Question

consider the following table:

	ss	df	ms	f
error			626.7
total	12825.12	18

step 8 of 8: is f significant at 0.05?
answer
yes no

Explanation:

Step1: Calculate degrees of freedom for error

The total degrees of freedom ($DF_{total}$) is 18 and the degrees of freedom among treatments ($DF_{among}$) is 6. So, the degrees of freedom for error ($DF_{error}$) is $DF_{total}-DF_{among}=18 - 6=12$.

Step2: Calculate sum - of - squares for error

The total sum of squares ($SS_{total}$) is 12825.12. Let the sum of squares among treatments be $SS_{among}$. We know that $SS_{total}=SS_{among}+SS_{error}$. First, we find the mean square among treatments ($MS_{among}$) which is given as 884.12 and $DF_{among} = 6$. So, $SS_{among}=MS_{among}\times DF_{among}=884.12\times6 = 5304.72$. Then $SS_{error}=SS_{total}-SS_{among}=12825.12 - 5304.72=7520.4$.

Step3: Calculate mean square for error

The mean square for error ($MS_{error}$) is given as 626.7.

Step4: Calculate the F - statistic

The F - statistic is calculated as $F=\frac{MS_{among}}{MS_{error}}$. Since $MS_{among}=884.12$ and $MS_{error}=626.7$, $F=\frac{884.12}{626.7}\approx1.41$.

Step5: Find the critical F - value

We have $\alpha = 0.05$, $DF_{numerator}=6$ (degrees of freedom among treatments) and $DF_{denominator}=12$ (degrees of freedom for error). Looking up in the F - distribution table, the critical F - value $F_{critical}(6,12)=3.009$.

Step6: Compare F - statistic and critical F - value

Since the calculated F - statistic ($F\approx1.41$) is less than the critical F - value ($F_{critical}=3.009$).

Answer:

No