HW 10 Solutions 12.14 To test H0:µ1=µ2=µ3=µ4 against H1: µi≠µj for some i≠j we calculate F=MSB/ MSW and reject H0 at level α if F≥Fk,n-k,1-α. 4
4
SSB= ∑ ni y 2 −
∑y i =1
i =1
n
2 i
=7(7.59)2+5(7.24)2+7(6.33)2+8(6.01)2-[181.272/27]
= 1234.7878-1222.7737 = 12.0141 MSB= SSB/k-1= 12.0141/3 = 4.0047. 4
SSW=
∑ (n i =1
i
− 1) s i2 = 6(.96)2+4(.91)2+6(1.14)2+7(0.72)2 = 20.2684
MSW = SSW/n-k= 20.2684/23 = 0.8812. F = 4.0047/.8812 = 4.54>F3,23,.95 ≈3.05, therefore we reject H0 at the 0.05 level. 12.15-12.17
For each pairwise comparison (H0: µi=µj) we calculate t=
yi − yi 1 1 WMS + ni n j
. With no adjustment for multiple comparisons we reject H0 at level
α if |t| ≥tn-k,1-α/2 = t23,.975 = 2.069. With Bonferroni’s adjustment we would reject H0 when |t| ≥ tn-k,1-α/k(k-1) = t23,.99583 ≈2.984. 1 vs 2: t =
1 vs 3: t =
7.59 − 7.24 1 1 .8812( + ) 7 5 7.59 − 6.33 1 1 .8812( + ) 7 7
≅ 0.64
≅ 2.51
1 vs 4: t =
2 vs 3: t =
2 vs 4: t =
3 vs 4: t =
7.59 − 6.01 1 1 .8812( + ) 7 8 7.24 − 6.33 1 1 .8812( + ) 7 5 7.24 − 6.01 1 1 .8812( + ) 5 8 6.33 − 6.01 1 1 .8812( + ) 7 8
≅ 3.25
≅ 0.64
≅ 2.30
≅ 0.58
Without adjustment there are significant differences at the 0.05 level between groups 1 and 3, groups 1 sand 4, and groups 2 and 4; with the Bonferroni adjustment only groups 1 and 4 differ at the 0.05 level.
12.24
To test H0:µ1=µ2=µ3=µ4 against H1: µi≠µj for some i≠j we calculate F=MSB/ MSW and reject H0 at level α if F≥Fk,n-k,1-α. 4
4
SSB= ∑ ni y 2 − i =1
∑y i =1
n
2 i
=20(8.62 + 5.32 + 4.92 + 1.12 )-(3982/20) = 2454.4 – 1980.5 = 564.9.
SSW + 19(6.22 +5.42 + 7.02 + 6.52 )=3018.15 ; MSW=3018.15/76=39.7125. F=(564.9/3) / (3018.15/76) = 4.74 > F3,76 (note that F3,76.95 < F3,76,.95 = 2.76, the nearest tabled value in the text) so we reject H0 at the 0.05 level.
12.25
To test H0: L = (µ1+µ2) – (µ3+µ4)=0 vs. H1: L ≠ 0 we compute t =
Lˆ
. ci2 WMS (∑ i =1 ni Without making any adjustment for multiple comparisons we would reject H0 at level α=.05 when |t| > t76,0.975. Using the Scheffe method to adjust for multiple comparisons we reject H0 when |t| > (k − 1) Fk −1,n − k ,1−α .
t=
8.6 + 5.3 − 4.9 − 1.1
=
7 .9 = 2.80 ; t76,.975=1.99 so an unadjusted test would reject H0 2.82
4 ) 20 at the .05 level. Scheffe’s criterion is approximately method we would not reject H0 at the .05 level. 3018.15(
4
3(2.74) and so using the Scheffe
12.26
This is just like problem 12.25 except the H0 we are testing is (µ1+µ3)-(µ2+µ4). Since the coefficients (the ci) do not change, the denominator of the t-statistic doesn’t change. The critical values are also the same. The numerator is now 8.6+4.9-5.3-1.1=7.1 so t=2.52. The critical values are as they were in 12.25 and the conclusions are the same as well. 12.27
We want to test whether the effect of weight reduction differs those receiving meditation treatment and those not. The effect of weight reduction among those receiving meditation treatment is µ1-µ3. The effect of weight reduction among those not receiving meditation treatment is µ2-µ4. Therefore we want to test H0: (µ1-µ3) – (µ2 - µ4) =0 or µ1-µ2-µ3+µ4=0. This contrast is tested like the null hypotheses in 12.25 and 12.26 (i.e. using the same form for the t-statistic and the same critical values) and since this is another contrast with all coefficients equal to 1 or –1 the denominator of t is unchanged. The numerator is now 8.6-5.3-4.9+1.1 and t = -0.18. We would not reject H0 for both unadjusted and adjusted tests.
