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# Appendix 17 - Appendix 17 A17.1a z-test of p1 p2(case 1 H0...

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Appendix 17 A17.1a z-test of p 1 – p 2 (case 1) H 0 : p 1 – p 2 = 0 H 1 : p 1 – p 2 < 0 + - - = 2 1 2 1 n 1 n 1 ) p ˆ 1 ( p ˆ ) p ˆ p ˆ ( z 1 2 3 4 5 6 7 8 9 10 11 A B C D z-Test: Two Proportions Allicin Cold? Placebo Cold? Sample Proportions 0.3288 0.8904 Observations 73 73 Hypothesized Difference 0 z Stat -6.96 P(Z<=z) one tail 0 z Critical one-tail 1.6449 P(Z<=z) two-tail 0 z Critical two-tail 1.96 z = −6.96; p-value = 0. There is enough evidence to conclude that garlic does help prevent colds? b. Equal-variances t-test of µ 1 − µ 2 H 0 : µ 1 – µ 2 = 0 H 1 : µ 1 – µ 2 < 0 + μ - μ - - = 2 1 2 p 2 1 2 1 n 1 n 1 s ) ( ) x x ( t 49

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Two-Sample Assuming Equal Variances Allicin Days Placebo Days Mean 6.29 8.11 Variance 2.39 4.25 Observations 24 65 Pooled Variance 3.76 Hypothesized Mean Difference 0 df 87 t Stat -3.92 P(T<=t) one-tail 0.0001 t Critical one-tail 1.66256 P(T<=t) two-tail 0.0002 t Critical two-tail 1.98761 t = −3.92; p-value = .0001. There is enough evidence to conclude that garlic reduces the number of days until recovery if a cold is caught A17.2 t-test of µ D H 0 : µ D = 0 H 1 : µ D > 0 D D D D n / s x t μ - = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C t-Test: Paired Two Sample for Means Eye-level Lower shelf Mean 302.4 290.8 Variance 2482.2 6262.7 Observations 40 40 Pearson Correlation 0.7334 Hypothesized Mean Difference 0 df 39 t Stat 1.35 P(T<=t) one-tail 0.0922 t Critical one-tail 1.6849 P(T<=t) two-tail 0.1845 t Critical two-tail 2.0227 t = 1.35; p-value = .0922. There is not enough evidence to conclude that placement of the product at eye level significantly increases sales? A17.3 Equal-variances t-test of 2 1 μ - μ ) ( : H 2 1 0 μ - μ = 0 50
) ( : H 2 1 1 μ - μ < 0 + μ - μ - - = 2 1 2 p 2 1 2 1 n 1 n 1 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 British American Mean 6344.5 6358.3 Variance 5084.0 3104.6 Observations 28 33 Pooled Variance 4010.4 Hypothesized Mean Difference 0 df 59 t Stat -0.84 P(T<=t) one-tail 0.2010 t Critical one-tail 1.6711 P(T<=t) two-tail 0.4019 t Critical two-tail 2.0010 t = –.84, p-value = .2010. There is not enough evidence to conclude that British courses are shorter than American courses. A17.4 Chi-squared test of a contingency table : H 0 The two variables are independent : H 1 The two variables are dependent = - = χ 6 1 i i 2 i i 2 e ) e f ( 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C D E Contingency Table Group Choice 1 2 TOTAL 1 7 19 26 2 8 17 25 3 11 14 25 TOTAL 26 50 76 chi-squared Stat 1.73 df 2 p-value 0.4206 chi-squared Critical 5.9915 51

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2 χ = 1.73, p-value = .4206. There is not enough evidence to infer that there is a relationship between choices students make and their level of intoxication. A17.5 t-test of ρ or β 1 ρ : H 0 = 0 0 : H 1 < ρ 2 r 1 2 n r t - - = 1 2 3 4 5 6 7 8 9 10 A B C D Correlation Repair cost and Credit score Pearson Coefficient of Correlation -0.3830 t Stat -6.11 df 217 P(T<=t) one tail 0 t Critical one tail 1.6519 P(T<=t) two tail 0 t Critical two tail 1.9710 t = −6.11; p-value = 0. There is enough evidence to infer that as the credit score increases the repair cost decreases.
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