Comparing Competing Models
Introduced| English | Meaning |
|---|---|
| linear model/ˈlɪnɪə ˈmɒdl/ | A model that describes a relationship using a linear equation. |
| quadratic model/kwɒˈdrætɪk ˈmɒdl/ | A model that describes a relationship using a polynomial of degree two. |
| exponential model/ˌekspəˈnenʃl ˈmɒdl/ | A model in which equal input increments multiply the output by a constant factor. |
| residuals/rɪˈsɪdʒuːəlz/ | An observed value minus the value predicted by a model. |
| residual plot/rɪˈsɪdʒuːəl plɒt/ | A graph of prediction errors against fitted values or an explanatory variable, used to assess a model. |
Which forecast would you use?
- A school video has $100,151,224,339,505$ views on days $0,1,2,3,4$. The counts rise by roughly the same factor, rather than the same number.
- A linear model 线性模型, quadratic model 二次模型 and exponential model 指数模型 can all follow part of this rising trend. Compare them on the same observations, then ask whether the forecast makes sense.
Fit all three families to the same data
- Fit all five pairs with technology. The linear and quadratic regressions minimise squared errors in the counts; here the exponential comes from linear regression of $\ln y$ on $x$, then converting back.
- The rules below give predicted counts, written $\hat y$. The exponential method minimises squared errors in $\ln y$, so another calculator method may give slightly different coefficients. Keep full precision for residuals; the displayed coefficients are rounded.
Give the residual its sign
- A residual 残差 is observed minus predicted: $e=y-\hat y$. A positive residual puts the observation above the curve: the model underestimates it.
- A negative residual puts the observation below the curve: the model overestimates it. The size $|e|$ tells how far off the prediction is, in the same units as the count.
On day $2$, the observed count is $224$. Calculate the prediction, then subtract it from the observation:
The line overestimates by $39.8$ views. The quadratic predicts about $224.94$, and the exponential about $225.19$: their residuals are about $-0.94$ and $-1.19$.
A signed residual is…
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Residual = observed − predicted. A negative residual means that the model predicts too high.
On day 2 the observed count is 224 and the line predicts 263.8. What is the signed residual?
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Observed minus predicted: $224-263.8=-39.8$. The line overestimates by 39.8 views.
Read the whole residual plot
- A residual plot 残差图 shows each residual against its input $x$, with a horizontal line at zero. A curve or a run that changes systematically from positive to negative and back suggests that the model misses a feature of the data.
- Small residuals scattered on both sides of zero, with no clear pattern, support a model. A few points cannot prove that a family is the true rule or that distant predictions will work.
To judge a fit, examine the size and pattern of its ____.
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Smaller residuals (and no obvious pattern in them) signal a better-fitting model.
Residuals are positive at both ends of the interval and negative in the middle. What does this suggest?
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This is a pattern, rather than a pattern-free scatter around zero. A different family may capture the shape better.
Compare size and pattern together
- For the line, the five residuals are $35.8,-13,-39.8,-24.6,41.6$: a curved pattern. Its total absolute error is $154.8$ views. Positive and negative errors must not cancel in this size comparison.
- The quadratic's total absolute error is about $18.34$ views; the exponential's is about $4.59$. Here the exponential has the smaller errors and no obvious residual pattern, consistent with the roughly constant growth factor.
Three fits, the same five observations
Compare fitted curves and signed residuals. Both graph scales stay fixed as the model changes; day 2 starts with observed 224, predicted 263.8 and residual −39.8.
Let the context decide what matters
- A model with small errors can still be unsuitable: view counts cannot be negative, and a short burst of interest may stop. These five days support a short-range comparison, not unlimited future growth.
- The cost of an error also matters. Underestimating demand can leave a school bus with too few seats; a cautious overestimate may be preferable. Name the direction and size of the error, then explain its consequence.
A polynomial can pass through every observed point yet behave badly between or beyond them. A perfect fit to a small sample does not guarantee a useful forecast. Use fit, context and a sensible domain together.
Positive data that doubles over every equal time interval suggests a…
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Constant-ratio doubling is the hallmark of an exponential model, not a polynomial one.
A bus-demand model predicts 48 passengers; 55 arrive. Which description is correct?
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$55-48=7$. A positive residual means actual demand exceeds the prediction. The direction matters when choosing capacity.
Separate eventual growth from fit
- If $a>0$ and $b>1$, the exponential $ab^x$ eventually exceeds any fixed polynomial as $x\to\infty$. This is a statement about the formulas over a long range, not evidence that an exponential fits a particular data set best.
- A decay model such as $100(0.8)^x$ approaches zero instead. A zero or negative leading multiplier also fails the positive-growth claim; always state the assumptions.
For $a>0$ and $b>1$, $ab^x$ eventually exceeds every fixed polynomial as $x$ increases without bound.
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These positive-growth conditions matter. The statement concerns the formulas as $x$ increases, and does not prove that an exponential fits a given data set best.
The decay model $100(0.8)^x$ must eventually exceed the growing line $10+x$ as $x$ increases.
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The exponential tends to zero, while the line increases. An exponential family alone does not imply positive eventual dominance.
Label the starting time
- For the earlier sales example $100,150,225,340$, label the observations as months $m=0,1,2,3$. The model $S(m)=100(1.5)^m$ predicts $337.5$ at month $3$, giving residual $340-337.5=2.5$.
- If the first observation is called month $1$, the same model is $S(m)=100(1.5)^{m-1}$. Changing the labels changes the exponent; it does not change the observations.
Sales first recorded at month 1 are 100 and grow by a factor of 1.5 each month. Using $S(m)=100(1.5)^{m-1}$, what is the prediction for month 3?
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Month 3 is two changes after month 1: $100(1.5)^2=225$.
Make a supported recommendation
- State the family and the interval, quote residual sizes and describe any pattern. Then add one contextual reason and one limit. For the video, an exponential is a reasonable description of days $0$–$4$; it is not a promise of growth forever.
- Compare all models on the same observations and units. The total of signed residuals can hide large errors, and the eventual winner of two formulas need not be the best fit now.
Construct candidates from the data, calculate observed minus predicted, and inspect the whole residual plot. Choose with both numerical evidence and context. No clear pattern supports a model; it does not prove that the model is true.
Select all true statements about comparing models.
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The first three statements are true. Signed errors can cancel; inspect their sizes and pattern, together with context and the comparison interval.
Write a recommendation from the evidence
- Model A has residuals $2,-1,3$ views; model B has $4,4,4$ on the same three days. Calculate both totals of absolute residuals and say which has the smaller error by this measure.
- Explain why this comparison alone does not establish that A is the best model for a long-range forecast. Include one further check you would make.
On the same three days, model A has residuals $2,-1,3$ views and model B has $4,4,4$. Find both totals of absolute residuals and recommend a model by this measure. Why is this insufficient for a long-range forecast? Name one further check.
Use both the numerical error measure and a limit to the recommendation. A smaller total alone does not validate a forecast.
Model A has residuals $2,-1,3$. What is its total of absolute residuals?
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Absolute residuals add to 6.