Question

The correlation between the height and weight of children aged 6 to 9 is found to be about r = 0.8. Suppose we use the height x of a child to predict the weight y of the child. We conclude that:
a. the least-squares regression line of y on x would have a slope of 0.8. about 80% of the time, age will accurately predict weight.
b. height is generally 80% of a child’s weight.
c. the fraction of variation in weights explained by the least-squares regression line of weight on height is 0.64.

Answer 1

Answer:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The value of r is always between $-1 \leq r \leq 1$

And we have another measure related to the correlation coefficient called the R square and this value measures the % of variance explained between the two variables of interest, and for this case we have:

$r^2 = 0.8^2 = 0.64$

So then the best conclusion for this case would be:

c. the fraction of variation in weights explained by the least-squares regression line of weight on height is 0.64.

Step-by-step explanation:

For this case we know that the correlation between the height and weight of children aged 6 to 9 is found to be about r = 0.8. And we know that we use the height x of a child to predict the weight y of the child

We need to rememeber that the correlation is a measure of dispersion of the data and is given by this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

The value of r is always between $-1 \leq r \leq 1$

And we have another measure related to the correlation coefficient called the R square and this value measures the % of variance explained between the two variables of interest, and for this case we have:

$r^2 = 0.8^2 = 0.64$

So then the best conclusion for this case would be:

c. the fraction of variation in weights explained by the least-squares regression line of weight on height is 0.64.