Solutions for 02402 exam 15. December 20111

Machine AverageI II III IVDay shift 30 35 59 38 40.5Evening shift 58 55 64 49 56.5Night shift 80 54 72 63 67.25Average 56.0 48.0 65.0 50.0 54.75We assume that the data follows a normal distribution.Below, the ANOVA table for the problem is given as it was produced by R. However,some of the numbers are substituted by letters: (In addition, the following informationcan be used: The ”total sums of squares”, SST , equals 2434.25)Analysis of Variance TableResponse: yDf Sum Sq Mean Sq F value Pr(>F)shift 2 1449.50 724.75 9.4430 Dmachine 3 524.25 174.75 2.2769 EResiduals A B CQuestion XI.1 (25) What is A, B and C? This is a two-way ANOVA situation as in12.3 in the book page 371-373 with a = 3 and b = 4 (OR a = 4 and b = 3) Thetable can be looked at/compared with the generic table on page 373. A is the residualdegrees of freedom. In any way the residual degrees of freedom is (a − 1)(b − 1) = 6.B is the residual SS, that can be found as: (page 372)SSE = SST − SS(T r) − SS(Bl) = 2434.25 − 1449.5 − 524.25 = 460.5And finally C is the MSE = A/B = 76.75. So the correct answer is 5: A = 6 , B =460.5 and C = 76.75Question XI.2 (26) The influence of the machines on ’down time’ is studied, a significancelevel of 5% is used. The conclusion and P-value become? (Both conclusion andargument must be correct) We must find the P-value for the relevant F-test (that is,E) which is for the F = 2.2769 using degrees of freedom (3, 6). From Table 6(a) we cansee that the P-value must be above 0.05. In R it could be found as 1-pf(2.2769,3,6)(=0.18). In any case, the correct answer can only be 5: Machines have no significantinfluence on the result, since the P-value= 0.18Question XI.3 (27) The influence of the shifts on ’down time’ is studied, a significancelevel of 5% is used. The critical value and conclusion become: (Both critical value andconclusion must be correct) We must find the critical value for the relevant F-test (thatis, D) which is for the F = 9.4430 using degrees of freedom (2, 6). From Table 6(b) wecan see that this value is 5.14. In R it could be found as qf(0.95,2,6). In any case,13