When a Fitted Line Predicts a Negative Height
In “Checking Whether a Prediction Is Reasonable,” we checked whether a predicted response fits the response’s possible values and the situation’s practical limits. Height provides a clear example: a height measured upward from a defined baseline cannot be negative. Yet a fitted regression line can calculate a negative height, particularly when it is extended beyond the data.
A negative result is not necessarily an arithmetic mistake. The equation of a least-squares regression line is a mathematical rule, and the rule is not automatically restricted to realistic response values. When the line predicts a negative height, report what the equation calculates, then explain why that output cannot be taken literally in context. Do not treat the prediction as evidence that an object will have negative height.
The input value matters too. If the requested \(x\)-value is outside the range of the data, the prediction is an extrapolation, as explained in “What Extrapolation Means.” A negative predicted height can therefore be both an extrapolation and an impossible response value. Those are separate concerns: one describes where the input lies relative to the data, while the other describes whether the predicted response is possible.
Finding Where the Fitted Line Reaches Zero
For a fitted line \(\hat{y}=a+bx\), a zero crossing is an \(x\)-value where the line predicts \(\hat{y}=0\). When \(b\ne 0\), set the fitted response equal to zero and solve for \(x\). This value can help identify where a line with a negative predicted height reaches the boundary between positive and negative predictions.
The zero crossing has the units of the explanatory variable. It is a feature of the fitted equation, not necessarily a measured point in the data or a real-world time when an object actually reaches zero height. Check whether it lies in the observed \(x\)-range and whether zero is meaningful in the context.
The formula requires a nonzero slope. If \(b=0\), the fitted line is horizontal, so it predicts the same response for every \(x\). If \(b=0\) and \(a\ne 0\), it never crosses zero. If \(a=0\) and \(b=0\), the fitted value is zero for every \(x\), so every \(x\) is a zero crossing. These cases cannot be handled by dividing by \(b\).
A zero crossing is most useful as a warning. If the line predicts a negative height at a requested \(x\)-value, the calculation can flag that the linear pattern should not be used literally for that prediction. It does not tell us what a better model would predict. In particular, do not automatically replace a negative prediction with zero and call that the regression prediction; the fitted line did not produce that adjusted value.
Worked Examples: Recognizing Impossible Height Predictions
Worked Example: A Snowbank’s Predicted Height Becomes Negative
Hypothetical setting. A park employee records the height \(y\), in centimeters, of a snowbank at different times \(x\), in hours, after a storm ends. The observed times ran from 0 to 30 hours. A fitted line is \(\hat{y}=120-3x\). The employee asks for the line’s prediction at \(x=45\) hours.
State. We need to calculate the predicted snowbank height at 45 hours and decide whether that value is a plausible height in this context.
Plan. Substitute \(x=45\) into the fitted line and keep the response units. The prediction is an extrapolation because 45 hours is beyond the observed range of 0 to 30 hours. Since snowbank height measured from the ground cannot be negative, we will also check the calculated height against zero.
Do. Substitution gives:
The arithmetic can be checked by finding the line’s zero crossing. Here \(a=120\) and \(b=-3\), so \(b\ne 0\) and:
At 40 hours the fitted line predicts zero centimeters. At 45 hours, which is 5 hours beyond that crossing, it predicts \(3(5)=15\) centimeters below zero, consistent with the calculation of \(-15\) centimeters.
Conclude in context. The fitted line calculates a snowbank height of \(-15\) cm at 45 hours, but a height measured upward from the ground cannot be negative. This is not a plausible prediction of the snowbank’s actual height. The requested time is also outside the observed range, so extending the line there is extrapolation. The zero crossing at 40 hours is a warning about the line’s output, not proof that the snowbank reaches exactly zero at that time.
Worked Example: A Negative Prediction at a Positive Time
Hypothetical setting. A gardening club measures the height \(y\), in centimeters, of a patch of snow remaining beside a path \(x\) days after a snowstorm. Measurements used to fit a line were taken from day 4 through day 12. The fitted equation is \(\hat{y}=-6+2.4x\). A member asks for the prediction on day 1.
State. We need to calculate and assess the fitted height at day 1, even though day 1 is earlier than the observed days.
Plan. Substitute \(x=1\). Because the observed \(x\)-values ran from 4 to 12, day 1 is outside the observed range, so this is an extrapolation. Then compare the predicted response with the fact that a measured height above the path cannot be negative.
Do. The model gives:
To locate the line’s zero crossing, use \(a=-6\) and \(b=2.4\). The slope is not zero, so:
The line predicts zero height at day 2.5 and negative heights for \(x<2.5\). Day 1 is 1.5 days before the crossing; multiplying \(1.5\) by the slope of 2.4 cm per day gives a fitted value 3.6 cm below zero, matching the substitution.
Conclude in context. At day 1, the line calculates a height of \(-3.6\) cm, which is impossible as a snow height measured above the path. Day 1 is also earlier than the first day used to fit the model. The calculation therefore should not be presented as the actual snow height; it shows that the fitted line does not provide a plausible prediction at this input.
Worked Example: Checking a Horizontal Fitted Line
Hypothetical setting. A workshop models the height \(y\), in centimeters, of a display platform using \(x\), the number of weeks since a renovation began. For a particular set of measurements, the fitted line is \(\hat{y}=18+0x\). A student wants to find the \(x\)-value where the line predicts zero height.
Check the slope before solving. The slope is \(b=0\), so the formula \(x=-a/b\) does not apply. In fact, substituting any \(x\) into this line gives:
The line is horizontal at 18 cm and never reaches zero. Since 18 cm is a possible platform height, this equation does not make an impossible-height prediction. The requested zero crossing simply does not exist.
Consider the other horizontal case. If a fitted line were instead \(\hat{y}=0+0x\), then:
In that case, every \(x\)-value is a zero crossing. The line predicts zero height everywhere, rather than crossing from positive to negative predictions at one particular \(x\).
Conclude in context. When the slope is zero, do not divide by zero to find a crossing. If the intercept is nonzero, the horizontal line never crosses zero; if both the intercept and slope are zero, the line is already at zero for every input. Checking these cases prevents an undefined calculation from being mistaken for a meaningful crossing.
Common Mistakes and AP Exam Tips
- Calling the negative value the real height. A regression calculation is not automatically a possible outcome. State that the line predicts a negative value, then explain why that value cannot represent the height in context.
- Using the zero-crossing formula when \(b=0\). The expression \(-a/b\) requires \(b\ne 0\). For a horizontal line, check the intercept instead: a nonzero intercept means no crossing, while an intercept of zero means every \(x\) is a crossing.
- Treating the zero crossing as an observed fact. It is a value calculated from the fitted equation. Unless the data include a measurement there, do not claim that the object actually reaches zero at that exact input.
- Confusing the input check with the response check. An input outside the observed range makes a prediction an extrapolation. A negative predicted height violates the response’s possible range. Mention both when both apply, without treating them as the same issue.
- Changing the prediction to zero without explanation. If the line predicts \(-15\) cm, do not report zero as the model’s prediction. Report the fitted value and say that it is not plausible. A boundary value may be meaningful in a different modeling approach, but it is not the value calculated by this fitted line.
- Forgetting context and units. A complete answer names the response and its units. A negative height above the ground is impossible in this setting, but a negative value for a variable that can take negative values would need a different assessment.
A clear AP-style conclusion might say: “The line predicts a height of \(-15\) cm at 45 hours, but a snowbank’s height above the ground cannot be negative, so this is not a plausible prediction in context. Also, 45 hours is outside the observed range.” This reports the calculation, identifies the response limit, and distinguishes the extrapolation concern.
Check Your Understanding
For each situation, calculate or assess the line’s prediction and explain what the result means in context.
- A fitted line for a snowbank’s height is \(\hat{y}=90-2x\), where \(x\) is hours after a storm and \(y\) is height in centimeters. Find the zero crossing and the predicted height at \(x=50\).
- A plant-height model is \(\hat{y}=-8+4x\), where \(x\) is weeks. Find the zero crossing and explain what a negative prediction at \(x=1\) would mean if the plant’s height is measured above the soil.
- A horizontal fitted line is \(\hat{y}=7+0x\). Does it have a zero crossing? Explain without dividing by zero.
- A horizontal fitted line is \(\hat{y}=0+0x\). How many \(x\)-values are zero crossings?
- Explain why a zero crossing calculated from a fitted line does not, by itself, prove that the actual object reaches zero height at that input.