Question

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Exercise 1 stated that ris found to be 0.499. Does that value change if the actual enrollment values of 53,000, 28,000, 27,000, 36,000, and 42,000 are used instead of 53, 28, 27, 36, and 42?

Short answer

The value of the correlation coefficient remains the same at 0.499.

Step-by-step solution

Given information

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.

\(r = 0.499\)

State the formula for the correlation coefficient

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.

Discuss the change in the measure

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.