Some of the examples are explained below to have a better understanding of the topic:Example 1: For the female athlete strength study, x = Number of 60 pound bench presses and y = Maximum bench press had
x : Mean = 11.0, standard deviation = 7 . 1
x : Mean = 79.9, standard deviation = 13.3
Regression Equation $\hat{y}$ = 63.5 + 1.49x
Find the correlation r between these two variables.
Solution: Slope of the regression equation is b = 1.49. Since s$_{x}$ = 7.1 and s$_{y}$ = 13.3
r = b(
$\frac{s_{x}}{s_{y}}$)
= 1.49 (
$\frac{7.1}{13.3}$)
= 0.80
The variables have a strong positive association.
Example 2: For the 64 female college athletes, the ANOVA table for the multiple regression
predicting y = Weight using x$_{1}$ = height, x$_{2}$ = % body fat, and x$_{3}$ = Age shows
Source

Degrees of freedom 
Sum of Squares 
Mean sum of Squares

F 
P 
Regression 
3 
12407.9 
4136.0 
40.48 
0.0000 
Residual Error  60  6131.0  102.2   
1) State and interpret the null hypothesis tested in this table.
2) From the F table, which F value would have a P value of 0.05 for these data?
3) Report the observed test statistic and P value. Interpret the P value and make a decision for a 0.05 significance level.
Solution: Since there are 3 explanatory variables, the null hypothesis is H$_{0}$ : $\beta_{1}$ = $\beta_{2}$ = $\beta_{3}$ = 0. It states that weight is independent of height, % body fat, and age.
2) In the degrees of freedom column, the ANOVA table shows df$_{1}$ = 3 and df$_{2}$ = 60. FRom the F tables we see that the F value with right tail probability of 0.05 is 2. 76.
3) From the ANOVA table, the observed F test statistic value is 40.5. Since this is well above 2.76, the p value is less than 0.05. The ANOVA table reports p value = 0.0000. If H$_{0}$ were true, it would be extremely unusual to get such a large F test statistic. We can reject H$_{0}$ at the 0.05 significance level.
In summary, we conclude that atleast one predictor has an effect on weight.