Question

Complete an ANOVA table given the following information. You must show all work to receive partial credit.

Question 1: How does a dog’s tail wag in response to seeing different people and other pets?

Owner: 69, 72, 65, 75, 70

Cats: 28, 32, 30, 29, 31

Other dogs: 45, 43, 47, 45, 44

1. Find the degrees of freedom for each group:

(Remember df = N -1)

Group 1:

Group 2:

Group 3:

2. Degrees of freedom (within):

3. Degrees of freedom (total):

4. Degrees of freedom (between):

5. Calculate the total sum of squares. Be sure to calculate the grand mean (GM) and the mean (M) for each group before calculating the sum of squares.

Sample	X	(X – GM)	(x – GM)²
Group 1 - owner M(group 1) =	69 72 65 75 70
Group 2 - cats M(group 2) =	28 32 30 29 31
Group 3 – other dogs M(group 3) =	45 43 47 45 44
GM =			SS(total) =

6. Calculate the Within-Groups Sum of Squares

Sample	X	(X – M)	(x – M)²
Group 1 - owner M(group 1) =	69 72 65 75 70
Group 2 - cats M(group 2) =	28 32 30 29 31
Group 3 – other dogs M(group 3) =	45 43 47 45 44
GM =			SS(within) =

7. Calculate the Between-Groups Sum of Squares

Sample	X	(M – GM)	(M – GM)²
Group 1 - owner M(group 1) =	69 72 65 75 70
Group 2 - cats M(group 2) =	28 32 30 29 31
Group 3 – other dogs M(group 3) =	45 43 47 45 44
GM =			SS(between) =

8. Complete the ANOVA table using the information you calculated above

Variance	SS	df	MS	F
Between
Within				---------
Total			---------	---------

Answer 1

	A	B	C
count, ni =	5	5	5
mean , x̅ i =	70.200	30.00	44.80
std. dev., si =	3.701	1.581	1.483
sample variances, si^2 =	13.700	2.500	2.200
total sum	351	150	224		725	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	48.33
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	478.151	336.111	12.484
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	2390.756	1680.556	62.422		4133.733
SS(within ) = SSW = Σ(n-1)s² =	54.800	10.000	8.800		73.6000

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 =    2066.8667
  
mean square within groups , MSW = SSW/N-k =    6.1333
  
F-stat = MSB/MSW =    336.9891

anova table
	SS	df	MS	F	p-value
Between:	4133.73	2	2066.87	336.99	0.0000
Within:	73.60	12	6.13
Total:	4207.33	14
					α =	0.05
			conclusion :	p-value<α , reject null hypothesis

THANKS

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