Question

w y 315 (VIS. JUMIM NO 3ldal) Sep2012 sec-2/31270 Question 3 Not yet answered Marked out of 3.00 p Flag question 1. Refer to the following ANOVA results from Excel, Test if there is a differences between means at - .05. ANOVA af MS F Source of Variation Between Groups Within Groups SS 75660.67 26715.33 Fcrit 5.143253 2 6 Total 102376 3 Paragraph - BI EE PRED

Answer 1

Solution:

Given:

ANOVA df MSF Fcrit Source of Variation Between Groups Within Groups SS 75660.67 26715.33 5.143253 Total 102376

We have to test if there is a differences between means at

= 0,05

Thus hypothesis are:

H₀: there is no differences between means, that is all three populations have equal means.

Vs

H₁: there is differences between means, that is at least one of the populations have different mean.

Thus we need to find F test statistic value.

F = MS_between / MS_within

where

MS_between = SS_between / df_between

MS_between =  75660.67 / 2

MS_between =  37830.335

and

MS_within = SS_within / df_within

MS_within = 26715.33 / 6

MS_within =  4452.555

thus

F = MS_between / MS_within

F =  37830.335 /  4452.555

F = 8.50

and we have F critical value = 5.143253

Decision Rule:

Reject null hypothesis H0, if F test statistic value  > F  critical value = 5.143253 , otherwise we fail to reject H0.

Since F test statistic value =  F = 8.50 > F  critical value = 5.143253 , thus we reject null hypothesis H0.

Conclusion:

At 0.05 level of significance , we have sufficient evidence to conclude that there is a differences between means.