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Consider the Following ANOVA Table The Null Hypothesis Is to Be Tested at the 5

question 30

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Consider the following ANOVA table.  Source  of Variation  Sum  of Squares  Degrees  of Freed om  Mean  Square F Between Treatments 2073.64 Between Blocks 600051200 Error 20288 Total 29\begin{array}{llll}\begin{array}{l}\text { Source } \\\text { of Variation }\end{array} & \begin{array}{l}\text { Sum } \\\text { of Squares }\end{array} & \begin{array}{l}\text { Degrees } \\\text { of Freed om }\end{array} & \begin{array}{l}\text { Mean } \\\text { Square }\end{array}&F \\\text { Between Treatments } & 2073.6 & 4 & \\\text { Between Blocks } & 6000 & 5 & 1200 \\\text { Error } & & 20 & 288 \\\text { Total } & & 29 &\end{array}
The null hypothesis is to be tested at the 5% level of significance.The p-value is


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The systematic, theoretical analysis of the methods applied to a field of study, encompassing the principles behind the choice and use of particular techniques.

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