Question

2. For the summary table for one-way ANOVA shown on the answers sheet in cells D79 to G81:
	a. fill in the missing items (highlighted in yellow)
	b. identify the null and alternative hypotheses
	c. use the 0.025 level of significance to find the critical value of F and to reach a conclusion regarding the null hypothesis.

2.	12	SS	df	MS	F
a.	Treatment	665.00	4
	Error		60
	Total	3736.30
b.	Define H0 :
	Define H1 :
c.	Fcrit
	Reject H0or not?

Answer 1

Answer 2

(A) SS error = SS(total) - SS(treatment)

where SS(total) = 3736.3 and SS(treatment) = 665

this implies, SS(error) = 3736.3 - 665 = 3071.3

df(total) = df(error) + df(treatment)

where df(error) = 60 and df(treatment) = 4

this implies, df(total) = 60+4 = 64

MS(treatment) = SS(treatment)/df(treatment)

where SS(treatment) = 665 and df(treatment) = 4

this implies, MS(treatment) = 665/4 = 166.25

MS(error) = SS(error)/df(error)

where SS(error) = 3071.3 and df(error) = 60

this implies, MS(error) = 3071.3/60 = 51.19

F statistics = MS(treatment)/MS(error)

where MS(treatment) = 166.25 and MS(error) = 51.19

this implies, F statistics = 166.25/51.19 = 3.25

(B) We use one way ANOVA to test if there is at least one pair of means which are different from each other. So, null hypothesis is that there is no difference or all means are equal and alternate hypothesis is that there is at least one significantly different pair of means.

(C) F critical with df1= 4 and df2 = 60

using F distribution table, for 0.025 level of significance, check 60 under "degree of freedom in denominator column" and 4 under the "degree of freedom in numerator row"

Select the intersecting cell, it gives us

F(4,60) = 3.01

Reject Ho because the F statistics is greater than F critical value, which means result is significant.