Question

CH13 Q3

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 g table.) a. Calculate 99% confidence intervals for μ 1-2, μ1-#3, and μ2-#3 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.) Population Mean Differences Can we conclude that the population means differ? Confidence Interval H1 -H3

Answer 1

a)

Fishers LSD critical value=t_α/2,df √(MSE(1/ni+1/nj))

Group	Mean	Size
group 1	31.8	8
group2	39.3	10
group3	46.4	5

confidence interval= mean difference±critical value

Level of significance	0.01
Numerator d.f. , k =	3
Denominator d.f. = N-k = 23-3 =	20
MSE	33.4
t-critical	2.84534

			confidence interval
population mean difference	mean difference	critical value	lower limit	upper limit	result
µ1-µ2	-7.5	7.8001	-15.30	0.30	Means are not different
µ1-µ3	-14.6	9.3745	-23.97	-5.23	Means are different
µ2-µ3	-7.1	9.0067	-16.11	1.91	Means are not different

b)

	Sample	Sample
Group	Mean	Size
group 1	31.8	8
group2	39.3	10
group3	46.4	5

critical value = q*√(MSE/2*[1/ni+1/nj])

Level of significance   0.01
Numerator d.f. , k =   3
Denominator d.f. = N-k = 23-3 =   20
MSE = 33.4
Q Statistic =3.58

confidence interval = mean difference ± critical value

if confidence interval contans zero, then means are not different.

			confidence interval
population mean difference	mean difference	critical value	lower limit	upper limit	result
µ1-µ2	-7.5	8.9943	-16.49	1.4943	Means are not different
µ1-µ3	-14.6	10.8098	-25.41	-3.79	Means are different
µ2-µ3	-7.1	10.3857	-17.49	3.29	Means are not different