Question

1.         A.        An educational psychologist is studying student motivation in elementary school.  A sample of n=5 students is followed over 3 years from fourth to sixth grade and measurements of motivation are taken.  The following data are obtained:

            Student            Fourth             Fifth                Sixth

            A                     4                      3                      1

            B                     8                      6                      4

            C                     5                      3                      3

            D                     7                      4                      2

            E                      6                      4                      0

Test the differences at .05.  

B.        Use Tukey’s HSD to report which groups are different.

C.        Use a post hoc t-test to look at the differences between fourth and sixth grade.  

2          A.        An educational psychologist is studying student motivation in elementary school.  A sample of n=15 students, 5 from each grade from fourth to sixth grade and measurements of motivation are taken.  The following data are obtained:

            Fourth             Fifth                Sixth

            4                      3                      1

            8                      7                      0

            3                      3                      2

            5                      5                      2

            5                      4                      0

Test the differences at .05.  

B.  Use Scheffe’s to compare fourth and sixth.  

C.  Use a post hoc t-test to look at the differences between fourth and sixth grade.  

Answer 1

	A	B	C
count, ni =	5	5	5
mean , x̅ i =	6.000	4.00	2.00
std. dev., si =	1.581	1.225	1.581
sample variances, si^2 =	2.500	1.500	2.500
total sum	30	20	10		60	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	4.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	4.000	0.000	4.000
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	20.000	0.000	20.000		40
SS(within ) = SSW = Σ(n-1)s² =	10.000	6.000	10.000		26.0000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   15
df within = N-k =   12
  
mean square between groups , MSB = SSB/k-1 =    20.0000
  
mean square within groups , MSW = SSW/N-k =    2.1667
  
F-stat = MSB/MSW =    9.2308

	SS	df	MS	F	p-value	F-critical
Between:	40.00	2	20.00	9.23	0.0037	3.89
Within:	26.00	12	2.17
Total:	66.00	14
					α =	0.05
			conclusion :	p-value<α , reject null hypothesis

.................

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	12
MSE	2.167
q-statistic value(α,k,N-k)	3.7700
critical value = q*√(MSE/n)

			confidence interval
population mean difference	critical value	lower limit	upper limit	result
µ1-µ2	2.00	2.48	-0.48	4.48	means are not different
µ1-µ3	4.00	2.48	1.52	6.48	means are different
µ2-µ3	2.00	2.48	-0.48	4.48	means are not different

........................

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