Question

The following statistics are computed by sampling from three normal populations whose variances are equal: (You may find it useful to reference the t table and the q table.)

x−1x−1 = 25.3, n₁ = 8; x−2x−2 = 31.5, n₂ = 10; x−3x−3 = 32.3, n₃ = 6; MSE = 27.2

a. Calculate 95% confidence intervals for μ₁ − μ₂, μ₁ − μ₃, and μ₂ − μ₃ to test for mean differences with Fisher’s LSD approach. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

b. Repeat the analysis with Tukey’s HSD approach. (If the exact value for n_T – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

Answer 1

Answer:

a) 'or (4 -605a (21df) = 2.079614 The 9o confidence interval among the means are Marginal Error=tal貳)xVMS, (1/A + 1/A)-2.0796 14.272(1/10+ 1,8) = 514468 1 x1-x2)tME (25.3-31.59)5.144681 (-11.3446805, 105531947) (Noi) no significance Marginal Error tor (dj.) /MS. ( 1 / M + 1 / %)-20796 14 /272 (1 / 6 + 1 / 8)-5857476 Xi-3)+ ME (25.3-32.3)+5.857476 (-12.8574765, 1.14252355) (Yes) signiicance Marginal Error=trit(d.)x /MS, (1/4 +11m)-2.0796 14 /27. 2(1/10+ 1/6)-5.600819 (2 -x3)E (31.5-32.3)+5.600819 (-6.40081875, 4.800818745) (No) No signijficance b) The 99% confidence interval between M1 and M2 is (XA-XB)±9a.MS, / 2,f/ni)+(1/n)-(25 3-31.59)±356¥27 212)(1/10 + 1/8)-(-12.4362031, (XA-Xe)±4w/MS /211 / )4(1/ぬ)-(253-323)±3.562,027 2/2)(1/6+1/8)-(-14 1002295, (хв-xe) q, as, / 2./(1/4)+(1/%)-(31.5-32 3) 3.565./(272 / 2)(1 / 10 + 1/6)-(-7.5891 179, 0.036203052) No significant The 95% confidence interval between M1 and M3 is 0.100229515) No significant The 95% confidence interval between M1 and M3 is 5.989 1 17959) No significant