Question

Given the following information obtained from four normally distributed populations, construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, and "F" to 3 decimal places.)

SST = 65.97; SSTR = 13.68; c = 4; n₁ = n₂ = n₃ = n₄ = 15

ANOVA
Source of Variation	ss	df	ms	f	p value
Between Groups					.004
Within Groups
Total

b. At the 5% significance level, what is the conclusion to the ANOVA test of mean differences?

• Reject H₀; we can conclude that some means differ.

• Do not reject H₀; we cannot conclude that some means differ.

• Do not reject H₀; we can conclude that some means differ.

• Reject H₀; we cannot conclude that some means differ.

Answer 1

Given the following information obtained from four normally distributed populations, construct an ANOVA table

SST = 65.97; SSTR = 13.68; c = 4; n₁ = n₂ = n₃ = n₄ = 15

Here we need to test weather some means differ or not .

Now we are incomplete ANOVA , but to test this we do not need to complete whole table as we are given P-value which is equal to P-Value = 0.004

{ Note that Since P-Value is provided , Complete ANOVA table is not required to give conclusion , but if in case it is required then solved ANOVA is provided below . If there is any doubt regarding completed ANOVA , can ask for that in comment box }

Calculation

SSE = SST - SSTR                   { Sum of square due to Error i.e Within-group variation }

         = 65.97 - 13.68

         = 52.29

Now , ms = ss / df

So after calculation

Completed Anova Table is as follow

ANOVA
Source of Variation	ss	df	ms	f	p value
Between Groups	13.68	3	4.56	4.883534	0.004
Within Groups	52.29	56	0.93375
Total	65.97	59

To Test

H0 : All means do not differ significantly from each other i.e all are same

vs

H1 : Some means ( atleast one mean ) differs significantly with each other

Test Statistics F :-

F = MSTR / MSE

   = 4.56/ 0.93375

F = 4.883534                               ( calculated using given information about SST ,SSTR )

We are also given P-value correspond to calculated F-value

i.e P-Value = 0.004                        ( Also given in ANOVA )

Rejection criteria : - We reject null hypothesis is P-Value is less than given significance level .

Now we are given 5% ( i.e 0.05 ) significance level,

thus   P-Value = 0.004   < 0.05

Since P-Value is less than 0.05 ( given significance ) , we reject null hypothesis .

Conclusion :- Since we reject null hypothesis H0 , we concluded that some means differ.

Thus correct option is

Reject H₀; we can conclude that some means differ.