JC2 H2 Mathematics: Correlation and Regression

Find a missing assessment score

Five students have practice scores $x=1,2,3,4,5$ and assessment scores $y=2,5,k,8,7$. The regression line of y on x is $y=1.1+1.3x$. Find k.

The mean point $(\bar x,\bar y)$ lies on the regression line:

The line is a model for the whole dataset. It need not pass through the observed point $(3,k)$; substituting x = 3 and claiming k = 5 would be wrong.

Estimate in each direction

Use the completed data $(1,2),(2,5),(3,3),(4,8),(5,7)$. Find both regression lines. Estimate y when x = 4, then estimate x when y = 6, treating neither score as a controlled independent variable.

The centred sums are:

For y on x, minimise vertical y residuals:

For x on y, minimise residuals in the x direction instead:

Rearranging the y-on-x line would give $(6-1.1)/1.3\approx3.77$, a different estimate. Choose the regression direction from the prediction required when neither variable is clearly controlled. If the experiment identifies a controlled predictor, keep the fitted response-on-predictor model and interpret any reverse use within that model.

Both regression slopes are positive, agreeing with the sign of r. Their product is $1.3(0.5)=0.65=r^2$, not 1.

Calculate the minimum sum of squared residuals

For the same five score pairs, calculate the residuals from $\widehat y=1.1+1.3x$. Let $S=\sum[y_i-(a+bx_i)]^2$ for any line y = a + bx. Find a lower bound for S.

Therefore $S\geq9.10$, with equality for the least-squares line. It minimises the sum of squared vertical residuals. It does not make each residual zero.

Every line through $(3,5)$ has residual sum zero, because $a=5-3b$. Only one of those lines minimises the sum of squared residuals:

Growth that slows down

A plant’s heights h cm at ages t weeks are shown below. Compare $h=c+d\sqrt t$ with $h=a+b\ln t$, then estimate its height at six weeks.

Height increases at a decreasing rate. A straight line in t would impose a constant increase; a positive $t^2$ term would impose an increasing rate. Both proposed transformed models have the observed slowing shape, so compare their transformed correlations.

The log model fits these data better. Put $u=\ln t$, regress h on u, then restore t:

For six weeks,

Six weeks lies in the data range 1–16 weeks and the correlation between h and $\ln t$ is very strong. These support interpolation, assuming the same growth conditions. In the long run the model is unsuitable for a plant with a limiting height: $\ln t$ grows without bound, although increasingly slowly.

The relevant r is between h and $\ln t$, not between h and t. A near-perfect transformed correlation does not prove that the model remains valid outside the observed ages.

Changing units in a transformed model

Rewrite the fitted plant model using age v in days, where v = 7t. Explain the effect on its correlation coefficient.

The transformed predictor $\ln v=\ln t+\ln7$ differs only by a constant, so the correlation between h and the transformed predictor is unchanged. The intercept changes; the slope in front of the logarithm stays the same.

For ordinary variables, positive rescaling or adding a constant leaves r unchanged. Multiplying one variable by a negative constant reverses its sign. Replacing a variable with its square, reciprocal or logarithm can change r because that is not generally a linear change of the original variable.

Predict renovation time, then minimise estimated wages

The observed working days y for crews of x workers are shown below. A model $y=a+b/x^2$, with a, b positive, is proposed. Fit it and find the crew size from 4 to 16 workers that minimises total wages at $120 per worker per day.

As the crew grows, completion time decreases more slowly toward a positive lower limit. A reciprocal-square model allows that limit; a decreasing logarithmic model would eventually predict negative working days. Regress y on $u=1/x^2$:

The transformed correlation is $r\approx0.999507$. It is positive because y rises with $1/x^2$, even though y decreases with x. For nine workers, the fitted time is about 20.1 days; this is interpolation within 4–16 workers.

Total estimated wages depend on both the crew size and duration:

The derivative is negative below this value and positive above it. Crew size is an integer, so compare the neighbouring integers:

Hire eight workers under this fitted cost model. The estimate assumes all workers are paid for the modelled duration and ignores scheduling or productivity changes not represented in these data. Choosing the smallest predicted duration alone would not minimise wages.

An exponential model transforms the response

A decaying signal has measured strengths 101, 79, 65 and 51 units at times t = 0, 1, 2 and 3 s. Fit $y=Ae^{bt}$ using linear regression on the logarithms, then predict its strength at 4 s.

Since all strengths are positive, take logarithms of the response:

Regress $\ln y$ on t. The fitted line is

This predicts outside the observed range 0–3 s, so it is extrapolation and depends on the decay law continuing. The log fit minimises squared residuals in $\ln y$, not squared residuals in y. Do not compare those two sums as if they used the same response scale.