A company has five branches and want to test if the mean employee's salaries are equal. A sample of 3 employees from each location was selected. You are given the ANOVA table below: Source of Sum of squares Degrees of Мean F variation freedom square Treatments 36,648 9,162 Y Error A C Total 57,095 D Using the critical value approach what is your conclusion at the %5 significance level (E. =3.48) : Select one: O a. F = 4.48 and we fail to reject the null hypothesis O b. F= 2.50 and we fail to reject the null hypothesis n c.F = 4.48 and we reject the null hypothesis O d. F = 2.50 and reject the null hypothesis A O Ps Lr JUN étv 12

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A company has five branches and want to test if the mean employee's salaries are equal. A sample of
3 employees from each location was selected. You are given the ANOVA table below:
Source of
Sum of squares
Degrees of
Мean
F
variation
freedom
square
Treatments
36,648
B
9,162
Y
Error
A
Total
57,095
Using the critical value approach what is your conclusion at the %5 significance level (Fa
=3.48) :
Select one:
n a. F = 4,48 and we fail to reject the null hypothesis
b. F= 2.50 and we fail to reject the null hypothesis
O c. F = 4,48 and we reject the null hypothesis
O d. F = 2.50 and reject the null hypothesis
11
JUN
12
étv
Transcribed Image Text:A company has five branches and want to test if the mean employee's salaries are equal. A sample of 3 employees from each location was selected. You are given the ANOVA table below: Source of Sum of squares Degrees of Мean F variation freedom square Treatments 36,648 B 9,162 Y Error A Total 57,095 Using the critical value approach what is your conclusion at the %5 significance level (Fa =3.48) : Select one: n a. F = 4,48 and we fail to reject the null hypothesis b. F= 2.50 and we fail to reject the null hypothesis O c. F = 4,48 and we reject the null hypothesis O d. F = 2.50 and reject the null hypothesis 11 JUN 12 étv
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