Question

1. State the null and alternative hypotheses. 2. Verify that the conditions are met 3. What is the P-value? 4. What is the Test Statistic? 5. State the conclusion in context

Answer 1

a)

Ho: mean reading score is same for all four meyhods

Ha: at least one mean is different

b)

conditios are met for anova test as samples taken are random and are normally distributed

c)

p-value

0.2501

d)

F

1.45

one way anova

	method 1	method 2	method 3	method4
count, ni =	8	8	7	7
mean , x̅ i =	67.500	71.63	80.86	71.14
std. dev., si =	15.866	12.883	10.761	10.01
sample variances, si^2 =	251.714	165.982	115.810	100.143
total sum	540	573	566	498	2177	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				72.57
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	25.671	0.887	68.732	2.027
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	205.369	7.094	481.124	14.191	707.7774
SS(within ) = SSW = Σ(n-1)s² =	1762.000	1161.875	694.857	600.857	4219.5893

no. of treatment , k =   4
df between = k-1 =    3
N = Σn =   30
df within = N-k =   26
  
mean square between groups , MSB = SSB/k-1 =    235.9258
  
mean square within groups , MSW = SSW/N-k =    162.2919
  
F-stat = MSB/MSW =    1.4537
P value =   0.2501
  
SST=Σ( x - x̅̅ )² =    3773.958

anova table	0.143642117
	SS	df	MS	F	p-value	F-critical
Between:	707.78	3	235.93	1.45	0.2501	2.98
Within:	4219.59	26	162.29
Total:	4927.37	29
					α =	0.05
			conclusion :	p-value>α , do not reject null hypothesis

e)

conclusion: there is enough evidence to say that mean of reading scores by all methods are same