Question

A college dean is interested in the exam performance of students in a biology course. After the final exam, students are randomly selected from three different section of the biology course. What can be conclude with an α of 0.05? The data are below.

section 1	section 2	section 3
74 68 74 65 41 75 64 77 40 78 69	90 82 82 77 75 82 81 91 70 90 82	94 81 87 82 72 77 81 87 67 86 77

a) What is the appropriate test statistic?
---Select--- na one-way ANOVA within-subjects ANOVA two-way ANOVA

b) Compute the appropriate test statistic(s) to make a decision about H₀.
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

c) Compute the corresponding effect size(s) and indicate magnitude(s).
η² =  ;  ---Select--- na trivial effect small effect medium effect large effect

d) Make an interpretation based on the results.

At least on section differs on the final exam.None of the sections differ on the final exam.    

e) Conduct Tukey's Post Hoc Test for the following comparisons:
1 vs. 2: difference =  ; significant:  ---Select--- Yes No
2 vs. 3: difference =  ; significant:  ---Select--- Yes No

f) Conduct Scheffe's Post Hoc Test for the following comparisons:
2 vs. 3: test statistic =  ; significant:  ---Select--- Yes No
1 vs. 2: test statistic =  ; significant:  ---Select--- Yes No

Answer 1

here we want to test the

Null hypothesis H0:meal of all sections are same

Alternate hypothesis Ha: atleast one section mean is different from others

(a) one-way ANOVA ,

A one-way ANOVA only involves one factor or independent variable and here only one factor Section is there.

(b1) critical value=F(0.05,2,30)=3.32

(b2)test statistic=F=9.59

(b3) Reject H0, as the critical value=3.32 is less than test statistic=9.59

					within SS		between SS
Group	nj	mean(xj-)	s2	nj*xj-	(n-1)s2	(xj--x-)	nj(xj--x-)2
advanced	11	65.90909	179.2909	725	1792.909091	-10.3939	1188.37374
Intermediate	11	82	42.8	902	428	5.69697	357.010101
Beginner	11	81	57.6	891	576	4.69697	242.676768
sum	33	228.9091	279.6909	2518	2796.909091	1.42E-14	1788.06061
grand mean(x-)	76.30303
		ANOVA
SOURCE	DF	SS	MS	F	CRITICAL F(0.05)	p-value
BETWEEN	2	1788.06	894.0303	9.59	3.32	0.0006
WITHIN(ERROR)	30	2796.91	93.2303
TOTAL	32	4584.97

(c)  η² =Between_SS/Totao_SS=1788.06/4584.97=0.3900

it is large effect

• .01 ~ small
• .06 ~ medium
• >.14 ~ large

(d)At least on section differs on the final exam.

(e)

comparison	Difference	HSD	Remarks
1 Vs 2	16.09	10.16	significant, Yes
2 Vs 3	1	10.16	significant, NO

since difference is more than HSD , so comparison 1 Vs 2 is significant but 2 Vs 3 is not significant

Tukey HSD=	q*sqrt((MSE/2)*(1/n1+1/n2)=	10.16
q(0.05, 3,30)=	3.49
MSE=	93.23
n1=	11
n2=	11

(f) since difference is less than Scheffe value , so comparison 1 Vs 2 and 2 Vs 3 is not significant

comparison	Difference	Scheffe	Remarks
2 Vs 3	1	18.03	significant, NO
1 Vs 2	16.09	18.03	not significant, NO

Scheffe= =sqrt(2*9.59*93.23*(1/11 + 1/11))=18.03

please look the anova for calcualtion