Module 3 · Fit a line and know what it means

Multiply inputs by weights. Add a baseline. Check the misses. Check the uncertainty. That is the path through Module 3.

Read each small idea, then cover the answer and try its calculation. Jump to Q1: matrices, Q2: averages, Q3: fit a line, or Q5: test slopes. Keep the formula sheet nearby.

Browse the original slides above each topic; click a slide to enlarge it. Practice-sheet previews include the original solutions.

Matrix = a grid of numbers

Original slides · Linear AlgebraOpen PDF ↗

Slide 2 of 4

• Scalar: one number.
• Vector: an ordered list, usually a column.
• Matrix: a rectangular grid.
• $A\in\mathbb R^{m\times n}$: matrix $A$ has $m$ rows, $n$ columns, and real entries.
• $A_{ij}$: the entry in row $i$, column $j$.
• Transpose $A^\mathsf T$: swap rows and columns.

For example, $[2,1]^\mathsf T$ is the column with 2 above 1.

Multiplication = row meets column

Shape trick: inner dimensions match; outer dimensions survive.

$$(m\times n)(n\times p)\longrightarrow(m\times p).$$

Multiply matching entries, then add:

$$(AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}.$$

$k$ runs along row $i$ of $A$ and down column $j$ of $B$. The sum sign $\sum$ means “add these terms.”

Invented example: use each row of $A$ with input column $x$.

$$A=\begin{bmatrix}1&2\\3&4\end{bmatrix}, \qquad x=\begin{bmatrix}2\\1\end{bmatrix}.$$ $$Ax=\begin{bmatrix}1(2)+2(1)\\3(2)+4(1)\end{bmatrix} =\begin{bmatrix}4\\10\end{bmatrix}.$$

Inner product = number; outer product = grid

For column vectors $u,v\in\mathbb R^n$:

$$u^\mathsf Tv=\sum_{k=1}^{n}u_kv_k, \qquad (uv^\mathsf T)_{ij}=u_iv_j.$$

Invented values: $u=[2,1]^\mathsf T$, $v=[3,4]^\mathsf T$.

• Inner: $2(3)+1(4)=10$.
• Outer: every entry of $u$ multiplies every entry of $v$:

$$uv^\mathsf T=\begin{bmatrix}2(3)&2(4)\\1(3)&1(4)\end{bmatrix} =\begin{bmatrix}6&8\\3&4\end{bmatrix}.$$

Other matrix rules to recognize

• Identity $I$: ones on the diagonal, zeros elsewhere; $AI=IA=A$ when sizes match.
• Order matters: usually $AB\ne BA$.
• Grouping works: $(AB)C=A(BC)$.
• Distribution works: $A(B+C)=AB+AC$.
• Transpose twice: $(A^\mathsf T)^\mathsf T=A$.
• Transpose a sum: $(A+B)^\mathsf T=A^\mathsf T+B^\mathsf T$.
• Transpose a product: $(AB)^\mathsf T=B^\mathsf TA^\mathsf T$ — reverse the order.

Practice 1: multiply matrices

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Given: calculate these three products. Rule: matrix multiplication.

$$\text{(a)}\quad \begin{bmatrix}1&0\\0&1\end{bmatrix} \begin{bmatrix}1&2\\3&4\end{bmatrix}$$ $$\text{(b)}\quad \begin{bmatrix}9&6\\3&2\end{bmatrix}^{2} \qquad \text{(c)}\quad \begin{bmatrix}1&1\\1&0\end{bmatrix}^{3}$$

Hint: $A^2=AA$ and $A^3=(AA)A$; these powers do not square or cube entries individually.

Work through all three products

(a) Use row times column. The four entries are

$$\begin{aligned} 1(1)+0(3)&=1,\\ 1(2)+0(4)&=2,\\ 0(1)+1(3)&=3,\\ 0(2)+1(4)&=4. \end{aligned}$$

So the answer is $\begin{bmatrix}1&2\\3&4\end{bmatrix}$, as expected when multiplying by the identity.

(b) Use the rows and columns of the given matrix twice.

$$\begin{aligned} 9(9)+6(3)&=99,\\ 9(6)+6(2)&=66,\\ 3(9)+2(3)&=33,\\ 3(6)+2(2)&=22. \end{aligned}$$

Thus $A^2=\begin{bmatrix}99&66\\33&22\end{bmatrix}$.

(c) Call the given matrix $F$. First multiply $F$ by itself:

$$F^2=\begin{bmatrix}1+1&1+0\\1+0&1+0\end{bmatrix} =\begin{bmatrix}2&1\\1&1\end{bmatrix}.$$

Then multiply those calculated entries by the original $F$:

$$\begin{aligned} (F^3)_{11}&=2(1)+1(1)=3,\\ (F^3)_{12}&=2(1)+1(0)=2,\\ (F^3)_{21}&=1(1)+1(1)=2,\\ (F^3)_{22}&=1(1)+1(0)=1. \end{aligned}$$

Thus $F^3=\begin{bmatrix}3&2\\2&1\end{bmatrix}$.

Eigenvector = an arrow that stays on its line

Original slides · Eigenvalues and EigenvectorsOpen PDF ↗

Slide 3 of 4

A matrix can turn an arrow. An eigenvector is a nonzero arrow that the matrix only scales:

$$Mv=\lambda v,\qquad v\ne0.$$

• $M$: a square matrix.
• $v$: the eigenvector.
• $\lambda$: the eigenvalue, or scaling factor.
• Negative $\lambda$: reverse the arrow. Zero $\lambda$: collapse it to zero.

Check trick: multiply, then compare.

For invented $M=\begin{bmatrix}3&0\\0&2\end{bmatrix}$ and $v=[1,0]^\mathsf T$:

$$Mv=\begin{bmatrix}3(1)+0(0)\\0(1)+2(0)\end{bmatrix} =\begin{bmatrix}3\\0\end{bmatrix}=3v.$$

So $v$ is an eigenvector and $\lambda=3$.

Symmetric matrix: $M=M^\mathsf T$. A real symmetric $n\times n$ matrix has real eigenvalues and admits $n$ mutually perpendicular eigenvectors. Perpendicular means $v_i^\mathsf Tv_j=0$ for different chosen vectors; general real matrices can have complex eigenvalues/eigenvectors.

Invented self-check: with the same $M$, is $v=[1,1]^\mathsf T$ an eigenvector? Rule.

Check your answer

$Mv=[3,2]^\mathsf T$. A single multiplier cannot turn $[1,1]^\mathsf T$ into $[3,2]^\mathsf T$: the first entry needs 3 and the second needs 2. So this $v$ is not an eigenvector.

Coefficient = multiplier

Original slides · Linear RegressionOpen PDF ↗

Slide 2 of 7

In a line, a coefficient tells you how much to add per input unit:

$$Y=\beta_0+\beta_1X+\varepsilon.$$

• $X$: input; $Y$: observed output.
• $\beta_1$: slope, the multiplier on $X$.
• $\beta_0$: intercept, the starting amount.
• $\varepsilon$: noise, the part the rule does not explain.

Intercept = baseline; hat = estimated

The intercept is the prediction when inputs equal zero. We do not know nature’s coefficients; fitting data gives estimates, marked with hats:

$$\widehat y_i=\widehat\beta_0+\widehat\beta_1x_i.$$

$i$ identifies one observation. Uppercase $X,Y$ are random variables; lowercase $x_i,y_i$ are recorded values. The prediction leaves out unknown noise.

Invented fitted rule: $\widehat y=10+3x$. At $x=4$:

1. Input contribution: $3(4)=12$.
2. Add baseline: $10+12=22$.

One extra $x$ unit adds 3; at $x=0$, the prediction is 10.

Multiple regression = more multipliers

With $p$ inputs, add one contribution per input:

$$Y=\beta_0+\sum_{j=1}^{p}\beta_jX_j+\varepsilon.$$

$j$ selects an input. $\beta_j$ is its per-unit effect holding the other inputs fixed. $p$ excludes the intercept.

The data-generating story is “draw inputs, then an output given those inputs”:

$$f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)f_X(x).$$

$f_X$ is the input density; $f_{Y\mid X}$ is the output density given an input; $f_{X,Y}$ is their joint density. The linear rule and noise distribution describe that conditional output.

Least squares = smallest total squared miss

Original slides · Linear RegressionOpen PDF ↗

Slide 6 of 7

Residual = observed − predicted. For observation $i$:

$$e_i=y_i-\widehat y_i, \qquad \mathrm{RSS}=\sum_{i=1}^{n}e_i^2.$$

$n$ counts observations. RSS means residual sum of squares. Least squares chooses coefficients that make RSS smallest.

Why square? Invented misses $+2$ and $-2$ should both count: $2^2+(-2)^2=8$, rather than canceling.

Slope recipe = center, multiply, add, divide

For observed pairs $(x_i,y_i)$, first calculate the means:

$$\bar x=\frac1n\sum_{i=1}^n x_i, \qquad\bar y=\frac1n\sum_{i=1}^n y_i.$$

Then subtract those means — center the values — and build two totals:

$$S_{xy}=\sum_{i=1}^{n}(x_i-\bar x)(y_i-\bar y),$$ $$S_{xx}=\sum_{i=1}^{n}(x_i-\bar x)^2.$$

• $S_{xy}$: how inputs and outputs move together.
• $S_{xx}$: how much the inputs spread out.

Divide to get the slope; use the means to position the line:

$$\widehat\beta_1=\frac{S_{xy}}{S_{xx}}, \qquad \widehat\beta_0=\bar y-\widehat\beta_1\bar x.$$

Require $S_{xx}>0$. Identical inputs cannot tell you the slope.

Memory cue: the intercept puts the line through its center, $(\bar x,\bar y)$.

Practice 2: why the line crosses the averages

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Given: the least-squares formulas above. Show that the fitted line passes through $(\bar x,\bar y)$. Rule: simple regression.

Hint: substitute $x=\bar x$ into the prediction; replace the intercept with its formula.

Work through the cancellation

The prediction rule is $\widehat y(x)=\widehat\beta_0+\widehat\beta_1x$. At the average input, use $x=\bar x$ and $\widehat\beta_0=\bar y-\widehat\beta_1\bar x$:

$$\begin{aligned} \widehat y(\bar x) &=(\bar y-\widehat\beta_1\bar x)+\widehat\beta_1\bar x\\ &=\bar y. \end{aligned}$$

The two slope terms cancel. The slope itself does not become zero. This property applies to ordinary least squares with an intercept.

TSS = the average-only guess’s error

Original slides · Hypothesis TestingOpen PDF ↗

Slide 7 of 9

Before fitting a line, predict the sample mean $\bar y$ for everyone. Its total squared miss is total sum of squares:

$$\mathrm{TSS}=\sum_{i=1}^{n}(y_i-\bar y)^2.$$

RSS grades the fitted model. TSS grades the average-only baseline.

R² = fraction of baseline error removed

$$R^2=1-\frac{\mathrm{RSS}}{\mathrm{TSS}}.$$

For example, $R^2=0.94$ means 94% less squared error than guessing the mean. It is not the percentage of individually correct predictions.

On training data, least squares with an intercept gives $0\le R^2\le1$, provided TSS $>0$.

RSE = scale of the misses

Residual standard error estimates noise size in output units.

$$\mathrm{RSE}=\sqrt{\frac{\mathrm{RSS}}{n-p-1}}.$$

• $n$: observations.
• $p$: predictors, excluding the intercept.
• $n-p-1$: observations minus fitted coefficients, called residual degrees of freedom.

Require $n>p+1$ and linearly independent model columns: no column is an exact combination of others.

Practice 3: fit and score a line

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Given: $Y=\beta_0+\beta_1X+\varepsilon$ with normal noise of mean zero and variance $\sigma^2$. The three observed pairs are $(-1,-2)$, $(0,0)$, and $(1,5)$.

Find $\widehat\beta_0$, $\widehat\beta_1$, RSS, RSE, TSS, and $R^2$. Fitting formulas · Scoring formulas.

Hint: follow this order: averages → centered totals → coefficients → predictions → squared residuals → scores. Normality is not needed to do the least-squares arithmetic.

Work through the complete calculation

1. Calculate the averages from the three given observations.

$$\begin{aligned} \bar x&=\frac{-1+0+1}{3}=0,\\ \bar y&=\frac{-2+0+5}{3}=1. \end{aligned}$$

2. Calculate the slope using centered totals. With calculated means $\bar x=0$, $\bar y=1$:

$$\begin{aligned} S_{xx}&=(-1)^2+0^2+1^2\\ &=1+0+1=2,\\ S_{xy}&=(-1)(-3)+(0)(-1)+(1)(4)\\ &=3+0+4=7. \end{aligned}$$

The centered outputs $-3,-1,4$ come from subtracting $\bar y=1$ from the given outputs $-2,0,5$. Therefore

$$\widehat\beta_1=\frac{S_{xy}}{S_{xx}}=\frac72=3.5.$$

3. Calculate the intercept. Use $\bar y=1$, $\bar x=0$, and the calculated slope $3.5$:

$$\widehat\beta_0=\bar y-\widehat\beta_1\bar x =1-3.5(0)=1.$$

The fitted rule is $\widehat y=1+3.5x$.

4. Predict, subtract, and square. Apply $\widehat y=1+3.5x$ at each given input; calculate $e=y-\widehat y$:

$x$	$y$	$\widehat y$	$e$	$e^2$
$-1$	$-2$	$1+3.5(-1)=-2.5$	$0.5$	$0.25$
$0$	$0$	$1+3.5(0)=1$	$-1$	$1$
$1$	$5$	$1+3.5(1)=4.5$	$0.5$	$0.25$

Add the calculated squared residuals:

$$\mathrm{RSS}=0.25+1+0.25=1.5.$$

5. Estimate noise size. There are $n=3$ observations and $p=1$ predictor. Use calculated RSS $=1.5$:

$$\mathrm{RSE}=\sqrt{\frac{1.5}{3-1-1}} =\sqrt{1.5}\approx1.225.$$

6. Calculate the mean-only baseline. Use given outputs $-2,0,5$ and calculated $\bar y=1$:

$$\begin{aligned} \mathrm{TSS}&=(-2-1)^2+(0-1)^2+(5-1)^2\\ &=9+1+16=26. \end{aligned}$$

7. Calculate the relative improvement. With RSS $=1.5$ and TSS $=26$:

$$R^2=1-\frac{1.5}{26} \approx0.9423\approx0.94.$$

Two quick checks: the calculated residuals sum to $0.5-1+0.5=0$, and RSS $=1.5\le26=$ TSS. Both hold for training-set least squares with an intercept.

Matrix fitting = all predictions at once

Original slides · Linear RegressionOpen PDF ↗

Slide 5 of 7

The design matrix $M$ stores the recipe’s inputs:

• One row per observation.
• One column per predictor.
• An initial column of ones to add the intercept every time.

$y$ holds observed outputs; $\beta$ holds coefficients $\beta_0,\ldots,\beta_p$. Predictions are $M\widehat\beta$.

$$\widehat\beta=\operatorname*{argmin}_{\theta} \lVert y-M\theta\rVert^2.$$

$\theta$ is a trial coefficient vector. Argmin = coefficients giving the smallest error. The squared length $\lVert r\rVert^2=\sum_i r_i^2$ adds squared entries of residual vector $r$.

With linearly independent columns, the reusable solution is

$$\widehat\beta=(M^\mathsf TM)^{-1}M^\mathsf Ty.$$

An inverse undoes a matrix: $A^{-1}A=I$. The displayed inverse exists when no design column is an exact combination of others.

Same Q3 data, using the matrix formula

Given pairs: $(-1,-2),(0,0),(1,5)$. Put ones beside their inputs:

$$M=\begin{bmatrix}1&-1\\1&0\\1&1\end{bmatrix}, \qquad y=\begin{bmatrix}-2\\0\\5\end{bmatrix}.$$

Take column dot products to form $M^\mathsf TM$:

• Ones with ones: $1^2+1^2+1^2=3$.
• Ones with inputs: $-1+0+1=0$.
• Inputs with inputs: $(-1)^2+0^2+1^2=2$.

$$M^\mathsf TM=\begin{bmatrix}3&0\\0&2\end{bmatrix}.$$

Multiply columns by outputs to form $M^\mathsf Ty$:

$$M^\mathsf Ty= \begin{bmatrix}-2+0+5\\(-1)(-2)+0(0)+1(5)\end{bmatrix} =\begin{bmatrix}3\\7\end{bmatrix}.$$

The diagonal matrix’s inverse has entries $1/3,1/2$. Substitute into the fitting rule:

$$\widehat\beta= \begin{bmatrix}(1/3)(3)\\(1/2)(7)\end{bmatrix} =\begin{bmatrix}1\\3.5\end{bmatrix}.$$

Same intercept, 1; same slope, 3.5.

Standard error = how much the estimate bounces

Original slides · Coefficient UncertaintyOpen PDF ↗

Slide 3 of 6

Collect a fresh dataset and refit. Noise changes, so the estimated weights change.

• True coefficient: fixed but unknown.
• Coefficient estimate: varies between samples.
• Standard error (SE): estimated standard deviation of those estimates.

Smaller SE = a more precise estimate.

Invented self-check: two fits estimate the same slope, 2. Fit A has SE $0.2$; Fit B has SE $0.8$. Which is more precise? Rule.

Check your answer

Fit A. Its relative SE is $0.2/0.8=1/4$: its estimated repeated-sample spread is one quarter as large.

RSE and SE describe different spreads

• RSE: scale of output misses.
• Coefficient SE: uncertainty in a fitted multiplier or intercept.

Assumptions for these uncertainty formulas: correct linear model; zero-mean noise given inputs; independent errors of common variance $\sigma^2$; linearly independent design columns.

For one predictor, define input spread $S_{xx}=\sum_i(x_i-\bar x)^2$. If noise standard deviation $\sigma$ is known:

$$\operatorname{SD}(\widehat\beta_1)=\frac{\sigma}{\sqrt{S_{xx}}},$$ $$\operatorname{SD}(\widehat\beta_0) =\sigma\sqrt{\frac1n+\frac{\bar x^2}{S_{xx}}}.$$

Here SD means standard deviation. If variance $\sigma^2$ is supplied, take its square root to get $\sigma$.

Unknown noise? Use RSS → RSE → slope SE:

$$\operatorname{SE}(\widehat\beta_1) =\frac{\mathrm{RSE}}{\sqrt{S_{xx}}}.$$

Replace $\sigma$ with RSE in the intercept formula too. Supplied SE? Use it directly.

More noise increases uncertainty. More input spread reduces slope uncertainty — widely separated points reveal the tilt more clearly.

Invented calculation: $n=5$, $p=1$, $\bar x=0$, $S_{xx}=10$, RSS $=3$.

1. Estimate noise: $\mathrm{RSE}=\sqrt{3/(5-1-1)}=1$.
2. Slope SE: $1/\sqrt{10}\approx0.316$.
3. Intercept SE: $1\sqrt{1/5+0^2/10}\approx0.447$.

Invented self-check: a fit reports RSE $=8$ and slope SE $=0.3$. Which number describes uncertainty in the slope? Rule.

Check your answer

$0.3$ describes estimated slope spread across repeated samples. The RSE of 8 describes estimated noise size in response units.

Unbiased = centered on the true coefficient

Under the assumptions above, conditional on the observed design $M$:

$$\mathbb E[\widehat\beta\mid M]=\beta,$$ $$\operatorname{Cov}(\widehat\beta\mid M) =\sigma^2(M^\mathsf TM)^{-1}.$$

• $\mathbb E$: average over repeated fits. Unbiased means this average hits the true coefficient.
• $\operatorname{Cov}$: covariance matrix, describing how estimates vary together.
• Diagonal entries: coefficient variances. Take square roots for standard deviations.

Unbiased does not mean every estimate is close. SE tells you about spread around that center.

Confidence interval = estimate ± uncertainty margin

Original slides · Confidence IntervalsOpen PDF ↗

Slide 3 of 6

An approximately normal estimate gives the quick approximately 95% interval:

$$\widehat\beta_j\pm2\operatorname{SE}(\widehat\beta_j).$$

$j$ selects the coefficient. Calculate margin = $2\times\mathrm{SE}$, then subtract/add it.

Q5(c) supplied values: estimate 2, SE $1/2$.

1. Margin: $2(1/2)=1$.
2. Lower endpoint: $2-1=1$.
3. Upper endpoint: $2+1=3$.
4. Interval: $[1,3]$.

Meaning: about 95% of intervals made this way across repeated samples contain the fixed true coefficient. This particular interval is not guaranteed to contain it.

Approximation: 2 rounds the normal cutoff 1.96. With normal errors and estimated noise, exact regression intervals use a Student $t$ cutoff with $n-p-1$ degrees of freedom. The practice uses the stated $\pm2\mathrm{SE}$ rule.

t = gap ÷ uncertainty

Original slides · Hypothesis TestingOpen PDF ↗

Slide 3 of 9

A gap matters more when the estimate is precise. Measure the gap in standard-error units:

$$t=\frac{\widehat\beta_j-b}{\operatorname{SE}(\widehat\beta_j)}.$$

• $\widehat\beta_j$: estimate.
• $b$: proposed true value.
• SE: uncertainty in the estimate.
• $|t|$: distance in SEs; ignore direction for a two-sided test.

Approximate decision rule

• $|t|>2$: reject the proposed value.
• $|t|\le2$: fail to reject; insufficient evidence against it.

Same trick, two views: the proposed value is over 2 SEs away exactly when it is outside the interval. Endpoints count as inside in these practice questions.

The formal hypotheses are $H_0:\beta_j=b$ (null) and $H_a:\beta_j\ne b$ (alternative). To test for a linear relationship, test slope $b=0$.

A nonzero fitted slope can arise from noise even when the true slope is zero. Rejection supports an association under the model; it does not establish causation.

Significance $\alpha\approx0.05$: the long-run rate of rejecting a true null, under the assumptions. It is not the probability the null is true. Failing to reject does not prove it true.

Invented self-check: estimate $1.2$, SE $0.4$, proposed slope $b=0$. Reject under the approximate rule? Rule.

Check your answer

Gap: $1.2-0=1.2$. Divide by SE: $t=1.2/0.4=3$. Since $|t|>2$, reject. Interval check: margin $2(0.4)=0.8$, giving $[1.2-0.8,1.2+0.8]=[0.4,2]$, which excludes 0.

Practice 5: test two proposed slopes

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Given: $Y=\beta_0+\beta_1X+\varepsilon$. Test the two separate claims $H_1:\beta_1=1$ and $H_2:\beta_1=2$ in each case below. Rule: confidence intervals and tests.

Case	Given $\widehat\beta_1$	Given SE
(a)	$-3$	$1$
(b)	$1/2$	$2$
(c)	$2$	$1/2$

What this asks: could the true slope be 1? Could it be 2? Test each value separately. Here $H_1$ and $H_2$ are names for two separate claims, not a null/alternative pair.

Hint: two claims, two distances. Calculate (estimate − claim) ÷ SE separately for each claim. Reject only past 2 in magnitude. The interval provides a second check.

Work through the three decisions

(a) Estimate $=-3$, SE $=1$. Use the gap divided by SE for each claim:

$$\begin{aligned} b=1:\quad t&=\frac{-3-1}{1}=-4,\\ b=2:\quad t&=\frac{-3-2}{1}=-5. \end{aligned}$$

The claims are 4 and 5 SEs away. Both distances exceed 2, so reject $H_1$ and reject $H_2$. Check with the interval: its margin is $2(1)=2$,

$$[-3-2,\,-3+2]=[-5,-1].$$

Both 1 and 2 are outside that range.

(b) Estimate $=1/2$, SE $=2$. Divide each gap by the supplied SE, 2:

$$\begin{aligned} b=1:\quad t&=\frac{1/2-1}{2}=-0.25,\\ b=2:\quad t&=\frac{1/2-2}{2}=-0.75. \end{aligned}$$

The claims are only 0.25 and 0.75 SEs away. Fail to reject either claim. The estimate is too uncertain to rule either out. The interval check uses margin $2(2)=4$:

$$[1/2-4,\,1/2+4]=[-7/2,9/2].$$

Both 1 and 2 are inside. This does not establish that both values are true; it means the data do not distinguish them well enough.

(c) Estimate $=2$, SE $=1/2$. Divide each gap by the supplied SE, $1/2$:

$$\begin{aligned} b=1:\quad t&=\frac{2-1}{1/2}=2,\\ b=2:\quad t&=\frac{2-2}{1/2}=0. \end{aligned}$$

The first claim is exactly 2 SEs away; the second matches our estimate. Fail to reject either under the stated rule, which rejects only distances greater than 2. The interval has margin $2(1/2)=1$:

$$[2-1,\,2+1]=[1,3].$$

The value 1 is on the boundary and 2 is inside. The boundary decision depends on the chosen approximation.

What to do when a new problem changes the numbers

• Matrix: inner shapes match; outer shapes survive. Row times column.
• Eigenvector: multiply; check for one common scaling factor.
• Line: center → multiply → add → divide; intercept puts it through the averages.
• Fit: RSS grades the line; TSS grades the mean; R² compares; RSE measures noise size.
• Uncertainty: SE measures estimate spread; $t$ is gap divided by SE.

Next: Module 4 turns categories, interactions, and curves into inputs for the same weighted recipe.