UCL = .1058 Total: LCL = 220,958,853(.0913) = 20,178,792, UCL = 220, 958,853(.1058) = 23,385,758 A13.35 Group 1: EDCL = 1, 2, and 3; Group 2: EDCL = 4) F = 2.50, p-value = 0. Use unequal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0 t = -8.81, p-value = 0. A13.36 z-Estimate of p LCL = .0274, UCL = .0359 Total: LCL = 220,958,853(.0274) = 6,046,709, UCL = 220, 958,853(.0359) = 7,929,390 A13.37 z-test of p1  p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 z = 9.54, p-value = 0.349 A13.38 z-test of p1  p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 z = 14.68, p-value = 0.387 Appendix 14 A14.1 One-way Analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 91.43 2 45.72 1.09 0.3441 3.16 Within Groups 2397.5 57 42.06 Total 2488.9 59 F = 1.09, p-value = .3441. There is no evidence to infer that sales of candy differ according to placement. A14.2 t-test of D H : 0 0 D  H : 0 1 D  D D D D s / n x t    1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means Price shown Price not shown Mean 56.15 60.31 Variance 243.68 467.71 Observations 100 100 Pearson Correlation 0.79 Hypothesized Mean Difference 0 df 99 t Stat -3.12 P(T<=t) one-tail 0.0012 t Critical one-tail 1.6604 P(T<=t) two-tail 0.0024 t Critical two-tail 1.9842 t = –3.12, p-value = .0012. There is overwhelming evidence to conclude that ads with no price shown are more effective in generating interest than ads that show the price. A14.3 Unequal-variances t-test of 1 2 H0 : (1  2 )  0 H1 : (1 2 )  0388 22 2 21 1 1 2 1 2 s n s n (x x ) ( ) t        1 2 3 4 5 6 7 8 9 10 11 12 13 A B C t-Test: Two-Sample Assuming Unequal Variances Leftover Returned Mean 61.71 70.57 Variance 48.99 203.98 Observations 14 53 Hypothesized Mean Difference 0 df 44 t Stat -3.27 P(T<=t) one-tail 0.0011 t Critical one-tail 1.6802 P(T<=t) two-tail 0.0021 t Critical two-tail 2.0154 t = –3.27, p-value = .0011. There is enough evidence to support the professor's theory. A14.4 a z-test of p1  p2 (case 1) H : 0 p1  p2 = 0 H1 : p1  p2 > 0       1 2 1 2 1 n 1 n pˆ(1 pˆ) (pˆ pˆ ) z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Topiramate Placebo Sample Proportions 0.2364 0.1042 Observations 55 48 Hypothesized Difference 0 z Stat 1.76 P(Z<=z) one tail 0.0390 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0780 z Critical two-tail 1.96 z = 1.76, p-value = .0390. There is enough evidence to conclude that topiramate is effective in causing abstinence for the first month. b z-test of p1  p2 (case 1) H : 0 p1  p2 = 0389 H1 : p1  p2 > 0       1 2 1 2 1 n 1 n pˆ(1 pˆ) (pˆ pˆ ) z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Topiramate Placebo Sample Proportions 0.5091 0.1667 Observations 55 48 Hypothesized Difference 0 z Stat 3.64 P(Z<=z) one tail 0.0001 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0002 z Critical two-tail 1.96 z = 3.64, p-value = .0001. There is enough evidence to conclude that topiramate is effective in causing alcoholics to refrain from binge drinking in the final month. A14.5a Equal-variances t-test of 1  2 H0 : (1  2 )  0 H1 : (1  2 )  0          1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Expenses MSA Expenses Regular Mean 347.24 479.25 Variance Observations 63 141 Pooled Variance 21102 Hypothesized Mean Difference 0 df 202 t Stat -6.00 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6524 P(T<=t) two-tail 0.0000 t Critical two-tail 1.9718 t = –6.00, p-value = 0. There is enough evidence to infer that medical expenses for those under the MSA plan are lower than those who are not.390 b z-test of p1  p2 (case 1) H : 0 p1  p2 = 0 H1 : p1  p2 < 0       1 2 1 2 1 n 1 n pˆ(1 pˆ) (pˆ pˆ ) z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Health MSA Health Regular Sample Proportions 0.7619 0.7801 Observations 63 141 Hypothesized Difference 0 z Stat -0.29 P(Z<=z) one tail 0.3867 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.7734 z Critical two-tail 1.9600 z = –.29, p-value = .3867. There is not enough evidence to support the critics of MSA. A14.6 a One-way analysis of variance 11 12 13 14 15 16 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups .2 85.98 0.0000 2.63 Within Groups .36 Total F = 85.98, p-value = 0. Two-factor analysis of variance 23 24 25 26 27 28 29 30 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample .85 0.0000 3.87 Columns .12 0.0000 3.87 Interaction 679 1 678.6 6.97 0.0087 3.87 Within .36 Total Interaction: F = 6.97; p-value = .0087. There is enough evidence to infer that differences are caused by interaction. There is no need to conduct the other two tests.391 A14.7 t-tests of  45 minutes: H0 :  = 45 H1 :  < 45 s / n x t   1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean 45 minutes Mean 41.75 Standard Deviation 3.63 Hypothesized Mean 45 df 19 t Stat -4.01 P(T<=t) one-tail 0.0004 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0008 t Critical two-tail 2.0930 60 minutes: H0 :  = 60 H1 :  < 60 s / n x t   1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean 60 minutes Mean 58.75 Standard Deviation 5.02 Hypothesized Mean 60 df 19 t Stat -1.11 P(T<=t) one-tail 0.1399 t Critical one-tail 1.7291 P(T<=t) two-tail 0.2798 t Critical two-tail 2.0930 80 minutes: H0 :  = 80 H1 :  < 80 s / n x t  392 1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean 80 minutes Mean 69.05 Standard Deviation 6.31 Hypothesized Mean 80 df 19 t Stat -7.76 P(T<=t) one-tail 0.0000 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0000 t Critical two-tail 2.0930 100 minutes: H0 :  = 100 H1 :  < 100 s / n x t   1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean 100 minutes Mean 90.40 Standard Deviation 12.35 Hypothesized Mean 100 df 19 t Stat -3.48 P(T<=t) one-tail 0.0013 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0026 t Critical two-tail 2.0930 125 minutes: H0 :  = 125 H1 :  < 125 s / n x t  393 1 2 3 4 5 6 7 8 9 10 11 12 A B C D t-Test: Mean 125 minutes Mean 110.05 Standard Deviation 17.11 Hypothesized Mean 125 df 19 t Stat -3.91 P(T<=t) one-tail 0.0005 t Critical one-tail 1.7291 P(T<=t) two-tail 0.0010 t Critical two-tail 2.0930 Overall Conclusion: p-values are .0004, .1399, 0, .0013, and .0005, respectively. In four of the jobs there is overwhelming evidence to conclude that the times specified by the schedule are greater than the actual times. A14.8 Two-way analysis of variance H0 : 1  2  3 H1 : At least two means differ. 35 36 37 38 39 40 41 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Rows 31,154,590 24 1,298,108 15.05 5.20E-15 1.75 Columns 913,217 2 456,608 5.29 0.0084 3.19 Error 4,141,276 48 86,277 Total 36,209,083 74 F = 5.29; p-value = .0084. There is sufficient evidence to infer that differences exist between the estimated repair costs from different appraisers. A14.9 One-way analysis of variance H : 0 1  2  3 H1 : At least two means differ 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 126.0 2 63.02 3.30 0.0439 3.16 Within Groups 1087.3 57 19.08 Total 1213.3 59 F = 3.30, p-value = .0439. There is evidence to infer that at least one rust-proofing method is different from the others.394 A14.10 z-test of p1  p2 (case 2) H0 : p1  p2  .15 H1 : p1  p2  .15 2 2 2 1 1 1 1 2 1 2 n pˆ (1 pˆ ) n pˆ (1 pˆ ) (pˆ pˆ ) (p p ) z        1 2 3 4 5 6 7 8 9 10 11 A B C z-Test: Two Proportions Comm 1 Comm 2 Sample Proportions 0.268 0.486 Observations 500 500 Hypothesized Difference -0.15 z Stat -2.28 P(Z<=z) one tail 0.0114 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0228 z Critical two-tail 1.9600 z = –2.28, p-value = .0114. There is evidence to indicate that the second commercial is viable. A14.11 t-test of D H : 0 0 D  H1 :D < 0 D D D D s / n x t    1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means Prior After Mean 24.91 26.24 Variance 48.65 87.88 Observations 100 100 Pearson Correlation 0.79 Hypothesized Mean Difference 0 df 99 t Stat -2.29 P(T<=t) one-tail 0.0121 t Critical one-tail 1.6604 P(T<=t) two-tail 0.0242 t Critical two-tail 1.9842 t = 2.29, p-value = .0121. There is enough evidence to conclude that company should proceed to stage 2.395 A14.12 Two-factor analysis of variance 29 30 31 32 33 34 35 36 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample 427.61 2 213.81 39.97 0.0000 3.06 Columns 20.17 1 20.17 3.77 0.0541 3.91 Interaction 17.77 2 8.89 1.66 0.1935 3.06 Within 770.32 144 5.35 Total 1235.87 149 Interaction: F = 1.66, p-value = .1935. There is no evidence of interaction. Gender (Columns) : F = 3.77, p-value = .0541. There is not enough evidence of a difference between men and women. Fitness (Sample): F = 39.97, p-value = 0. There is overwhelming evidence of differences among the three levels of fitness. A14.13 Equal-variances t-test of 1 2 H0 : (1  2 )  0 H1 : (1 2 )  0         1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances ABS speed No ABS speed Mean 34.72 33.94 Variance 25.27 25.63 Observations 100 100 Pooled Variance 25.45 Hypothesized Mean Difference 0 df 198 t Stat 1.09 P(T<=t) one-tail 0.1394 t Critical one-tail 1.6526 P(T<=t) two-tail 0.2788 t Critical two-tail 1.9720 t = 1.09, p-value = .2788. There is not enough evidence that operating an ABS-equipped car changes a driver's behavior. A14.14 One-way analysis of variance396 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1813.7 2 906.87 6.46 0.0030 3.16 Within Groups 7998.0 57 140.32 Total 9811.7 59 Multiple comparisons 1 2 3 4 5 6 7 A B C D E Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05 Price: $34 Price: $39 -12.9 9.24 9.01 Price: $44 -3.1 9.24 9.01 Price: $39 Price: $44 9.8 9.24 9.01 Sales with $34 and $44 dollar prices do not differ. Sales with $39 differ from sales with $34 and $44 prices. A14.15 Equal-variances t-test of 1 2 H0 : (1  2 )  0 H1 : (1 2 )  0 22 2 21 1 1 2 1 2 s n s n (x x ) ( ) t        1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Echinacea Placebo Mean 7.02 7.06 Variance 2.51 2.42 Observations 262 262 Pooled Variance 2.47 Hypothesized Mean Difference 0 df 522 t Stat -0.31 P(T<=t) one-tail 0.3799 t Critical one-tail 1.6478 P(T<=t) two-tail 0.7598 t Critical two-tail 1.9645 t = –.31, p-value = .3799. There is not enough evidence to conclude that Echinacea is effective. A14.16 Two-way analysis of variance397 24 25 26 27 28 29 30 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Rows 335.17 13 25.78 16.86 0.0000 2.12 Columns 10.90 2 5.45 3.57 0.0428 3.37 Error 39.76 26 1.53 Total 385.83 41 a H0 : 1  2  3 H1 : At least two means differ. F = 3.57, p-value = .0428. There is enough evidence to conclude that there are differences in waiting times between the three resorts. b The waiting times are required to be normally distributed with the same variance at all three resorts. c Histograms are used to check the normality requirement. A14.17a There are 4 levels of ranks and 4 levels of faculties for a total of 16 treatments. b 23 24 25 26 27 28 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1091.15 15 72.74 2.84 0.0019 1.83 Within Groups 1638.40 64 25.60 Total 2729.55 79 F = 2.84, p-value = .0019. There is enough evidence to infer that at least two treatment means differ. c Factor A (columns) is the faculty. The levels are business, engineering, arts, and science. Factor B (samples) is the rank. The levels are professor, associate professor, assistant professor, and lecturer. 35 36 37 38 39 40 41 42 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample 46.85 3 15.62 0.61 0.6109 2.75 Columns 344.65 3 114.88 4.49 0.0064 2.75 Interaction 699.65 9 77.74 3.04 0.0044 2.03 Within 1638.40 64 25.60 Total 2729.55 79 d F = 3.04, p-value = .0044. There is evidence to conclude that ranks and faculties interact. e F =.61, p-value = .6109. There is not enough evidence to conclude that differences exist between the ranks. The answer to Part d indicates that this test is irrelevant. f F = 4.49, p-value = .0064. There is enough evidence to conclude that differences exist between the faculties. The answer to Part d indicates that this test is irrelevant.398 A14.18 a z-estimate of p 1 2 3 4 5 6 A B C D E z-Estimate of a Proportion Sample proportion 0.3569 Confidence Interval Estimate Sample size 1328 0.357 ± 0.026 Confidence level 0.95 Lower confidence limit 0.331 Upper confidence limit 0.383 Estimate of the number of households with at least one dog LCL = 112 million × .331 = 37.072 million UCL = 112 million × .383 = 42.896 million b 1 2 3 4 5 6 A B C D E z-Estimate of a Proportion Sample proportion 0.316 Confidence Interval Estimate Sample size 1328 0.316 ± 0.025 Confidence level 0.95 Lower confidence limit 0.291 Upper confidence limit 0.341 Number of households with at least one cat LCL = 112 million × .291 = 32.592 million UCL = 112 million × .341 = 38.192million c t-estimate of µ n s x t   / 2 1 2 3 4 5 6 7 A B C D t-Estimate: Mean Dogs Mean 247.19 Standard Deviation 133.16 LCL 235.17 UCL 259.20 Estimate of the total amount spent on dogs LCL = 40 million ×235.17 = $9.407 billion UCL = 40 million ×259.29 = $10.368 billion, d t-estimate of µ n s x t   / 2399 1 2 3 4 5 6 7 A B C D t-Estimate: Mean Cats Mean 158.07 Standard Deviation 88.94 LCL 149.53 UCL 166.61 Estimate of the total amount spent on cats LCL = 35 million ×149.53 = $5.234 billion UCL = 35 million ×166.61 = $5.831 billion A14.19 One-way Analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ F = 7.85, p-value = 0. A14.20 z-test of p (proportion of PARSOL 1 or 2) H0: p = .5 H1: p > .5 z = 7.02, p-value = 0. A14.21 t-estimate of µ400 LCL = 1.82, UCL = 1.97 A14.22 One-way Analysis of variance H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6 = µ7 = µ8 H1: At least two means differ F = 2.25, p-value = .0282. A14.23 F = .889, p-value = .2990. Use Equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) ≠ 0 t = .899, p-value = .3691. A14.24 z estimate of p LCL = .0861, UCL = . Total: LCL = 221,963,000(.0861) = 19,110,672, UCL = 221,963,000(.1125) = 24,967,087 A14.25 z estimate of p LCL = .1816, UCL = .2180 Total: LCL = 221,963,000(.1816) = 40,308,268, UCL = 221,963,000(.2180) = 48,381,310 A14.26 One-way Analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ F = 21.05, p-value = 0. A14.27 z estimate of p LCL = .0146, UCL = .0311 Total: LCL = 221,963,000(.0146) = 3,250,089, UCL = 221,963,000(.0311) = 6,902,793 A14.28 t test of p (proportion of HAPPY 1 and 2) H0: p = .75 H1: p > .75402 z = 31.82, p-value = 0. A14.29 F = 2.09, p-value = 0. Use Unequal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) >0 t = 7.50, p-value = 0. A14.30 z test of p1 – p2 (Case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z = -.83, p-value = .2031. A14.31 One-way Analysis of variance H0: µ1 = µ2 = µ3 = µ4 = µ5 H1: At least two means differ403 F = 11.22, p-value = 0. A14.32 One-way Analysis of variance H0: µ1 = µ2 = µ3 = µ4 H1: At least two means differ F = 47.46, p-value = 0. A14.33 t estimate of µ LCL = 372,397, UCL = 450,277 A14.34 t test of µ H0: µ = 2505 H1: µ < 2505 t = -8.14, p-value = 0. A14.35 F = .75, p-value = .0054. Use unequal-variances t-estimate of µ1 - µ2404 A14.36 z estimate of p LCL = .4356, UCL = .4598 Total: LCL = 220,958,853(.4356) = 96,248,803, UCL = 220,958,853(.4598) = 101,598,340 A14.37 z estimate of p LCL = .2129, UCL = .2332 Total: LCL = 220,958,853(.2129) = 47,051,990, UCL = 220,958,853(.2332) = 51,530,701 A14.38 t estimate of µ LCL = 1,455,724, UCL = 1,660,339 A14.39 z estimate of p405 LCL = .6646, UCL = .6874 Total: LCL = 220,958,853(.6646) = 146,856,572, UCL = 220,958,853(.6874) = 151,891,251 A14.40 F = 1.10, p-value = .3298. Use equal-variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 t = 1.24, p-value = .1072. A14.41 z test of p1 – p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0 z = 16.29, p-value = 0. A14.42 z test of p1 – p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) > 0406 z = 9.45, p-value = 0. Appendix 15 A15.1 Chi-squared goodness-of-fit test H : 0 p1  .50 , p2  .20 , p3  .15 p4  .10 p5  .05 H1 : At least one pi is not equal to its specified value. 1 2 3 4 5 6 7 C D E Actual Expected 183 175 63 70 55 52.5 29 35 20 17.5 p-value= 0.6321 p-value = .6321. There is not enough evidence to conclude that applicants to WLU's MBA program are different in terms of their undergraduate degrees from the population of MBA applicants? A15.2 t-test of D H0 : D = 0 H1 :D < 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means First Sat Second SAT Mean Variance Observations 40 40 Pearson Correlation 0.91 Hypothesized Mean Difference 0 df 39 t Stat -1.20 P(T<=t) one-tail 0.1182 t Critical one-tail 1.6849 P(T<=t) two-tail 0.2365 t Critical two-tail 2.0227 t = –1.20, p-value = .1182. There is not enough evidence to indicate that repeating the SAT produces higher exam scores. A15.3 Time to solve the 48 problems: Equal-variances t-test of 1  2 H0 : (1  2 )  0 H1 : (1  2 )  0436 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Diet Not Mean 581.95 551.5 Variance 2716.6 2221.5 Observations 20 20 Pooled Variance 2469.1 Hypothesized Mean Difference 0 df 38 t Stat 1.94 P(T<=t) one-tail 0.0300 t Critical one-tail 1.6860 P(T<=t) two-tail 0.0601 t Critical two-tail 2.0244 t = 1.94, p-value = .0300. There is enough evidence to conclude that dieters take longer to solve the 48 problems than do nondieters. Successfully repeat string of five letters: z-test of p1  p2 (case 1) H : 0 p1  p2 = 0 H1 : p1  p2 < 0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Diet Not Sample Proportions 0.50 0.80 Observations 20 20 Hypothesized Difference 0 z Stat -1.99 P(Z<=z) one tail 0.0234 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0468 z Critical two-tail 1.96 z = –1.99, p-value = .0234. There is enough evidence to conclude that dieters are less successful at repeating string of five letters. Successfully repeat string of five words: z-test of p1  p2 (case 1) H : 0 p1  p2 = 0 H1 : p1  p2 > 0437 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Diet Not Sample Proportions 0.35 0.60 Observations 20 20 Hypothesized Difference 0 z Stat -1.58 P(Z<=z) one tail 0.0567 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.1134 z Critical two-tail 1.96 z = –1.58, p-value = .0567. There is not enough evidence to conclude that dieters are less successful at repeating string of five words. A15.4 a t-estimate of  1 2 3 4 5 6 7 A B C D t-Estimate: Mean Overdue Mean 7.09 Standard Deviation 6.97 LCL 6.40 UCL 7.77 LCL = 6.40, UCL = 7.77 b LCL = 50,000($.25)(6.40) = $80,000 UCL = 50,000($.25)(7.77) = $97,125 It does appear that not all fines are collected A15.5 One-way analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups .23 0.0468 3.16 Within Groups 57 8894 Total 59 F = 3.23; p-value = .0468. There is enough evidence to conclude that differences in sales exist between the three advertising strategies.438 A15.6 Chi-squared test of a contingency table H : 0 The two variables (income category and mutual fund ownership) are independent H1 : The two variables are dependent 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 A B C D E Contingency Table Income category Mutual fund 1 2 TOTAL 1 71 13 84 2 59 28 87 TOTAL chi-squared Stat 196.77 df 5 p-value 0 chi-squared Critical 11.0705 2  = 196.77;p-value = 0. There is overwhelming evidence to infer that household income and ownership of mutual funds are related A15.7 Two-factor analysis of variance 23 24 25 26 27 28 29 30 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample .42 0.2387 4.02 Columns .33 0.0077 3.17 Interaction .09 0.9171 3.17 Within 54 9280 Total 59 Interaction: F = .09; p-value = .9171. There is no evidence of interaction. Advertising strategy (Columns): F = 5.33; p-value = .0077. There is sufficient evidence to conclude that advertising strategies differ with respect to sales. Media (Sample): F = 1.42; p-value = .2387. There is not enough evidence to conclude that differences in the medium for advertising differ in terms of sales. A15.8 z-test of p1  p2 (case 1) Code 3 results were omitted.439 H0 : (p1  p2 )  0 H1 : (p1  p2 )  0 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Folic acid Placebo Sample Proportions 0.0101 0.0343 Observations 597 612 Hypothesized Difference 0 z Stat -2.85 P(Z<=z) one tail 0.0022 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.0044 z Critical two-tail 1.9600 z = –2.85, p-value = .0022. There is overwhelming evidence to conclude that folic acid reduces the incidence of spina bifida. A15.9 Unequal-variances t-test of 1  2 H0 : (1  2 ) = 0 H1 : (1  2 ) < 0 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C t-Test: Two-Sample Assuming Unequal Variances British American Mean 238.0 252.0 Variance 149.9 220.2 Observations 263 279 Hypothesized Mean Difference 0 df 531 t Stat -12.00 P(T<=t) one-tail 7.64E-30 t Critical one-tail 1.6477 P(T<=t) two-tail 1.53E-29 t Critical two-tail 1.9644 z = –12.00, p-value = 0. There is enough evidence to infer that British golfers play golf in less time than do American golfers. A15.10 one-way analysis of variance H : 0 1  2  3 H1 : At least two means differ440 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 2 58.37 0.0000 3.03 Within Groups 284 5363 Total 286 F = 58.37, p-value = 0. There is enough evidence to conclude that there are differences between the three groups. Multiple Comparisons 1 2 3 4 5 6 7 A B C D E Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05 Before 1976 After 1986 122.62 28.03 25.46 Canadian 78.04 24.67 25.46 After 1986 Canadian -44.58 25.75 25.46 All three groups differ from each other. A15.11 Chi-squared test of a contingency table H : 0 The two variables (year and sport) are independent H1 : The two variables are dependent 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A B C D E Contingency Table Sport Year 1 2 TOTAL 3 29 58 87 4 52 39 91 5 48 34 82 6 16 33 49 7 26 29 55 8 24 21 45 TOTAL chi-squared Stat 23.8101 df 8 p-value 0.0025 chi-squared Critical 15. 2  = 23.8101, p-value = .0025. There is enough evidence to infer that North Americans changed their favorite sport between 1985 and 1992. A15.12 t-estimate of  1 2 3 4 5 6 7 A B C D t-Estimate: Mean Cars Mean 165.79 Standard Deviation 51.59 LCL 157.17 UCL 174.41 Five minute interval: LCL = 157.17, UCL = 174.41 Twenty-four hour day (12 5-minute intervals, 24 hours per day): LCL = 12  24 157.17 = 45,265 UCL = 12  24 174.41 = 50,230 A15.13 z-estimate of p 1 2 3 4 5 6 A B z-Estimate: Proportion Exercise? Sample Proportion 0.551 Observations 671 LCL 0.514 UCL 0.589 Total number of adults who exercise: LCL = 205.9 million (.514) = 105.8 million UCL = 205.9 million (.589) = 121.3 million A15.14 Equal-variances t-test of 1  2 H0 : (1  2 )  0 H1 : (1 2 ) < 0442 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Activity Usual Mean 57.06 87.28 Variance 296.18 215.42 Observations 67 67 Pooled Variance 255.80 Hypothesized Mean Difference 0.00 df 132 t Stat -10.94 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6565 P(T<=t) two-tail 0.0000 t Critical two-tail 1.9781 t = –10.94, p-value = 0. There is enough evidence to indicate to infer that graded activity is effective. A15.15 Chi-squared test of a contingency table H : 0 The two variables (group and improvement) are independent H1 : The two variables are dependent 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C D E Contingency Table Group Improvement 1 2 TOTAL 1 42 8 50 2 32 18 50 3 13 37 50 TOTAL 87 63 150 chi-squared Stat 35.63 df 2 p-value 0 chi-squared Critical 5.9915 2  = 35.63, p-value = 0. There is sufficient evidence to infer there are differences between the three groups. A15.16 z-estimate of p443 1 2 3 4 5 6 A B C D E z-Estimate of a Proportion Sample proportion 0.774 Confidence Interval Estimate Sample size 780 0.774 ± 0.0294 Confidence level 0.95 Lower confidence limit 0.7446 Upper confidence limit 0.8034 Total number of on-time departures: LCL = 7,140,596(.7446) = 5,316,888 UCL = 7,140,596(.8034) = 5,736,755 A15.17 One-way analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ F = 119.3, p-value = 0. A15.18 z test of p1 – p2 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z = -1.56, p-value = .0591. A15.19 z test of p H0: p = .5 H1: p > .5444 z = 3.56, p-value = .0002. A15.20 t estimate of µ LCL = 23.79, UCL = 24.37 A15.21 z estimate of p LCL = .1924, UCL = .2892 Total: LCL = 115,219,000(.1924) = 22,171,028, UCL = 115,219,000(.2892) = 33,320,475 A15.22 F = 1.04, p-value = .6422. Use equal variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0445 t = 7.29, p-value = 0. A15.23 z-test of p H0: p = .5 H1: P > .5 z = 9.38, p-value = 0. A15.24 One-way analysis of variance H0: µ1 = µ2 = µ3 = µ4 = µ5 H1: At least two means differ F = 9.38, p-value = 0. A15.25 z estimate of p LCL = .0398, UCL = .0724 Total: LCL = 221,963,000(.0398) = 8,841,895, UCL = 221,963,000(.0724) = 16,078,233 A15.26 Chi-squared test of a contingency table H0: The two variables (race and work for oneself) are independent H1: The two variables are dependent446 2χ = 5.94, p-value = .0514. A15.27 F = .95, p-value = .5794. Use equal variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) > 0 t = 8.93, p-value = 0. A15.28 One-way analysis of variance H0: µ1 = µ2 = µ3 H1: At least two means differ F = 21.38, p-value = 0. A15.29 One-way analysis of variance H0: µ1 = µ2 = µ3 = µ4447 H1: At least two means differ F = 16.55, p-value = 0. A15.30 z estimate of p LCL = .1031, UCL = .1472 Total: LCL = 106,744,000(.1031) = 11,005,306, UCL = 106,744,000(.1472) = 15,712,717 A15.31 Chi-squared test of a contingency table H0: The two variables (party and support for capital punishment) are independent H1: The two variables are dependent 2χ = 116.5, p-value = 0. A15.32 Chi-squared test of a contingency table H0: The two variables (sex and party) are independent H1: The two variables are dependent448 2χ = 10.04, p-value = .0066. A15.33 F = .51, p-value = 0. Use unequal variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0 t = -2.58, p-value = .0052. A15.34 t estimate of µ LCL = 69,627, UCL = 104,904 A15.35 F = 2.37, p-value = 0. Use unequal variances t-test of µ1 - µ2 H0: (µ1 - µ2) = 0 H1: (µ1 - µ2) < 0449 t = 8.09, p-value = 0. A15.36 t estimate of µ LCL = 398,801, UCL = 481.080. A15.37 One-way analysis of variance H0: µ1 = µ2 = µ3 = µ4 H1: At least two means differ A15.38 z estimate of p A15.39 z estimate of p450 LCL = .1534, UCL = .1714 Total: LCL = 220,958,853(.1534) = 33,905,345, UCL = 220,958,853(.1714) = 37,873,052 A15.40 z estimate of p LCL = .2561, UCL = .2776 Total: LCL = 220,958,853(.2561) = 56,584,216, UCL = 220,958,853(.2776) = 61,342,304 A15.41 z test of p1 – p2 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z = -1.70, p-value = .0447. A15.42 z test of p1 – p2 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0451 z = -4.83, p-value = 0. A15.43 Chi-square test of a contingency table 2χ = 94.53, p-value = 0. Case A15.1 One-way analysis of variance H : 0 1  2  3  4 H1 : At least two means differ Weight loss: 11 12 13 14 15 16 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 189.4 3 63.14 1.78 0.1532 2.66 Within Groups 5533.6 156 35.47 Total 5723.0 159 F = 1.78, p-value = .1532. There is not enough evidence to conclude that there are differences in weight loss between the four diets.452 Percent LDL decrease: 11 12 13 14 15 16 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 2569.4 3 856.48 32.56 0.0000 2.66 Within Groups 4104.0 156 26.31 Total 6673.4 159 F = 32.56, p-value = 0. There is enough evidence to conclude that there are differences in bad cholesterol reduction between the four diets. Percent HDL Increase: 11 12 13 14 15 16 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 5595.0 3 1864.99 92.62 0.0000 2.66 Within Groups 3141.3 156 20.14 Total 8736.2 159 F = 96.62, p-value = 0. There is enough evidence to conclude that there are differences in good cholesterol increase between the four diets.491 Appendix 16 A16.1 t-test of ρ H : 0 0   H1 :   0 1 2 3 4 5 6 7 8 9 10 A B Correlation Weight and B/A Level Pearson Coefficient of Correlation 0.4177 t Stat 3.19 df 48 P(T<=t) one tail 0.0013 t Critical one tail 1.6772 P(T<=t) two tail 0.0026 t Critical two tail 2.0106 r = .4177, t = 3.19, p–value = .0026. There is sufficient evidence to infer that weight and blood–alcohol level are related. A16.2 Two-way analysis of variance H : 0 1  2  3  4 H1 : At least two means differ 40 41 42 43 44 45 46 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Rows 3708.8 28 132.46 8.63 6.52E-15 1.61 Columns 997.0 3 332.33 21.64 1.77E-10 2.71 Error 1289.8 84 15.35 Total 5995.5 115 F = 21.64; p-value = 0. There is enough evidence to conclude that there are differences in the decrease in test scores between the four types of breakfast meals. A16.3a Chi-squared goodness-of-fit test (the percentages must be converted to actual and expected values and we must include those who did not have cancer) p 57 / 420,000, p 12 / 420,000, p 13 / 420,000, p 419.634 / 420,000 H : p 143 / 420,000, p 9 / 420,000, p 80 / 420,000, p 52 / 420,000, 5 6 7 8 0 1 2 3 4         H1 :At least one pi is not equal to its specified value.492     8 i 1 i 2 2 i i e (f e ) 1 2 3 4 5 6 7 8 9 A B C D Actual Expected 135 143 7 9 77 80 32 52 42 57 8 12 13 13 p-value = 0.0515 p-value = .0515. There is not enough evidence to conclude that there is a relationship between cell phone use and cancer. b The data are observational. Even if we regard the statistical result as significant we cannot automatically infer that cell phone use causes cancer. Additionally, an examination of the actual and expected values reveals that in all 7 types of cancers the actual values are less than or equal to the expected values, indicating that (if anything) cell phone use prevents cancer. A16.4 t-test of ρ or t-test of β1 H : 0 0   H1 :   0 2 1 r n 2 t r    1 2 3 4 5 6 7 8 9 10 A B C D Correlation Age and Duration Pearson Coefficient of Correlation 0.558 t Stat 7.90 df 138 P(T<=t) one tail 0 t Critical one tail 1.6560 P(T<=t) two tail 0 t Critical two tail 1.9773 t = 7.90; p-value = 0. There is overwhelming evidence to infer that the older the patient the longer it takes for the symptoms to disappear?493 A16.5a Equal-variances t-test of 1  2 H0 : (1  2 )  0 H1 : (1 2 ) > 0         1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Home Outside Mean 59.21 54.91 Variance 102.03 88.28 Observations 196 152 Pooled Variance 96.03 Hypothesized Mean Difference 0 df 346 t Stat 4.06 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6493 P(T<=t) two-tail 0.0001 t Critical two-tail 1.9668 t = 4.06, p-value = 0. There is enough evidence to infer that men whose wives stay at home earn more than men whose wives work outside the home. b It may be that men whose wives stay at home work harder, and thus earn more. A16.6 Question 1: Equal-variances t-test of 1 2 H0 : (1  2 )  0 H1 : (1 2 ) < 0         1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t494 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances US Days Canada Days Mean 26.98 29.44 Variance 55.90 56.82 Observations 300 300 Pooled Variance 56.36 Hypothesized Mean Difference 0 df 598 t Stat -4.00 P(T<=t) one-tail 0.0000 t Critical one-tail 1.6474 P(T<=t) two-tail 0.0001 t Critical two-tail 1.9639 t = –4.00, p-value = 0. There is enough evidence to indicate that recovery is faster in the United States. Question 2: z-tests of p1  p2 (case 1) H0 : (p1  p2 )  0 H1 : (p1  p2 )  0       1 2 1 2 1 n 1 n pˆ(1 pˆ) (pˆ pˆ ) z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions U.S. Canada Sample Proportions 0.6267 0.6867 Observations 300 300 Hypothesized Difference 0 z Stat -1.55 P(Z<=z) one tail 0.0609 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.1218 z Critical two-tail 1.9600 z = –1.55, p-value = .0609. There is not enough evidence to infer that recovery is faster in the United States.495 6 months after heart attack: 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions U.S. Canada Sample Proportions 0.1867 0.1733 Observations 300 300 Hypothesized Difference 0 z Stat 0.43 P(Z<=z) one tail 0.3354 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.6708 z Critical two-tail 1.9600 z = .43, p-value = 1 - .3354 = .6646. There is no evidence to infer that recovery is faster in the United States. 12 months after heart attack 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions U.S. Canada Sample Proportions 0.1167 0.1100 Observations 300 300 Hypothesized Difference 0 z Stat 0.26 P(Z<=z) one tail 0.3984 z Critical one-tail 1.6449 P(Z<=z) two-tail 0.7968 z Critical two-tail 1.9600 z = .26, p-value = 1 – .3984 = .6016. There is no evidence to infer that recovery is faster in the United States. A16.7 Chi-squared test of a contingency table H : 0 The two variables are independent H1 : The two variables are dependent     12 i 1 i 2 2 i i e (f e )496 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A B C D E F Contingency Table Favored Result 1 2 3 TOTAL 4 16 2 3 21 TOTAL chi-squared Stat 13.4477 df 6 p-value 0.0365 chi-squared Critical 12.5916 2  = 13.4477, p-value = .0365. There is enough evidence to infer that Pro-Line's forecasts are related to outcomes and thus, can be useful to bettors. A16.8 t-test of  D H0 : D = 0 H1 :D < 0 D D D D s / n x t    1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means No-Slide Slide Mean 3.73 3.78 Variance 0.0653 0.0727 Observations 25 25 Pearson Correlation 0.96 Hypothesized Mean Difference 0 df 24 t Stat -3.04 P(T<=t) one-tail 0.0028 t Critical one-tail 1.7109 P(T<=t) two-tail 0.0057 t Critical two-tail 2.0639 t = –3.04, p-value = .0028. There is overwhelming evidence to indicate that sliding is slower.497 A16.9 z-test of p1  p2 (case 1) (The data were unstacked prior to applying the z-test.) H0 : (p1  p2 )  0 H1 : (p1  p2 ) > 0       1 2 1 2 1 n 1 n pˆ(1 pˆ) (pˆ pˆ ) z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Optimist Pessimist Sample Proportions 0.9499 0.8797 Observations 1478 241 Hypothesized Difference 0 z Stat 4.26 P(Z<=z) one tail 0 z Critical one-tail 1.6449 P(Z<=z) two-tail 0 z Critical two-tail 1.9600 z = 4.26, p-value = 0. There is sufficient evidence that pessimists are less likely to survive than optimists. A16.10 Simple linear regression with cholesterol reduction (Before – After) as the dependent variable a t-test of β1 or test of ρ H0 :1  0 H1 :1  0 We used the t-test of β1 because parts (b) and (c) use the regression equation to predict and estimate.498 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 A B C D E F SUMMARY OUTPUT Regression Statistics Multiple R 0.7138 R Square 0.5095 Adjusted R Square 0.4993 Standard Error 10.53 Observations 50 ANOVA df SS MS F Significance F Regression .5 49.87 5.92E-09 Residual .9 Total 49 10850 Coefficients Standard Error t Stat P-value Intercept 2.05 3.94 0.52 0.6051 Exercise 0.0909 0.0129 7.06 5.92E-09 t = 7.06; p-value = 0. There is overwhelming evidence to infer that exercise and cholesterol reduction are related. b. Prediction interval 2 2x g / 2,n 2 (n 1)s (x x) 1 n yˆ t s 1         1 2 3 4 5 6 7 8 9 10 11 12 13 A B C Prediction Interval Reduction Predicted value 11.14 Prediction Interval Lower limit -10.76 Upper limit 33.05 Interval Estimate of Expected Value Lower limit 5.54 Upper limit 16.75 The cholesterol reduction is predicted to fall between −10.76 and 33.05. c Confidence interval estimator of the expected value of y 2 2x g / 2,n 2 (n 1)s (x x) 1 n yˆ t s       499 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C Prediction Interval Reduction Predicted value 12.96 Prediction Interval Lower limit -8.83 Upper limit 34.76 Interval Estimate of Expected Value Lower limit 7.79 Upper limit 18.14 We estimate that the mean reduction in cholesterol lies between 7.79 and 18.14. A16.11 Instructors: Unequal-variances t-test of 1  2 H0 : (1 2 )  0 H1 : (1 2 ) < 0          22 2 21 1 1 2 1 2 s n s n (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C t-Test: Two-Sample Assuming Unequal Variances Public-Instructors Private-Instructors Mean 39.40 41.23 Variance 18.38 24.66 Observations 56 130 Hypothesized Mean Difference 0 df 120 t Stat -2.54 P(T<=t) one-tail 0.0061 t Critical one-tail 1.6577 P(T<=t) two-tail 0.0122 t Critical two-tail 1.9799 t = −2,54; p-value = .0061. There is enough evidence to conclude that the salaries of instructors at publicly-funded colleges and universities are less than the salaries of instructors at private colleges and universities.500 Assistant professors: Equal-variances t-test of 1  2 H0 : (1 2 )  0 H1 : (1 2 ) < 0                1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Public-Assistant Private-Assistant Mean 54.08 59.37 Variance 26.10 32.84 Observations 137 130 Pooled Variance 29.38 Hypothesized Mean Difference 0 df 265 t Stat -7.98 P(T<=t) one-tail 2.18E-14 t Critical one-tail 1.6506 P(T<=t) two-tail 4.36E-14 t Critical two-tail 1.9690 t = −7.98; p-value = 0. There is overwhelming evidence to conclude that the salaries of assistant professors at publicly-funded colleges and universities are less than the salaries of assistant professor at private colleges and universities. Associate professors: Equal-variances t-test of 1  2 H0 : (1 2 )  0 H1 : (1 2 ) < 0                1 2 2p 1 2 1 2 1 n 1 n s (x x ) ( ) t501 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Public-Associate Private-Associate Mean 64.45 71.07 Variance 30.96 30.96 Observations 162 160 Pooled Variance 30.96 Hypothesized Mean Difference 0 df 320 t Stat -10.69 P(T<=t) one-tail 2.65E-23 t Critical one-tail 1.6496 P(T<=t) two-tail 5.31E-23 t Critical two-tail 1.9674 t = −10.69; p-value = 0. There is overwhelming evidence to conclude that the salaries of associate professors at publicly-funded colleges and universities are less than the salaries of associate professor at private colleges and universities. Professors: Unequal-variances t-test of 1  2 H0 : (1 2 )  0 H1 : (1 2 ) < 0          22 2 21 1 1 2 1 2 s n s n (x x ) ( ) t 1 2 3 4 5 6 7 8 9 10 11 12 13 A B C t-Test: Two-Sample Assuming Unequal Variances Public-Professor Private-Professor Mean 88.89 107.39 Variance 49.14 74.99 Observations 268 172 Hypothesized Mean Difference 0 df 310 t Stat -23.51 P(T<=t) one-tail 3.61E-71 t Critical one-tail 1.6498 P(T<=t) two-tail 7.22E-71 t Critical two-tail 1.9676 t = −23.51; p-value = 0. There is overwhelming evidence to conclude that the salaries of professors at publiclyfunded colleges and universities are less than the salaries of professors at private colleges and universities.502 A16.12a One-way analysis of variance H : 0 1  2  3 H1 : At least two means differ 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1.65 2 0.823 0.1332 0.8753 3.0259 Within Groups 1847.2 299 6.18 Total 1848.9 301 F = .1332; p-value = .8753. There is no evidence to infer that there are differences between the three groups of patients. b One-way analysis of variance H : 0 1  2  3 H1 : At least two means differ 10 11 12 13 14 15 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 247.0 2 123.48 15.81 2.96E-07 3.03 Within Groups 2334.6 299 7.81 Total 2581.6 301 F = 15.81; p-value = 0. There is overwhelming evidence to conclude that there are differences between the three groups of patients. Multiple comparisons 1 2 3 4 5 6 7 A B C D E Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.0167 Alpha = 0.05 Group 1 After Group 2 After 0.099 0.949 0.922 Group 3 After -1.867 0.942 0.922 Group 2 After Group 3 After -1.965 0.954 0.922 Group 3 differs from both group 1 and group 2. Groups 1 and 2 do not differ. c. The test assures researchers that the three groups of patients were very similar prior to treatments.503 A16.13 a. One-way analysis of variance H : 0 1  2  3  4  5  6 H1 : At least two means differ 13 14 15 16 17 18 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Between Groups 527,465 5 105,493 4.43 0.0015 2.3538 Within Groups 1,571,667 66 23,813 Total 2,099,132 71 F = 4.43; p-value = .0015. There is enough evidence to infer that differences exist between the six groups. b. Two-factor analysis of variance 29 30 31 32 33 34 35 36 A B C D E F G ANOVA Source of Variation SS df MS F P-value F crit Sample 303,247 2 151,623 6.37 0.0030 3.1359 Columns 190,139 1 190,139 7.98 0.0062 3.9863 Interaction 34,080 2 17,040 0.72 0.4927 3.1359 Within 1,571,667 66 23,813 Total 2,099,132 71 Test for interaction: F = .72; p-value = .4927. There is no evidence of interaction. Test for gender (columns): F = 7.98; p-value = .0062. There is enough evidence to conclude that there are differences in cash offers between males and females. Test for age: F = 6.37; p-value = .0030. There is enough evidence to conclude that there are differences in cash offers between the three age groups. A16.14 a Chi-squared test of a contingency table H : 0 The two variables (year and party) are independent H1 : The two variables are dependent     8 i 1 i 2 2 i i e (f e )504 1 2 3 4 5 6 7 8 9 10 11 12 A B C D E F Contingency Table TOTAL Democrats Republicans Other TOTAL chi-squared Stat 19.27 df 6 p-value 0.0037 chi-squared Critical 12.5916 2  = 19.27; p-value = .0037. There is overwhelming evidence to infer that party affiliation in Broward County changed over the four years. b Chi-squared test of a contingency table H : 0 The two variables (year and party) are independent H1 : The two variables are dependent     8 i 1 i 2 2 i i e (f e ) 1 2 3 4 5 6 7 8 9 10 11 12 A B C D E F Contingency Table TOTAL Democrats Republicans Other TOTAL chi-squared Stat 22.65 df 6 p-value 0.0009 chi-squared Critical 12.5916 2  = 22.65; p-value = .0009. There is overwhelming evidence to infer that party affiliation in Miami-Dade changed over the four years. c Chi-squared test of a contingency table H : 0 The two variables (County and party in 2004) are independent H1 : The two variables are dependent505     6 i 1 i 2 2 i i e (f e ) 1 2 3 4 5 6 7 8 9 10 11 12 A B C D Contingency Table Broward Miami-Dade TOTAL Democrats Republicans Other 60 63 123 TOTAL chi-squared Stat 4.44 df 2 p-value 0.1085 chi-squared Critical 5.9915 2  = 4.44; p-value = 1095. There is not enough evidence to infer that party affiliation differ between Broward County and Miami-Dade County. A16.15 t-test of ρ H : 0 0   H1 :   0 1 2 3 4 5 6 7 8 9 10 A B Correlation CO and NO3 Pearson Coefficient of Correlation 0.8913 t Stat 13.62 df 48 P(T<=t) one tail 0 t Critical one tail 1.6772 P(T<=t) two tail 0 t Critical two tail 2.0106 r = .8913, t = 13.62, p–value = 0; there is enough evidence to infer that the belief is correct. A16.16 t-estimator of µ n s x t   / 2506 1 2 3 4 5 6 7 A B C D t-Estimate: Mean Commute Mean 24.54 Standard Deviation 11.63 LCL 23.64 UCL 25.45 Total time spent commuting by all workers: LCL = 129,142,000 (23.64) = 3,052,916,880 minutes UCL = 129,142,000 (25.45) = 3,286,663,900 minutes A16.17 One-way analysis of variance H0: µ1= µ2 = µ3 = µ4 = µ5 H1: At least two means differ F = 13.91, p-value = 0. A16.18 z estimate of p LCL = .2070, UCL = .2440 Total: LCL = 221,963,000(.2070) = 45,948,235, UCL = 221,963,000(.2440) = 54,161,608 A16.19 One-way analysis of variance H0: µ1= µ2 = µ3 H1: At least two means differ507 F = 14.88, p-value = 0. A16. 20 t test of ρ H0: ρ = 0 H1: ρ ≠ 0 r = .1158, p-value = 0. A16.21 One-way analysis of variance H0: µ1= µ2 = µ3 H1: At least two means differ F = 73.84, p-value = 0. A16.22 z estimate of p508 LCL = .1223, UCL = .1527 Total: LCL = 221,963,000(.1223) = 27,144,017, UCL = 221,963,000(.1527) = 33,892,992 A16.23 z estimate of p LCL = .6783, UCL = .7189 Total: LCL = 221,963,000(.6783) = 150,567,518, UCL = 221,963,000(.7189) = 159,560,426 A16.24 z test of p H0: p = .5 H1: p > .5 z = 16.33, p-value = 0. A16.25 t estimate of µ LCL = 56,120, UCL = 60,659 A16.26 t test of ρ H0: ρ = 0 H1: ρ > 0509 r = .5331, p-value = 0. A16.27 F = 1.56, p-value = .0003. Use unequal-variances t test of µ1- µ2 H0: (µ1- µ2) = 0 H1: (µ1- µ2) > 0 t = .253, p-value = .4003. A16.28 F = 1.22, p-value = .0944. Use equal-variances t test of µ1- µ2 H0: (µ1- µ2) = 0 H1: (µ1- µ2) > 0 t = .502, p-value = .3078. A16.29 z estimate of p510 LCL = .2688, UCL = .3596 Total: LCL = 221,963,000(.2688) = 59,659,268, UCL = 221,963,000(.3596) = 79,828,703 A16.30 z test of p1 – p2 H0: (p1 – p2) = 0 H1: (p1 – p2) < 0 z = -.66, p-value = .2545. A16.31 t test of ρ H0: ρ = 0 H1: ρ > 0 A16.32 Chi-square test of a contingency table H0: The two variables are independent H1: The two variables are dependent511 2χ = 41.29, p-value = 0. A16.33 F = 1.54, p-value = 0. Use unequal-variances t test of µ1- µ2 H0: (µ1- µ2) = 0 H1: (µ1- µ2) ≠ 0 t = 11.38, p-value = 0. A16.34 z test of p1 – p2 (case 1) H0: (p1 – p2) = 0 H1: (p1 – p2) ≠ 0 z = 1.95, p-value = .0506.512 A16.35 z test of p1 – p2 (case 1) H0: (p