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In: Math

A professor has a class with four recitation sections. Each section has 16 students (rare, but...

A professor has a class with four recitation sections. Each section has 16 students (rare, but there are exactly the same number in each class...how convenient for our purposes, yes?). At first glance, the professor has no reason to assume that these exam scores from the first test would not be independent and normally distributed with equal variance. However, the question is whether or not the section choice (different TAs and different days of the week) has any relationship with how students performed on the test.

Group-1	Group-2	Group-3	Group-4
73.5	76.7	75	65.7
81	66.4	77.8	50.5
61.8	60.3	66.7	83
69.5	81	70.3	81.4
77.4	57.9	77.7	74.9
91.2	59.2	68.1	82.9
70.6	67.9	83.5	85.4
64	54.9	87.8	63.6
73.2	63.2	80.6	67.6
77.7	69.8	58.9	73.6
73.6	69.1	86.7	81.5
77.2	51.8	74.7	80.5
54	60.5	66.9	71.8
65.4	55.4	76.7	68.1
77.8	68.2	76.3	55.8
81.6	64.8	69.5	70.4

First, run an ANOVA with this data and fill in the summary table. (Report P-values accurate to 4 decimal places and all other values accurate to 3 decimal places.

Source	SS	df	MS	F-ratio	P-value
Between
Within

To follow-up, the professor decides to use the Tukey-Kramer method to test all possible pairwise contrasts.

What is the Q critical value for the Tukey-Kramer critical range (alpha=0.01)?
Use the website link in your notes (http://davidmlane.com/hyperstat/sr_table.html) to locate the Q critical value to 4 decimal places.
Q =

Using the critical value above, compute the critical range and then determine which pairwise comparisons are statistically significant?

group 1 vs. group 2
group 1 vs. group 3
group 1 vs. group 4
group 2 vs. group 3
group 2 vs. group 4
group 3 vs. group 4
none of the groups are statistically significantly different

Expert Solution

	Group-1	Group-2	Group-3	Group-4	Total
Sum	1169.5	1027.1	1197.2	1156.7	4550.5
Count	16	16	16	16	64
Mean, sum/n	73.09375	64.19375	74.825	72.29375
Sum of square, Ʃ(xᵢ-x̅)²	1195.049	941.0294	917.31	1551.929

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3 = µ4

H1: At least one mean is different

Number of treatment, k = 4

Total sample Size, N = 64

df(between) = k-1 =3

df(within) = k(n-1) =60

df(total) = N-1 =63

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 + (Sum4)²/n4 - (Grand Sum)²/ N =1071.552

SS(within) = SS1 + SS2 + SS3 + SS4 =4605.318

SS(total) = SS(between) + SS(within) =5676.870

MS(between) = SS(between)/df(between) =357.184

MS(within) = SS(within)/df(within) =76.755

F = MS(between)/MS(within) =4.654

p-value = F.DIST.RT(4.6535, 3, 60) = 0.0054

Decision:

Reject the null hypothesis.

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	1071.552	3	357.184	4.654	0.0054
Within Groups	4605.318	60	76.755
Total	5676.870	63

At α = 0.01, k = 4, N-K = 60, Q value = 4.5942

Critical Range, CV = Q*√(MSW/n) = 4.5942*√(76.7553/16) = 10.0625

Comparison	Absolute Diff. = (xi - xj)	Critical Range	Results
x̅1-x̅2	8.9	10.0625	Means are not different
x̅1-x̅3	-1.73125	10.0625	Means are not different
x̅1-x̅4	0.8	10.0625	Means are not different
x̅2-x̅3	-10.6313	10.0625	Means are different
x̅2-x̅4	-8.1	10.0625	Means are not different
x̅3-x̅4	2.53125	10.0625	Means are not different

milcah answered 10 months ago

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