ANOVA table

\begin{array}{lrrrr}\hline\text { Source } & S S & d f & M S & F &p \text {-value }\\\hline \text { Treatment } & 6.000 & 3 & 1.9998 & 18.85&3.4E-05\\\hline\text { Error } & 1.486 & 14 & 0.1061 & \\\hline\text { Total } & 7.485 & 17 & &\end{array}

Post hoc analysis
Tukey simultaneous comparison t-values

(

d.f.

=14)

\begin{array}{ccccc}&\text { Brand 3 } &nbsp; \text { Brand 2 }&nbsp; \text { Brand 4 } &nbsp;\text { Brand 1 } \\&\quad1.40 \quad 2.28 \quad 2.58 \quad 2.95\\\begin{array}{ll}\text { Brand 3 } & 1.40 \\\text { Brand 2 } & 2.28 \\\text { Brand 4 } & 2.58 \\\text { Brand 1 } & 2.95\end{array}&\begin{array}{|l|l|l|l|}\hline & & & \\\hline 4.27 & & & \\\hline 5.38 & 1.35 & & \\\hline 7.09 & 3.07 & 1.63 &\quad\quad \\\hline\end{array}\end{array}

The Excel/Mega-Stat output given above summarizes the results of a one-way analysis of variance in an attempt to compare the performance characteristics of four brands of vacuum cleaners. The response variable is the amount of time it takes to clean a specific size room with a specific amount of dirt.
-At a significance level of 0.05,we would:
A)Not be able to reject the null hypothesis of equal treatment group means.
B)Reject the null hypothesis of equal treatment group means.
C)Reject or fail to reject depending on the value of the t statistic.
D)Not be able to decide whether or not to reject the null hypothesis due to insufficient information.
E)Reject the significance level of the error term.

Consider the Following Partial Analysis of Variance Table from a Randomized

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