Regression with Mean, Standard Deviation, Range and Correlation A research team collected data on students in a statistics course. Their dependent variable
was the student’s score on the final examination, which ranged from 200 to 800 points.
The observed average final examination score was 467, with an observed standard
deviation of 107.2 (the divisor in the estimated variance was 249). Their independent variable was the score on the first examination in the course, which also ranged from 200 to 800.  The average was 424, with an observed standard deviation of 81.7. The correlation
coefficient between the first examination score and the final examination score was 0.58.
a. Report the analysis of variance table and the test of the null hypothesis that the
slope of the regression line is zero against the alternative that it is not. Use the 0.10,
0.05, and 0.01 levels of significance.
I'm having trouble getting this problem started.  To create an ANOVA table for regression I'm required to determine the SSR and SSE and thus MSR and MSE and finally the Fstat.  Doesn't this require the individual values of y and x?  I'm unsure of how to determine this using only mean, standard deviation, correlation and range.
 A: This CANNOT be done with only the means, standard deviations, correlation, and range.  The range is not needed, but one other thing is needed that was mentioned in the first paragraph but not in the question in the last paragraph: the sample size, which is $250$.
The equation of the least-squares line is:
$$
\frac{y - 478}{107.2} = 0.58\left(\frac{x-424}{81.7}\right).
$$
The sum of squares of deviations from the average $y$-value is $249\cdot 107.2^2$.  The proportion of that that is explained by variability in $x$-values is $0.58^2$, so the explained sum of squares is $0.58^2\cdot249\cdot 107.2^2\approx 962597.889$.  The unexplained sum of squares, which is the sum of squares of residuals, is $(1-0.58^2)\cdot249\cdot 107.2^2\approx 1898870.271$.  So we have this ANOVA table:
$$
\begin{array}{c|c|c|c|c}
\text{source of variability} & \text{sum of squares} & \text{degrees of freedom} & \text{mean square} & F & p \\
\hline
\text{first test} & 962597.889 & 1 & 962597.889 & 125.719 & \approx 0 \\
\text{error} & 1898870.271 & 248 & 7656.734964
\end{array}
$$
Finding the value of $p$ is generally the only part that exceeds elementary arithmetic.  But in this case the $F$ value is so extreme that it's trivial to say that for all practical purposes, $p=0$.
