A2 June 2022 Q1

EdexcelCurrent spec7 marksCorrelationLinear Regression

1. Kwame is investigating a possible relationship between average March temperature, \(t\,{}^\circ\mathrm{C}\), and tea yield, \(y\) kg/hectare, for tea grown in a particular location.
He uses 30 years of past data to produce the following summary statistics for a linear regression model, with tea yield as the dependent variable.

\[\text{Residual Sum of Squares (RSS)} = 1\,666\,567 \qquad \mathrm{S}_{tt} = 52.0 \qquad \mathrm{S}_{yy} = 1\,774\,155\]\[\text{least squares regression line:} \qquad \text{gradient} = 45.5 \qquad y\text{-intercept} = 2080\]
(a) Use the regression model to predict the tea yield for an average March temperature of \(20\,{}^\circ\mathrm{C}\) (1)

He also produces the following residual plot for the data.

Residual plot: residual (from –600 to 800) against temperature (t °C, from 17 to 23) for the 30 years; residuals are mostly negative between about 20 and 21.5 °C and all positive and rising above 21.5 °C
(b) Explain what you understand by the term residual. (1)
(c) Calculate the product moment correlation coefficient between \(t\) and \(y\) (2)
(d) Explain why the linear model may not be a good fit for the data
(i) with reference to your answer to part (c)
(ii) with reference to the residual plot.
(2)

Kwame also collects data on total March rainfall, \(w\) mm, for each of these 30 years.

For a linear regression model of \(w\) on \(t\) the following summary statistic is found.

\[\text{Residual Sum of Squares (RSS)} = 86\,754\]

Kwame concludes that since this model has a smaller RSS, there must be a stronger linear relationship between \(w\) and \(t\) than between \(y\) and \(t\) (where RSS = 1 666 567)

(e) State, giving a reason, whether or not you agree with the reasoning that led to Kwame’s conclusion. (1)
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