Lecture notes. Written from the material of GTPB Psls20, licensed CC BY 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.
9. Simple Linear Regression¶
What this covers¶
This chapter builds simple linear regression from the ground up: how a straight-line model for a conditional mean is fit by least squares, what extra assumptions turn that fit into something you can attach a confidence interval or p-value to, how to check those assumptions and what to do when they fail, and how the same machinery gives prediction intervals, the sum-of-squares decomposition behind \(R^2\) and the F-test, and the two-sample t-test as a special case. It assumes you already have expectation and conditional expectation, the normal and \(t\) distributions, and confidence intervals and hypothesis tests for a single mean. The running example is a set of 32 breast-cancer patients whose tumour biopsies were profiled for gene expression.
The breast cancer dataset¶
The data are a subset of a published study (https://doi.org/10.1093/jnci/djj052): 32 patients with an estrogen-receptor-positive tumour who were treated with tamoxifen chemotherapy. For each patient the dataset records the histological grade of the tumour (grade 1 vs. grade 3), lymph node status (node = 0 if unaffected, 1 if lymph nodes were affected and removed), tumour size, and the expression level of two genes, ESR1 and S100A8, measured in the tumour biopsy by microarray.
The biology motivates the question. ESR1 (the estrogen receptor gene) is expressed in roughly 75% of breast tumours; a tumour that expresses it responds to hormone therapy, and tamoxifen works by interacting with the receptor and modulating gene expression. Proteins of the S100 family are often dysregulated in cancer, and S100A8 in particular represses the immune system in the tumour and creates an inflammatory environment that promotes tumour growth. So it is natural to ask whether ESR1 and S100A8 expression are associated in this sample — and, if so, how strongly.
For didactic reasons the first pass through the analysis removes three outliers in the S100A8 measurements and works with this trimmed subset; the chapter returns later to how to handle the full data properly, without throwing points away.
The linear regression model¶
Linear regression is a way to assess the association between two variables \((X_i, Y_i)\), measured on each of \(n\) subjects \(i = 1, \dots, n\) — here \(X\) is ESR1 expression and \(Y\) is S100A8 expression. For a fixed value of \(X\), \(Y\) does not have to take the same value every time: there is signal and there is noise, $\(Y_i = g(X_i) + \epsilon_i,\)$ where \(g(x)\) is defined as the expected outcome for subjects with \(X_i = x\), $\(E[Y_i \mid X_i = x] = g(x).\)$ It follows that \(\epsilon_i\) averages to zero among subjects who share the same \(X_i\): \(E[\epsilon_i \mid X_i] = 0\).
To get results that are both accurate and interpretable, \(g(x)\) is usually chosen to be linear in an unknown intercept and slope: $\(E(Y \mid X = x) = \beta_0 + \beta_1 x.\)$ This is a genuine assumption about the joint distribution of \(X\) and \(Y\), and it can be wrong. Its payoff is that it is an efficient way to analyse the data, because every observation contributes to learning the single line rather than to a separate estimate at each value of \(x\).
The model supports two uses. For prediction, once the line is fit, \(Y\) can be forecast from \(X\) via \(E(Y \mid X=x) = \beta_0 + \beta_1 x\). For association, the parameters have a direct biological reading: the intercept is the mean response at \(X=0\), \(E(Y\mid X=0) = \beta_0\), and the slope is the effect of a unit increase in the predictor, $\(E(Y \mid X = x+\delta) - E(Y \mid X = x) = \beta_0 + \beta_1(x+\delta) - \beta_0 - \beta_1 x = \beta_1 \delta.\)$ So \(\beta_1\) is the difference in mean outcome between subjects whose predictor differs by one unit.
Parameter estimation: least squares¶
\(\beta_0\) and \(\beta_1\) are unknown population parameters and have to be estimated from the sample. The natural criterion is to choose the line that fits best: for a given \(x_i\), the point on the line \((x_i, \beta_0+\beta_1 x_i)\) should sit as close as possible to the observed \((x_i, y_i)\). Concretely, choose \(\beta_0, \beta_1\) to minimise the sum of squared vertical distances between the observations and the fitted line, $\(SSE = \sum_{i=1}^n (y_i - \beta_0 - \beta_1 x_i)^2 = \sum_{i=1}^n e_i^2,\)$ where the residuals \(e_i\) are exactly those vertical distances. Minimising \(SSE\) gives closed-form estimators, $\(\hat\beta_1 = \frac{\sum_{i=1}^n (y_i-\bar y)(x_i-\bar x)}{\sum_{i=1}^n (x_i-\bar x)^2} = \frac{\mathrm{cor}(x,y)\, s_y}{s_x}, \qquad \hat\beta_0 = \bar y - \hat\beta_1 \bar x.\)$ Notice that the least-squares slope is proportional to the correlation between response and predictor — regression and correlation are two views of the same fact. Once fitted, the line can be used exactly as before, with hats on the parameters: to predict the response at a given \(x\), or to compare the mean response between two groups that differ by \(\delta\) in the predictor, $\(E[Y\mid X=x+\delta] - E[Y \mid X=x] = \hat\beta_1 \delta.\)$
Fitting lm(S100A8 ~ ESR1, brcaSubset) on the outlier-trimmed data gives a negative \(\hat\beta_1\):
expected S100A8 expression falls as ESR1 expression rises. Two cautions come with the fitted line.
First, extrapolation is dangerous: the linearity assumption can only be checked within the range of
the observed data, so predictions for ESR1 values far outside that range are not justified by
anything the fit has shown. Second — and this is the subject of the next two sections — a fitted
line is not yet evidence of an association; that requires a statistical model on top of the
geometry of least squares.
Statistical inference: what has to be assumed¶
Least squares gives a best-fitting line for any data, whether or not there is a real association. To draw conclusions — how much the estimators would vary from sample to sample, and how they behave under the null hypothesis of no association — requires an explicit statistical model for the distribution of \(Y\) given \(X\). Besides linearity, this needs three further assumptions:
- Independence: the pairs \((X_1,Y_1),\dots,(X_n,Y_n)\) come from \(n\) independent subjects. This is needed to estimate the variance at all.
- Homoscedasticity (equal variances): observations vary with the same spread around the regression line everywhere — \(\mathrm{var}(Y\mid X=x) = \sigma^2\) for every \(x\), where \(\sigma\) is called the residual standard deviation.
- Normality: the residuals \(\epsilon_i\) are normally distributed.
Given independence, homoscedasticity and normality, \(\epsilon_i\) are i.i.d. \(N(0,\sigma^2)\); together with linearity this means $\(Y_i \mid X_i \sim N(\beta_0 + \beta_1 X_i,\ \sigma^2).\)$
Under these assumptions one can show that the parameter estimators are themselves normally distributed, with variances $\(\sigma^2_{\hat\beta_0} = \frac{\sum_{i=1}^n X_i^2}{\sum_{i=1}^n (X_i-\bar X)^2} \times \frac{\sigma^2}{n}, \qquad \sigma^2_{\hat\beta_1} = \frac{\sigma^2}{\sum_{i=1}^n (X_i-\bar X)^2},\)$ $\(\hat\beta_0 \sim N\!\left(\beta_0, \sigma^2_{\hat\beta_0}\right), \qquad \hat\beta_1 \sim N\!\left(\beta_1, \sigma^2_{\hat\beta_1}\right).\)$
The formula for \(\sigma^2_{\hat\beta_1}\) says something useful on its own: the more spread out the \(X\) values are, the larger \(\sum(X_i-\bar X)^2\) is, and the smaller the variance of the slope estimator — a widely spread predictor pins the slope down much more precisely than a tightly clustered one.
The conditional variance \(\sigma^2\) is itself unknown and is estimated by the mean squared error, $\(\hat\sigma^2 = MSE = \frac{\sum_{i=1}^n (y_i - \hat\beta_0 - \hat\beta_1 x_i)^2}{n-2} = \frac{\sum_{i=1}^n e_i^2}{n-2}.\)$ This estimator relies on independence and equal variance, and it divides by \(n-2\) rather than \(n\) because two parameters, \(\beta_0\) and \(\beta_1\), were already estimated from the same data. Plugging \(MSE\) into the formulas above gives standard errors, $\(\mathrm{SE}_{\hat\beta_0} = \sqrt{\frac{\sum_i X_i^2}{\sum_i (X_i-\bar X)^2}\times\frac{MSE}{n}}, \qquad \mathrm{SE}_{\hat\beta_1} = \sqrt{\frac{MSE}{\sum_i(X_i-\bar X)^2}},\)$ and confidence intervals and tests follow the familiar pattern, $\(T = \frac{\hat\beta_k - \beta_k}{\mathrm{SE}(\hat\beta_k)}, \qquad k=0,1.\)$ If all four assumptions hold, \(T\) follows a \(t\)-distribution with \(n-2\) degrees of freedom. If normality fails but independence, linearity and equal variance hold and the sample is large, the central limit theorem rescues the same conclusion approximately.
Breast cancer example. The regression of S100A8 on ESR1 shows a negative association. A 95%
confidence interval for \(\beta_1\),
$\(\left[\hat\beta_1 - t_{n-2,\alpha/2}\,\mathrm{SE}_{\hat\beta_1},\ \hat\beta_1 + t_{n-2,\alpha/2}\,\mathrm{SE}_{\hat\beta_1}\right],\)$
obtained in R with confint(lm1), excludes zero: the negative association is significant at the 5%
level.
The hypothesis test¶
The research question — is there an association between S100A8 and ESR1 expression — translates
into a statement about \(\beta_1\). Under the null hypothesis of no association, \(H_0: \beta_1 = 0\);
under the alternative, \(H_1: \beta_1 \neq 0\). The test statistic is
$\(T = \frac{\hat\beta_1 - 0}{\mathrm{SE}(\hat\beta_1)},\)$
which, under \(H_0\), follows a \(t\)-distribution with \(n-2\) degrees of freedom. For the brca dataset,
summary(lm1) shows the association between S100A8 and ESR1 expression is extremely significant
(\(p \ll 0.001\)). But a significant test is only as trustworthy as the assumptions it rests on — they
have to be checked before the conclusion is believed.
Checking the assumptions¶
Four things need checking: independence is a property of how the data were collected (a matter of design, not something visible in a plot), linearity, homoscedasticity, and normality.
Linearity is assessed with a residual plot: put the fitted values \(\hat\beta_0+\hat\beta_1 x\)
on the horizontal axis and the residuals
$\(e_i = y_i - \hat g(x_i) = y_i - \hat\beta_0 - \hat\beta_1 x_i\)$
on the vertical axis (plot(lm1) in R). A residual plot is especially important once the model has
more than one covariate, where a simple scatterplot of \(Y\) against \(X\) no longer shows the whole
picture directly.
Homoscedasticity is checked the same way: residuals and squared residuals carry information about how variable the response is, so any association between the (squared) residuals and the predictor or the fitted values is a sign of heteroscedasticity.
Normality matters least when the sample is large — the central limit theorem makes the estimators approximately normal even when the observations themselves are not, though exactly how large "large" needs to be depends on the shape and size of the departure from normality. The assumption to check is that \(Y_i \mid X_i \sim N(\beta_0+\beta_1 X_i, \sigma^2)\) — note that a QQ-plot of the raw response \(Y\) is misleading and useless here, because the \(Y_i\) genuinely have different distributions (they have different conditional means at different \(X_i\)) even when the model is exactly right. The assumption is about the residuals, so the QQ-plot has to be of the residuals \(e_i\), not of \(Y\) itself.
A useful calibration when reading a residual plot on a small sample: simulate several batches of pure, independent normal noise against the same predictor and look at what those residual plots look like. With only \(n=32\) observations, apparently patterned scatter can appear in plots of noise that has no structure at all, so a real diagnostic plot is only worth acting on if it shows more than this baseline level of apparent pattern.
When the assumptions fail: transformations¶
Transforming the predictor does not change the distribution of \(Y\) given \(X\) — it is not useful for fixing homoscedasticity or non-normality — but it can fix non-linearity, when normality and homoscedasticity already hold, typically by adding higher-order terms \(X^2, X^3,\dots\), $\(Y_i = \beta_0 + \beta_1 X_i + \beta_2 X_i^2 + \cdots + \epsilon_i.\)$ Transforming the response \(Y\), by contrast, can fix both non-normality and heteroscedasticity — common choices are \(\sqrt{Y}\), \(\log(Y)\), or \(1/Y\).
Breast cancer example. The straightforward linear model has problems on several fronts: heteroscedasticity, a right-skewed departure from normality, predictions that can theoretically go negative even though a gene expression level cannot, and non-linearity. This combination is typical of concentration and intensity measurements, which are often log-normally distributed — normal once log-transformed. In gene expression this is standard practice, using a base-2 logarithm, because then differences on the log scale are exactly fold changes on the original scale. Rather than discarding the three outlying points as in the first pass, fitting $\(\log_2(\text{S100A8}) \sim \log_2(\text{ESR1})\)$ on the full dataset (not the trimmed subset) handles the skew, the impossible negative predictions, and much of the heteroscedasticity in one move — this is the promised "properly dealing with all the data" that the outlier removal at the start of the chapter deferred.
Fitting this model (lm2) gives \(\hat\beta_0 = 23.401\) and \(\hat\beta_1 = -1.615\), so
$\(\log_2\hat\mu = 23.401 - 1.615 \times \log_2(\text{ESR1}).\)$
This coefficient admits three different, complementary readings.
Interpretation 1 (log scale, direct). A patient whose ESR1 expression is one unit higher on the \(\log_2\) scale than another's has, on average, a \(\log_2\) S100A8 expression that is \(1.615\) units lower — this is just the slope read directly off the fitted line, since $\(\log_2\hat\mu_2 - \log_2\hat\mu_1 = -1.615(\log_2\text{ESR1}_2 - \log_2\text{ESR1}_1).\)$
Interpretation 2 (back-transformed, geometric mean). A model fit on the log scale, once back-transformed, describes geometric means, because $\(\frac{1}{n}\sum_{i=1}^n \log x_i = \frac{\log x_1 + \cdots + \log x_n}{n} = \frac{\log(x_1\times\cdots\times x_n)}{n} = \log\left(\sqrt[n]{\textstyle\prod_i x_i}\right).\)$ The population mean \(\mu\) is therefore estimated as a geometric mean, and because the logarithm is monotone, confidence intervals can be back-transformed too. Taking \(2\) to the power of the fitted slope gives \(2^{-1.615} = 0.326\) (equivalently \(2^{1.615} = 3.06\)): a patient whose ESR1 expression is double another's has, on average, an S100A8 expression that is \(0.326\) times as large — about a threefold decrease.
Interpretation 3 (percentage change). For a small relative change in the predictor, the effect translates into an approximate percentage change in the response. A 1% higher ESR1 expression is associated with an S100A8 expression that is \(1.01^{-1.615} = 0.984\), i.e. about \(1.6\%\) lower. More generally, for \(-10 < \beta_1 < 10\), $\(1.01^{\beta_1} - 1 \approx \frac{\beta_1}{100},\)$ which is why "\(\beta_1\) percent" is a serviceable shorthand for moderate slopes on the log scale.
Inference on the mean outcome vs. prediction for a new observation¶
The fitted line can be used to estimate the average response at a given predictor value, or to predict the actual response of one new, as-yet-unobserved subject. These are numerically the same quantity, \(\hat g(x) = \hat\beta_0 + \hat\beta_1 x\), but their sampling distributions are different, and conflating them understates uncertainty about a single future patient.
The mean. \(\hat g(x)\) estimates the conditional mean \(E[Y\mid X=x]\); since the parameter
estimators are unbiased and normal, so is \(\hat g(x)\), with
$\(\mathrm{SE}_{\hat g(x)} = \sqrt{MSE\left\{\frac{1}{n} + \frac{(x-\bar X)^2}{\sum_i(X_i-\bar X)^2}\right\}}, \qquad T = \frac{\hat g(x) - g(x)}{\mathrm{SE}_{\hat g(x)}} \sim t_{n-2}.\)$
In R, predict(model, newdata=..., interval="confidence") returns exactly this — the mean response
and its confidence interval at chosen predictor values.
A new observation. For an independent new observation \(Y^* = g(x) + \epsilon^*\), with
\(\epsilon^* \sim N(0,\sigma^2)\) independent of the sample, the uncertainty has two sources rather
than one: uncertainty about the estimated model parameters, exactly as before, plus the
irreducible extra spread of any individual observation around its own mean. This gives a larger
standard error,
$\(\mathrm{SE}_{\hat Y(x)} = \sqrt{\hat\sigma^2 + \hat\sigma^2_{\hat g(x)}} = \sqrt{MSE\left\{1 + \frac{1}{n} + \frac{(x-\bar X)^2}{\sum_i(X_i-\bar X)^2}\right\}}, \qquad \frac{\hat Y(x) - Y}{\mathrm{SE}_{\hat Y(x)}} \sim t_{n-2}.\)$
A prediction interval, obtained with predict(model, newdata=..., interval="prediction"), is
exactly this wider interval — an improved version of a plain reference interval that additionally
accounts for the fact that the model's own parameters are estimated, not known.
NHANES cholesterol example. Compare a reference interval for direct cholesterol built from a
large sample of women, \(\exp\!\left(\overline{\log(\text{DirectChol})} \pm z_{0.975}\, sd(\log(\text{DirectChol}))\right)\),
against the prediction interval from predict(lmChol, interval="prediction") on the same data. With
a large sample the two intervals are almost identical, because the model parameters are estimated
precisely enough that the extra term contributes little. Repeating the comparison on a subsample of
only 10 women changes this: the prediction interval becomes much wider than the naive reference
interval, especially at the upper limit once back-transformed, because now the uncertainty in the
mean and standard deviation is no longer negligible. In small samples, ignoring parameter
uncertainty and using a plain reference interval understates how uncertain a single new
observation really is.
Sum of squares and the ANOVA table¶
Three sums of squares organise how much of the variability in \(Y\) the model explains.
The total sum of squares, $\(SSTot = \sum_{i=1}^n (Y_i - \bar Y)^2,\)$ estimates the variance of the marginal distribution of the response — ignoring \(X\) altogether — whereas the chapter has otherwise been about the conditional distribution \(f(Y\mid X=x)\), whose variance \(MSE\) estimates.
The regression sum of squares, $\(SSR = \sum_{i=1}^n (\hat Y_i - \bar Y)^2 = \sum_{i=1}^n (\hat g(x_i) - \bar Y)^2,\)$ measures how far the fitted predictions move away from the overall mean. Equivalently, it compares the fitted model \(\hat g(x) = \hat\beta_0+\hat\beta_1 x\) against the model with no predictor at all, \(g(x)=\beta_0\) — in which case \(\beta_0\) would just equal \(\bar Y\) — so \(SSR\) measures the size of the predictor's effect.
The error sum of squares, $\(SSE = \sum_{i=1}^n (Y_i - \hat Y_i)^2 = \sum_{i=1}^n \{Y_i - \hat g(x_i)\}^2,\)$ is exactly the least-squares criterion from before: the smaller \(SSE\), the better the fit.
Squaring and summing the decomposition shown in the figure over every observation gives $\(SSTot = \sum_i (Y_i - \bar Y)^2 = \sum_i (Y_i - \hat Y_i + \hat Y_i - \bar Y)^2 = \sum_i(Y_i-\hat Y_i)^2 + \sum_i(\hat Y_i - \bar Y)^2 = SSE + SSR.\)$ The total variability in the data is partly explained by the predictor (\(SSR\)) and partly left over as residual variability that the regression model cannot account for (\(SSE\)).
The coefficient of determination, $\(R^2 = 1 - \frac{SSE}{SSTot} = \frac{SSR}{SSTot},\)$ is the fraction of the total variability in the sample explained by the model. A large \(R^2\) means the model has the potential to make good predictions (small \(SSE\) relative to the total spread), but \(R^2\) is not a good guide to the p-value of the test \(H_0: \beta_1=0\): the p-value is driven mainly by \(SSE\) and the sample size \(n\), but not by \(SSTot\), while \(R^2\) is driven by \(SSE\) and \(SSTot\) but not by \(n\). A model with a low \(R^2\) can still be the right, useful way to study an association, as long as the association is modelled correctly.
Sums of squares are also the basis of an F-test. Define $\(MSR = \frac{SSR}{1}, \qquad MSE = \frac{SSE}{n-2}, \qquad F = \frac{MSR}{MSE},\)$ where the denominators \(1\) and \(n-2\) are the degrees of freedom of \(SSR\) and \(SSE\) respectively. Under \(H_0: \beta_1 = 0\), \(F = MSR/MSE \sim F_{1,n-2}\), and the F-test is always a two-sided test of exactly the same null hypothesis, \(H_1: \beta_1 \neq 0\), that the t-test above addressed, with p-value \(p = P_0[F \geq f] = 1 - F_F(f; 1, n-2)\). These pieces are usually collected into an ANOVA table:
| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| Regression | degrees of freedom of \(SSR\) | \(SSR\) | \(MSR\) | \(f\)-statistic | p-value |
| Error | degrees of freedom of \(SSE\) | \(SSE\) | \(MSE\) |
produced in R with anova(lm2).
Dummy variables: regression as a two-sample comparison¶
Linear regression can also be used to compare the means of two groups — for instance, whether
average age differs between the brca patients with unaffected and with affected lymph nodes. Define
a dummy variable
$\(x_i = \begin{cases} 1 & \text{affected lymph nodes} \\ 0 & \text{unaffected lymph nodes} \end{cases}\)$
and the group coded \(x_i=0\) is the reference group. The regression model is unchanged,
\(Y_i = \beta_0 + \beta_1 x_i + \epsilon_i\) with \(\epsilon_i\) i.i.d. \(N(0,\sigma^2)\), but since \(x_i\)
can only take two values, the model splits into two separate statements:
$\(Y_i = \beta_0 + \epsilon_i \ \ (x_i=0), \qquad Y_i = \beta_0 + \beta_1 + \epsilon_i \ \ (x_i=1),\)$
so that \(E[Y_i\mid x_i=0] = \beta_0 = \mu_0\) and \(E[Y_i\mid x_i=1] = \beta_0+\beta_1 = \mu_1\), and
hence
$\(\beta_1 = E[Y_i \mid x_i=1] - E[Y_i \mid x_i=0] = \mu_1 - \mu_0.\)$
The least-squares estimators turn out to be exactly the familiar two-sample quantities: \(\hat\beta_0\)
is the sample mean of the reference group, \(\hat\beta_1\) is the difference between the two group
means, and \(MSE\) is the pooled sample variance \(S_p^2\). Testing \(H_0: \beta_1 = 0\) against
\(H_1: \beta_1 \neq 0\) in this regression is therefore exactly the two-sample \(t\)-test of
\(H_0: \mu_0 = \mu_1\) under equal variances — in R, t.test(age ~ node, brca, var.equal=TRUE) and
summary(lm(age ~ node, brca)) give the same test.
Association is not causation¶
The brca data are observational, not experimental, and that limits what can be concluded from either regression in this chapter. Patients were not randomly assigned to have affected or unaffected lymph nodes, so an age difference between the two groups cannot be read as age causing a higher risk of node involvement — the groups may differ in other, unmeasured ways too (confounding). The most that can be said is that lymph node status and age are associated.
The same caveat applies to the ESR1–S100A8 model. ESR1 expression was not experimentally manipulated, so the negative association cannot be read as ESR1 expression causing a decrease in S100A8 expression — only that the two are negatively associated. Establishing that one gene's expression actually causes a change in another's typically requires an experimental intervention, such as a knockout mutant in the lab, rather than an observational regression.
Sources¶
All of this chapter comes from the GTPB PSLS20 course, theory unit 6, "Simple linear regression"
(theory/06-linearRegression.Rmd, CC BY 4.0), converted losslessly into the eight linked notes
files:
01-breast-cancer-dataset.md— the dataset, its biological motivation, and the outlier-trimming step.02-lineair-regression.md— the model, \(g(x)\), and the interpretation of intercept and slope.03-parameter-estimation.md— least squares, the closed-form estimators, and the first fitted model on the trimmed data.04-statistical-inference.md— the four assumptions, the sampling distribution of the estimators, standard errors, confidence intervals, and the hypothesis test.05-assess-assumptions.md— residual analysis for linearity, homoscedasticity, and normality.06-invalid-assumptions.md— transformations, the log-log model on the full data, and the three interpretations of a log-scale slope.07-prediction-intervals.md— confidence intervals for the mean response vs. prediction intervals for a new observation, and the NHANES cholesterol comparison.08-sum-of-squares-and-anova-table.md— the sum-of-squares decomposition, \(R^2\), the F-test, the ANOVA table, dummy-variable coding, and the closing note on confounding.
No slide deck, transcript, or problem set was supplied for this lecture; the notes above are the
only source. Two figures referenced in the original slides (an image of the model's assumptions,
RegModel3.png, and an image about the effect of the spread of \(X\) on precision, spread.png) were
not included in the conversion and are redrawn here from the surrounding text rather than copied.
The underlying dataset (breastcancer.csv) and the original study
(https://doi.org/10.1093/jnci/djj052) are referenced by the notes but were not supplied or used
directly — no numeric results beyond those stated in the notes are reported here.
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