MATH 38061/MATH48061/MATH68061: MULTIVARIATE STATISTICS Solutions to Problems on One-Sample Inference 1. Consider testing H0 : µT = [7, 11] using the data " X= 2 8 6 8 12 9 9 10 # . For this data, " µ0 = " ¯ = X " S= 7 11 # 6 10 # , , # 8 −10/3 −10/3 2 , and S −1 9 = 44 " # 2 10/3 10/3 8 . So, T 2 = n X − µ0 T S−1 X − µ0 = 13.63636. The distribution of T 2 is (n − 1)p Fp,n−p = 3F2,2 . n−p At α = 0.05, we have 3F2,2 (0.05) = 57. So, there is no evidence against the hypothesis that H0 : µT = [7, 11]. 2. Measurements of cranial length (x1 ) and cranial breadth (x2 ) on a sample of 35 mature female frogs led to the following statistics: ¯ T = [22.860, 24.397] x and " S= 17.178 19.710 23.710 1 # . To test the hypothesis that µ1 = µ2 , set l = [1, −1]T . Then s T l x− p(n − 1) Fp,n−p (α)lT Sl, lT x + n(n − p) s 22.860 − 24.397 − = s ! p(n − 1) Fp,n−p (α)lT Sl n(n − p) 2 × 34 × F2,33 (0.05) × 1.468, 35 × 33 s 22.860 − 24.397 + ! 2 × 34 × F2,33 (0.05) × 1.468 . 35 × 33 This interval does not contain zero. So, there is evidence against the hypothesis that µ1 = µ2 at the five percent level. 3. A physical anthropologist performed a mineral analysis of nine ancient Peruvian hairs. The result for the chromium (x1 ) and strontium (x2 ) levels, in parts per million (ppm) were as follows. x1 x2 0.48 12.57 40.53 73.68 2.19 11.13 0.55 20.03 0.74 20.29 0.66 0.78 0.93 4.64 0.37 0.43 0.22 1.08 For this data, " ¯ = X " S= 5.185556 16.070000 # , 176.0042 287.2412 287.2412 527.8493 # , and " S −1 = 0.05077341 −0.02762951 −0.02762951 0.01692971 # . Then a 90% joint confidence ellipse for the population mean vector µT = (µ1 , µ2 ) is " 9 ≤ 5.185556 − µ1 16.070000 − µ2 #T " 0.05077341 −0.02762951 −0.02762951 0.01692971 #" 5.185556 − µ1 16.070000 − µ2 # 2×8 F2,7 (0.1) = 21.82075 7 which can be rewritten as 9 × {0.05077341(5.185556 − µ1 )2 + 0.01692971(16.070000 − µ2 )2 − 2 × 0.02762951(5.185556 − µ1 )(16.070000 − µ2 )} ≤ 21.82075. The individual simultaneous 90% confidence intervals for µ1 and µ2 are s x1 − p(n − 1) Fp,n−p (α) (n − p) ≡ 5.185556 − r q s11 , x1 + n s p(n − 1) Fp,n−p (α) (n − p) 21.82075 × (176.0042/9), 5.185556 + 2 q r s11 n ! 21.82075 × (176.0042/9) and s x2 − p(n − 1) Fp,n−p (α) (n − p) r ≡ 16.070000 − s22 , x2 + n s p(n − 1) Fp,n−p (α) (n − p) q 21.82075 × (527.8493/9), 16.070000 + q r s22 n ! 21.82075 × (527.8493/9) . 4. Perspiration from 10 healthy females were analyzed. Three components, X1 = sweat rate, X2 = sodium content, X3 = potassium content, were measured, and the results are presented in the table below. Individual 1 2 3 4 5 6 7 8 9 10 X1 3.7 5.7 3.8 3.2 3.1 4.6 2.4 7.2 6.7 5.4 X2 48.5 65.1 47.2 53.2 55.5 36.1 24.8 33.1 47.4 54.1 X3 9.3 8.0 10.9 12.0 9.7 7.9 14.0 7.6 8.5 11.3 For this data, 4.58 ¯ = X 46.50 , 9.92 0.2612889 0.1787778 −0.2422889 S = 0.1787778 14.3724444 −0.5726667 , −0.2422889 −0.5726667 0.4381778 and S−1 7.94150203 0.08036875 4.4962623 = 0.08036875 0.07421317 0.1414309 . 4.49626227 0.14143086 4.9532113 Then a 90% joint confidence ellipse for the population mean vector µT = (µ1 , µ2 , µ3 ) is T 4.58 − µ1 7.94150203 0.08036875 4.4962623 4.58 − µ1 10 46.50 − µ2 0.08036875 0.07421317 0.1414309 46.50 − µ2 9.92 − µ3 4.49626227 0.14143086 4.9532113 9.92 − µ3 9×3 ≤ F3,7 (0.1) = 32.59781 7 which can be rewritten as 10×{7.94150203(4.58−µ1 )2 +0.07421317(46.50−µ2 )2 +4.9532113(9.92− µ3 )2 + 2 × 0.08036875(4.58 − µ1 )(46.50 − µ2 ) + 2 × 4.4962623(4.58 − µ1 )(9.92 − µ3 ) + 2 × 0.1414309(46.50 − µ2 )(9.92 − µ3 )} ≤ 32.59781. 3 The individual simultaneous 90% confidence intervals for µ1 , µ1 and µ3 are s x1 − ≡ 4.58 − s x2 − ≡ 46.5 − p(n − 1) Fp,n−p (α) (n − p) r s11 , x1 + n s p(n − 1) Fp,n−p (α) (n − p) r q 32.59781 × (0.2612889/10), 4.58 + p(n − 1) Fp,n−p (α) (n − p) r q s22 , x2 + n s ! q 32.59781 × (0.2612889/10) , p(n − 1) Fp,n−p (α) (n − p) 32.59781 × (14.3724444/10), 46.5 + s11 n r s22 n ! q 32.59781 × (14.3724444/10) , and s x3 − ≡ 9.92 − p(n − 1) Fp,n−p (α) (n − p) r s33 , x3 + n s p(n − 1) Fp,n−p (α) (n − p) q r 32.59781 × (4.9532113/10), 9.92 + s33 n ! q 32.59781 × (4.9532113/10) . 5. Many one-sample and two-sample tests involve the statistic dT S−1 d, where S is a p × p matrix. Suppose S is a diagonal. Then d T = = d T p X s11 0 · · · 0 0 s22 · · · 0 .. .. .. .. . . . . 0 0 · · · spp −1 s−1 0 ··· 0 11 0 s−1 ··· 0 22 .. .. .. .. . . . . −1 0 0 · · · spp d d 2 s−1 ii di i=1 = p X t2i , i=1 √ where ti = di / sii . Suppose now p = 2. Then S−1 = " 1 s11 s22 − s212 s22 −s12 −s12 s11 # and T −1 d S d = = = 1 s11 s22 − s212 " T d s22 −s12 −s12 s11 s22 d21 + s11 d22 − 2s12 d1 d2 s11 s22 − s212 t21 + t22 − 2t1 t2 r12 , 2 1 − r12 4 # d √ where r12 = s12 / s11 s22 . Suppose now p = 3. Then S−1 s33 s22 − s32 s23 − (s33 s12 − s32 s13 ) s23 s12 − s22 s13 1 − (s s − s s ) s s − s s − (s23 s11 − s21 s13 ) , = 33 12 32 13 33 11 31 13 D s23 s12 − s22 s13 − (s23 s11 − s21 s13 ) s22 s11 − s21 s12 where D = s11 (s33 s22 − s32 s23 ) − s21 (s33 s12 − s32 s13 ) + s31 (s23 s12 − s22 s13 ). So, dT S−1 d = 1n (s33 s22 − s32 s23 ) d21 − 2 (s33 s12 − s32 s13 ) d1 d2 + 2 (s23 s12 − s22 s13 ) d1 d3 D o + (s33 s11 − s31 s13 ) d22 − 2 (s23 s11 − s21 s13 ) d2 d3 + (s22 s11 − s21 s12 ) d23 which, after considerable simplification, reduces to dT S−1 d = 2 ) + t2 (1 − r 2 ) + t2 (1 − r 2 ) − 2t t u t21 (1 − r23 1 2 12·3 − 2t1 t3 u13·2 − 2t2 t3 u23·1 2 13 3 12 , 2 2 − r2 1 + 2r12 r13 r23 − r23 − r13 12 where u12·3 = r12 − r13 r23 , etc. 5
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