Question

Four brands (A, B, C and D) of hand-held calculators are to be compared for the time it takes to compute the sum of 25 numbers. Thirty-two (32) students were selected at random with 8 assigned to each brand of calculator. A partial ANOVA table is as follows. f Source | d.f. Sum of Squares Mean Squares Between Calculator Brands Within Brands 15.40 Total (a) Write a model for this experiment and define all terms. (b) The brand averages in minutes are YA = 4.52, YB = 3.08, YC = 3.92, and Yp = 4.75. Complete the ANOVA table above and test the equality of the 4 brands at the a = 0.05 level. (c) Check whether the following contrasts are orthogonal to each other. wi = Ma - 1 (uc + up), and we = HB - (uc + up). (d) It is claimed that, on the average, the time for brand A is more than the mean of the times for brands C and D. On the basis of the given data, is there sufficient evidence to support this claim? Use a significance level of 0.05.

Answer 1

(a) Let Yij = time taken by jth student using ith brand of hand- held calculator; i = A, B, C, D, j = 1(1)8. Then model : Yij = pi + eij where, Mi= true mean time taken by ith brand of hand-held calculator: i= A, B, C, D Cij = random error corresponding to (ij)th observation. Assume, e follows independently normal distribution with mean = 0 and variance = oº, where o' is unknown. (6) Null hypothesis, Ho: MA= B = c = HD Us. Alternative hypothesis, H, at least one pas is different from other s. SYA +8YB +8°c +SYD Grand mean = G = ? = (4.52 + 3.08 + 3.92 + 32 4.75)/4 = 4.0675 SS due to Between Calculator Brands = 8* (YX+Y+Y? +Y) - 32Gº = 13.3398

Source Sum of squares Mean Squares Between Calculator Brands 4-1=3 13.3398 13.3398/3 = 4.4466 4.4466/0.55 = 8.0847 Within Brands 31 -3 = 28 15.40 15.40/28 = 0.55 Total 32-1= 31 13.3398 + 15.40 = 28.7398 Critical value = F0.05,3,28 = 2.9467. Since F-value = 8.0847 > F0.05.3.28 = 2.9467 so we reject Ho at 5% level of significance and conclude that at least one mean of 4 brands is significantly different from means of other brands. 1 1 (c) Coefficient vector of w1 = ( 6) = a1, say 1 Coefficient vector of we = (0,1, --) = ay, say aſa2 = +0 the contrasts wi and w, are not orthogonal.

(d) Null hypothesis, H, : W1 = 0 vs. Alternative hypothesis, Ha: W1 > 0 W = YA (+YD) = 4.52 – (3.92 + 4.75)/2 = 0.185, Ôwi = Var(W1) = 3MSE/16 = 0.3211 , where MSE = mean sum of square due to error = 0.55 Test statistic = t = "1 = 0.5761. Critical value = 0.05,28 = 1.7011. Since value oft < Critical value = to.05.28 = 1.7011 so we fail to reject Ho at 5% level of significance and conclude that there is insufficient evidence to support this claim. OW