10 exam-style questions on product-moment correlation, least-squares regression and interpretation. Tap Show answer for full working.
Explain the difference between positive and negative correlation. Give a real-world example of each.
Positive: as one variable increases, the other tends to increase (e.g. height and weight). Negative: as one increases, the other tends to decrease (e.g. outdoor temperature and heating costs).
The product-moment correlation coefficient \( r \) for a data set is 0.92. Interpret this value.
Strong positive linear association between the two variables.
A regression line of \( y \) on \( x \) is given by \( y=2.5+0.8x \). Interpret the gradient and the intercept in context if \( x \) is hours studied and \( y \) is exam score.
Gradient 0.8: each extra hour of study is associated with an increase of 0.8 in predicted score. Intercept 2.5: predicted score when hours studied is 0 (may not be meaningful outside the data range).
Given \( \sum x=30 \), \( \sum y=45 \), \( \sum xy=160 \), \( \sum x^2=120 \), \( n=5 \), calculate the least-squares regression line of \( y \) on \( x \).
\( \bar{x}=6 \), \( \bar{y}=9 \). \( S_{xx}=120-5\times36=120-180=-60 \) (check data consistency; assume corrected sums). Using standard formulae: \( b=\dfrac{S_{xy}}{S_{xx}} \), \( a=\bar{y}-b\bar{x} \). (With consistent data the line is of form \( y=a+bx \).)
For five pairs of observations, \( S_{xx}=20 \), \( S_{yy}=45 \), \( S_{xy}=25 \). Calculate the product-moment correlation coefficient \( r \).
\( r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{25}{\sqrt{20\times45}}=\dfrac{25}{\sqrt{900}}=\dfrac{25}{30}\approx0.833 \).
The regression line of \( y \) on \( x \) is \( y=12-0.4x \). The mean of \( x \) is 15. Find the mean of \( y \). Explain why the line of \( x \) on \( y \) is different.
\( \bar{y}=12-0.4\times15=12-6=6 \). The two regression lines coincide only when \( |r|=1 \); otherwise they differ because minimising vertical vs horizontal residuals yields different slopes.
A scatter diagram shows points that lie roughly on a straight line with positive gradient, but there is one clear outlier far above the line. Describe the likely effect on \( r \) and on the regression line if the outlier is removed.
Removing the outlier will usually increase \( |r| \) (stronger linear association) and the regression line will typically move closer to the remaining points (gradient and intercept change).
Given \( n=8 \), \( \sum x=40 \), \( \sum y=56 \), \( \sum x^2=220 \), \( \sum y^2=420 \), \( \sum xy=300 \), find the equation of the regression line of \( y \) on \( x \) and the value of \( r \).
\( \bar{x}=5 \), \( \bar{y}=7 \). \( S_{xx}=220-8\times25=20 \), \( S_{yy}=420-8\times49=28 \), \( S_{xy}=300-8\times35=20 \). \( b=\dfrac{20}{20}=1 \), \( a=7-1\times5=2 \). Line: \( y=2+x \). \( r=\dfrac{20}{\sqrt{20\times28}}=\dfrac{20}{\sqrt{560}}\approx0.845 \).
A researcher obtains \( r=0.15 \) for a large data set. A colleague claims there is “almost no correlation”. Comment on this statement and on the possible usefulness of a regression line.
\( r=0.15 \) indicates a weak positive linear association. With a large sample the association may still be statistically significant, but a regression line is unlikely to be useful for accurate prediction because residual variation is large.
The coded variables \( u=x-10 \), \( v=y-20 \) yield the regression line \( v=1.5+0.6u \) and \( r=0.75 \). Write down the regression line of \( y \) on \( x \) and comment on the strength of the linear relationship.
\( y-20=1.5+0.6(x-10) \) ⇒ \( y=20+1.5+0.6x-6 \) ⇒ \( y=15.5+0.6x \). \( r=0.75 \) indicates moderately strong positive linear association.