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

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

A) greater than .10.
B) between .05 to .10.
C) between .025 to .05.
D) less than .01.