Question

Need to perform an anova test

om C1-T C2 C3 Field Cost SocialSci... 77 1 2. NaturalS... 231 3 Naturals... 189 SocialSci... 85 4 5 Naturals... 113 6 Humanities 132 7 SocialSci... 199 8 SocialSci... 161 9 Humanities 213 10 Arts 131 11 Arts 133 12 Arts 118 13 SocialSci... 197 14 SocialSci... 95 15 SocialSci... 73 16 SocialSci... 115 17 Naturals... 110 18 SocialSci... 87 19 Humanities 223 20 Humanities 74

Answer 1

treatment	SS	NS	HUM	ARTS
count, ni =	10	10	10	10
mean , x̅ i =	118.300	170.80	120.30	94.60
std. dev., si =	48.9	48.5	58.1	44.9
sample variances, si^2 =	2391.122	2351.511	3380.900	2020.267
total sum	1183	1708	1203	946			5040	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	126.00
( x̅ - x̅̅ )²	59.290	2007.040	32.490	985.960
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	592.900	20070.400	324.900	9859.600			30847.8
SS(within ) = SSW = Σ(n-1)s² =	21520.100	21163.600	30428.100	18182.400			91294.2000

no. of treatment , k =   4  
df between = k-1 =    3  
N = Σn =   40  
df within = N-k =   36  
      
mean square between groups , MSB = SSB/k-1 =    30847.8/3=   10282.6000
mean square within groups , MSW = SSW/N-k =    91294.2/36=   2535.9500
      
F-stat = MSB/MSW =    10282.6/2535.95=   4.05
      
P value =   0.0140  

anova table
	SS	df	MS	F	p-value	F-critical
Between:	30847.8	3	10282.6	4.05	0.014	2.866
Within:	91294.2	36	2536.0
Total:	122142.0	39

α =    0.05
Decision:   p-value<α , reject null hypothesis

conclusion :            there is enough evidence of significant mean difference among FOUR treatments  

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

Please let me know in case of any doubt.

Thanks in advance!

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