100 STATISTICAL TESTS phần 1 pdf

[...]... 1. 96 1. 96 2 0.05 1. 64 Data Conclusion H0 : µ0 = 4.0 n = 9, x = 4.6 ¯ σ = 1. 0 ∴ Z = 1. 8 1 Do not reject H0 [see Table 1] 2 Reject H0 6 10 0 STATISTICAL TESTS Test 3 Z-test for two population means (variances known and unequal) Hypotheses and alternatives 1 2 H0 : 1 − µ2 H1 : 1 − µ2 H0 : 1 − µ2 H1 : 1 − µ2 Z= Test statistics = µ0 = µ0 = µ0 > µ0 (¯ 1 − x2 ) − µ0 x ¯ 2 σ2 1 + 2 n1 n2 1 2 When used 2... degrees of freedom 2 2 (n1 − 1) s1 + (n2 − 1) s2 n1 + n2 − 2 1 DF = n1 + n2 – 2 0.025 0.025 –tn1 + n2 – 2; 0.025 tn1 + n2 – 2; 0.025 2 0.05 tn1 + n2 – 2; 0.05 H0 : 1 − µ2 = 0 n1 = 16 , n2 = 16 ¯ x1 = 5.0, x2 = 4 ¯ s = 2.0 ∴ t = 1. 414 Data Conclusion 1 2 t30; 0.025 = ±2.042 [see Table 2] Do not reject H0 t30; 0.05 = 1. 697 [see Table 2] Do not reject H0 EXAMPLES OF TEST PROCEDURES Test 10 9 Method of paired... freedom 1 DF = n – 1 0.025 –tn – 1; 0.025 0.025 tn – 1; 0.025 2 0.05 tn – 1; 0.05 n1 = 16 , d = 1. 0 s = 1. 0 ∴ t = 4.0 Data Conclusion 1 2 t15; 0.025 = ±2 .13 1 [see Table 2] Reject H0 t15; 0.05 = 1. 753 [see Table 2] Raject H0 10 10 0 STATISTICAL TESTS Test 15 χ 2 -test for a population variance Hypotheses and 1 alternatives 2 2 = σ0 2 = σ0 2 = σ0 2 > σ0 (n − 1) s2 2 σ0 Given a sample from a normal population... 1 2 H1 : 1 F= Test statistics 2 s1 2 s2 2 = σ2 2 = σ2 2 = σ2 2 > σ2 , 2 2 (s1 > s2 ) 2 2 where s1 and s2 are sample variances 2 2 (If, in 2, s1 < s2 , do not reject H0 ) 2 Given two sample with unknown variances 1 2 and normal populations and σ2 When used Critical region and degrees of freedom 1 DF = n1 – 1 and n2 – 1 0.025 Fn1 – 1, n2 – 1; 0.025 2 0.05 Fn1 – 1, n2 – 1; 0.05 2 2 H0 : 1 = σ2 n1... H1 : σ 2 H0 : σ 2 H1 : σ 2 1 DF = n – 1 0.025 2 χn – 1; 0.975 0.025 2 χn – 1; 0.025 2 0.05 2 χn – 1; 0.05 H0 : σ 2 = 4.0 n1 = 17 , s2 = 7.0 ∴ χ 2 = 28.0 Data Conclusion 2 16 ; 0.025 = 28.85 [see Table 5] ∴ Do not reject H0 2 2 16 ; 0.05 = 26.30 [see Table 5] ∴ Reject H0 1 EXAMPLES OF TEST PROCEDURES 11 Test 16 F -test for two population variances Hypotheses and 1 alternatives 2 2 H0 : 1 2 H1 : 1. .. 0.025 = ±2.306 [see Table 2] Reject H0 t8; 0.05 = 1. 860 (left-hand side) [see Table 2] Reject H0 8 10 0 STATISTICAL TESTS Test 8 t-test for two population means (variance unknown but equal) Htypotheses and alternative H0 : 1 − µ2 H1 : 1 − µ2 2 H0 : 1 − µ2 H1 : 1 − µ2 1 t= Test statistics = µ0 = µ0 = µ0 > µ0 (¯ 1 − x2 ) − ( 1 − µ2 ) x ¯ 1 1 + s n2 n1 1 2 where s2 = Given two samples from normal populations... that the mean µ of a population with known variance has the value µ0 rather than the value 1 Test 61 To test the null hypothesis that the standard deviation σ of a population with a known mean has the value σ0 rather than the value 1 86 88 89 91 93 94 95 96 97 98 99 10 1 10 2 10 4 10 6 10 7 10 8 10 9 11 0 11 2 11 4 ... = σ2 n1 = 11 , n2 = 2 2 s1 = 6.0, s2 = Data ∴ F = 2.0 Conclusion 16 3.0 F10, 15 ; 0.025 = 3.06 Do not reject H0 2 F10, 15 ; 0.05 = 2.54 [see Table 3] Do not reject H0 1 12 10 0 STATISTICAL TESTS Test 37 χ 2 -test for goodness of fit Hypotheses and alternatives Goodness of fit for Poisson distribution with known mean λ Test statistics χ2 = (Oi − Ei )2 Ei Oi is the ith observed frequency, i = 1 to k; Ei... ¯ 2 σ2 1 + 2 n1 n2 1 2 When used 2 When the variances of both populations, 1 2 , are known Populations are normally and σ2 distributed Critical region Using α = 0.05 [see Table 1] 1 0.025 0.025 1. 96 1. 96 2 0.05 1. 64 Data Conclusion H0 : 1 − µ2 = 0 n1 = 9, n2 = 16 ¯ x1 = 1. 2, x2 = 1. 7 ¯ 2 2 1 = 1, σ2 = 4 ∴ Z = −0.832 1 Do not reject H0 2 Do not reject H0 EXAMPLES OF TEST PROCEDURES Test 7 7 t-test... H1 : µ = µ0 2 H0 : µ = µ0 H1 : µ > µ0 1 t= Test statistics x − µ0 ¯ √ s/ n where s2 = If σ 2 is not known and the estimate s2 of σ 2 is based on a small sample (i.e n < 20) and a normal population When used Critical region and degrees of freedom (x − x )2 ¯ n 1 1 DF = n 1 0.025 0.025 –tn 1; 0.025 tn 1; 0.025 2 0.05 tn 1; 0.05 H0 : µ0 = 4.0 n = 9, x = 3 .1 ¯ s = 1. 0 ∴ t = −2.7 Data Conclusion 1 . GOKA: “FM” — 2006/6 /15 — 15 : 21 — PAGEi— #1 10 0 STATISTICAL TESTS GOKA: “FM” — 2006/6 /15 — 15 : 21 — PAGE ii — #2 GOKA: “FM” — 2006/6 /15 — 15 : 21 — PAGE iii — #3 10 0 STATISTICAL TESTS 3rd Edition Gopal. 1 and n 2 – 1 F n 1 – 1, n 2 – 1; 0.025 2. 0.05 F n 1 – 1, n 2 – 1; 0.05 Data H 0 : σ 2 1 = σ 2 2 n 1 = 11 , n 2 = 16 s 2 1 = 6.0, s 2 2 = 3.0 ∴ F = 2.0 Conclusion 1. F 10 , 15 ; 0.025 = 3.06. Do. µ 2 = µ 0 2. H 0 : µ 1 − µ 2 = µ 0 H 1 : µ 1 − µ 2 >µ 0 Test statistics t = (¯x 1 −¯x 2 ) −(µ 1 − µ 2 ) s  1 n 1 + 1 n 2  1 2 where s 2 = (n 1 − 1) s 2 1 + (n 2 − 1) s 2 2 n 1 + n 2 − 2 . When