Question

please show all work and explain steps. thank you!

The following table shows the results of four samples from four different normal populations. Assume that it is suitable for the ANOVA Sample Sample Size Sample Mean Sample Variance A 6 54 95 B 6 64 30 72 75 D 62 55 6 The sample grand mean is 63. We want to test whether there are differences in the four population means at the 1% level of significance. What is the value of error mean square

Answer 1

treatment	G1	G2	G3	L4
count, ni =	6	6	6	6
mean , x̅ i =	54.000	64.00	72.00	62.00
std. dev., si =	9.7	5.5	8.7	7.4
sample variances, si^2 =	95.000	30.000	75.000	55.000
total sum	324	384	432	372			1512	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	63.00
( x̅ - x̅̅ )²	81.000	1.000	81.000	1.000
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	486.000	6.000	486.000	6.000			984
SS(within ) = SSW = Σ(n-1)s² =	475.000	150.000	375.000	275.000			1275.0000

no. of treatment , k =   4  
df between = k-1 =    3  
N = Σn =   24  
df within = N-k =   20  
      
mean square between groups , MSB = SSB/k-1 =    984/3=   328.0000

mean square within groups , MSE = SSW/N-k =    1275/20=   63.7500

anova table
	SS	df	MS	F	p-value
Between:	984.0	3	328.0	5.1	0.008
ERROR	1275.0	20	63.75
Total:	2259.0	23

MSE = 63.75

α =    0.01  
Decision:   p-value<α , reject null hypothesis    

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

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