91影视

The table representing the number of enrolled students (in thousands) and the number of burglaries for randomly selected large colleges in recent years is provided.

\(r = 0.499\)

The formula for the correlation coefficient is

\(r = \frac{{n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{x^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\).

The measure is computed as 0.499 using the original data set given in Exercise 1.

All the values for variable x are multiplied by a fixed constant of 1000.

Change observation x as 1000x in the formula.

\(\begin{array}{c}{r_{new}} = \frac{{n\left( {\sum {1000xy} } \right)--\left( {\sum {1000x} } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{{\left( {1000x} \right)}^2}} } \right)--{{\left( {\sum {1000x} } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{1000\left( {n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)} \right)}}{{1000\sqrt {\left( {\left( {n\sum {{{\left( x \right)}^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = \frac{{n\left( {\sum {xy} } \right)--\left( {\sum x } \right)\left( {\sum y } \right)}}{{\sqrt {\left( {\left( {n\sum {{{\left( x \right)}^2}} } \right)--{{\left( {\sum x } \right)}^2}} \right)\left( {\left( {n\sum {{y^2}} } \right)--{{\left( {\sum y } \right)}^2}} \right)} }}\\ = r\end{array}\)

Therefore, there will be no change in the value of the correlation coefficient.