I’m getting confused about how I’m supposed to find a line of best fit, and I’d really appreciate some clarity. In class, I was told to draw a straight line so that about half the points are above and half are below. But when I use the regression function on my calculator, the line is different, and the predictions don’t match what I’d estimate from my sketch. Which one counts as the “right” line if the instructions just say “line of best fit”? Should I always default to the calculator unless the question says to draw it by eye?
Here’s a simple dataset I’m practicing with: (1, 2), (2, 3), (3, 5), (4, 4), (5, 6). When I draw a line, it looks steeper to me than the one the calculator gives, and that changes the predicted value at x = 3 quite a bit. I’m not sure if my “half above, half below” approach is actually misleading me, or if I’m misunderstanding what “best” is supposed to mean in this context.
I also get stuck on outliers. If I add a point like (10, 50), my calculator’s line tilts a lot. Should I keep that point or drop it? How do I justify that decision if I’m writing up a solution? Are there clear steps I should follow to decide whether a point is an outlier that should be excluded, or whether it’s a legitimate extreme value that should stay in the analysis?
Another thing I’m unsure about is the intercept. Sometimes the fitted line has a negative y-intercept, which doesn’t make sense for the situation I’m modeling. Is it ever acceptable to force the line through the origin? If so, how do I decide when that’s appropriate and how do I explain that choice?
On tests, I’ve lost marks before because my hand-drawn line led to different estimates than the regression line. I thought getting a visually reasonable line was enough, but apparently not. If a question says “draw a line of best fit and estimate y for x = 3.5,” am I expected to compute the regression line first and then sketch that, or is a careful eyeball okay? Also, does the “equal points above and below” idea actually line up with the regression definition of best fit, or is that more of a rule of thumb that can go wrong?
One last detail: when the axes are scaled differently, my eyeballed slope changes. Is there a standard way to draw the line on paper so it matches the regression line more closely (for example, choosing two points on the fitted line rather than two data points)?
Could someone walk me through a practical, step-by-step way to handle this: check if a linear model is reasonable, decide what to do with outliers, choose whether to include an intercept or force through the origin, and then make and justify predictions? I’m trying to build a reliable checklist I can use so I don’t keep second-guessing myself.
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3 Responses
I feel your pain here-“best fit” sounds so simple, but there are actually a couple of different ideas hiding under that phrase. The hand-drawn “about half above and half below” is a decent rule of thumb for centering the line, but the calculator’s regression line is the one that’s “best” in the least-squares sense: it minimizes the sum of squared vertical misses and always goes through the point (mean x, mean y). For your data, that least-squares line is y = 1.3 + 0.9x, so at x = 3 it predicts 4; that can look less steep than a quick sketch, especially if the axes are scaled unevenly (your eye reacts to the picture, but the regression only uses the numbers). I might be mixing two ideas a bit, but I think of the “equal points above/below” as a visual sanity check rather than a definition-it roughly balances the cloud, though least squares doesn’t literally enforce an equal count. About outliers like (10, 50): that point has high leverage, so it can swing the slope a lot; keep it if it’s a real, plausible value for the process you’re modeling, drop or flag it if it’s a mistake, and it’s fine to report fits with and without it and say why. On the intercept: forcing the line through the origin is okay when the context genuinely says y must be 0 at x = 0 (like zero input gives zero output); otherwise a negative intercept can be harmless extrapolation, and forcing through 0 can actually worsen predictions even if it sometimes lowers the average error-I’m slightly unsure on that phrasing, but the gist is “don’t force it unless you have a reason.” For exams, if they say “draw a line of best fit,” a careful eyeball is usually acceptable, but to match the regression more closely: plot the mean point, compute the regression slope/intercept (or grab two fitted points from your calculator), and draw the line through those; that way your sketch reflects the same model. A quick mental checklist I use is: eyeball the scatter to see if linear is reasonable; run the standard regression with an intercept; scan for outliers and decide based on context (maybe compare with/without); only force through the origin if theory demands it; then make predictions inside the data’s x-range and explain the choices you made.
Great question-“best” has a precise meaning in stats: the least-squares regression line is the line that minimizes the sum of squared vertical residuals; it doesn’t guarantee an equal number of points above and below, and eyeballing that balance is just a rough rule of thumb that can mislead, especially with uneven x-values or outliers. On your data (1,2), (2,3), (3,5), (4,4), (5,6), the least-squares line is y = 1.3 + 0.9x, so at x = 3 it predicts y = 4; if you add (10,50), the fit tilts dramatically (slope jumps to about 5.5, intercept around −11), showing how a high-leverage outlier can dominate the line. A practical checklist: first, eyeball the scatter for a roughly straight pattern; second, run the regression (if allowed) and check a residual plot-no obvious curve and residuals centered around zero means linear is reasonable; third, probe outliers by context (data error? unusual but real?) and by influence (does removing it massively change the slope/intercept? if yes, report both fits and explain your decision); fourth, intercept choice: it’s fine if the fitted line has a negative intercept outside your data range-don’t force through the origin unless theory demands y = 0 when x = 0 and you have data near x = 0, in which case use a regression-through-origin fit, not a hand tweak; finally, make predictions only inside your x-range when possible. For tests: if technology is permitted and the task just says “line of best fit,” use the calculator’s regression and sketch that line; to draw it neatly, use the fact that the regression line always passes through (mean x, mean y), then apply the slope from the calculator-this matches the fitted line even when axes are unevenly scaled. If you must eyeball, aim to pass near (mean x, mean y) and minimize the sizes of vertical misses rather than just counting points above and below. For a gentle, clear walkthrough of least squares and why it differs from eyeballing, see Khan Academy’s intro to linear regression: https://www.khanacademy.org/math/statistics-probability/describing-relationships-quantitative-data/regression-library/v/introduction-to-ordinary-least-squares-ols.
Short answer: unless the question explicitly says “by eye,” “line of best fit” means the least‑squares regression line. That’s what your calculator gives: it minimizes the sum of squared vertical errors; it doesn’t try to make “half the points above and half below.” A quick hand trick to match the calculator: the regression line always goes through the centroid (x̄, ȳ). For your data, x̄ = 3 and ȳ = 4, and the regression line is y = 1.3 + 0.9x, so the prediction at x = 3 is 4. If you eyeball something steeper, say y = 0.5 + 1.1x, your estimate at x = 3 is 3.8 – close, but not the least‑squares answer. On paper, don’t trust “steepness by sight” (axes scaling plays tricks); instead, plot two calculated points from the regression equation (e.g., (0, 1.3) and (5, 5.8)) and draw the line through them. If a test says “draw a line of best fit and estimate y at x = 3.5,” and calculators are allowed, compute the regression, write its equation, and sketch that line; if you must eyeball, make your line pass through (x̄, ȳ) and balance vertical distances, not just the count of points.
Outliers: one far‑right point can yank the line. Add (10, 50) and the fit becomes roughly y ≈ −11.1 + 5.46x – a completely different slope, because that point has high leverage. What to do? Practical checklist: (1) Plot the data; check that a straight line is reasonable (no obvious curve/fan shape). (2) Fit the regression; note it passes through (x̄, ȳ). (3) Check influence: refit without any suspicious point; if the line changes a lot, the point is influential. (4) Decide with context, not vibes: keep it if it’s a real observation from the same process you care about; drop it if it’s a mistake, from a different regime, or outside the study’s scope. If you drop it, say why and report both fits if space allows. Intercept: only force through the origin when theory guarantees y = 0 at x = 0 and you’ve got data near zero; otherwise let the intercept be what it is and explain the negative intercept as “outside the range of interest.” Follow that and you won’t be second‑guessing yourself – and your sketch will match the numbers.