Question

******Show work, no Exel/Spss please!!!!******

1. To answer this question, refer to the following hypothetical data collected using replicated measures design.

Subject	1	2	3	4	5	6	7	8	9	10
Pre	50	49	37	16	80	42	40	58	31	21
Post	56	50	30	25	90	44	60	71	32	22

a) In a two-tailed test of H 0 using α=0.05, what is p(obtained) for the results shown? *Hint: "two-tailed" means the same thing as nondirectional. ANSWER = 0.0216

b) What would you conclude regarding H0 using a = 0.05 2-tail? ANSWER = Reject H0

2. Conduct a one-way ANOVA on the following data.

Group 1	4	9	10
Group 2	11	11	10
Group 3	1	6	4

What is your conclusion (use an alpha)? ANSWER = Reject H0

3. What is the value of r for the following relationship between height and weight? ANSWER = 0.87

Height: 60, 64, 65, 68 Mean=62.25

Weight: 103, 122, 137, 132 Mean=123.5

4. Given the following data, what is the value of F crit for the column effect (variable b) *Hint: don't forget to evaluate a numerator (for variable B) and denominator df (within): ANSWER= 3.40

	Variable B
Variable A	1	2	3
1	5	4	4
	2	6	9
	1	3	7
	4	1	5
	2	3	4
2	6	6	5
	2	7	7
	3	4	5
	3	3	4
	2	3	8

5. Given the following data, what is the value of the SS total? ANSWER = 121.8667

	Variable B
Variable A	1	2	3
1	5	4	4
	2	6	9
	1	3	7
	4	1	5
	2	3	4
2	6	6	5
	2	7	7
	3	4	5
	3	3	4
	2	3	8

6. What is the value of the MS rows in the following ANOVA table? ANSWER = 225.25

Source	SS	DF	MS	F obt
Rows	450.5	2	225.25	0.11
Columns	116.4	1	116.40
Interaction	2.3	2	1.15
Within-Cells	829.6		10.85
Total		29

Answer 1

1)

a)

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
50	56	-6.000	0.160
49	50	-1.000	21.160
37	30	7.000	158.760
16	25	-9.000	11.560
80	90	-10.000	19.360
42	44	-2.000	12.960
40	60	-20.000	207.360
58	71	-13.000	54.760
31	32	-1.000	21.160
21	22	-1.000	21.160

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	424	480	-56.000	528.400

mean of difference ,    D̅ =ΣDi / n = -56/10 = -5.6000
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) = √(528.4/9) = 7.6623

std error , SE = Sd / √n =    7.6623   / √   10   =   2.4230      
                          
t-statistic = (D̅ - µd)/SE = (   -5.6   -   0   ) /    2.4230   =   -2.3111
                          
Degree of freedom, DF=   n - 1 =    9                  
  
p-value =        0.0461   [excel function: =t.dist.2t(t-stat,df) ]               

b) Decision:   p-value <α=0.05, Reject null hypothesis

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2)

treatment	I	II	III
count, ni =	3	3	3
mean , x̅ i =	7.667	10.67	3.667
std. dev., si =	3.215	0.577	2.517
sample variances, si^2 =	10.333	0.333	6.333
total sum	23	32	11		66	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				7.33
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.111	11.111	13.444
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	0.333	33.333	40.333		74
SS(within ) = SSW = Σ(n-1)s² =	20.667	0.667	12.667		34.000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   9
df within = N-k =   6
  
mean square between groups , MSB = SSB/k-1 =    37.0000
  
mean square within groups , MSW = SSW/N-k =    5.6667
  
F-stat = MSB/MSW =    6.5294

	SS	df	MS	F	p-value
Between:	74.000	2	37.000	6.529	0.0312
Within:	34.000	6	5.667
Total:	108.000	8

Ho: µ1=µ2=µ3
H1: not all means are equal

p value<α=0.05, Reject Ho

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