Distinguish between interpolation and extrapolation when using a regression line, and assess the reliability of predictions

(a) $x = 10$ is inside the data range $[2, 20]$, so this is interpolation. Predictions inside the range are generally reliable, assuming the regression line is a good fit.

(b) $x = 25$ is outside the data range, so this is extrapolation. The prediction is less reliable because the linear relationship may not extend beyond the observed range.

(c) $x = -3$ is well outside the data range and may also be physically implausible depending on context. This is extrapolation and is the least reliable prediction.

Markers reward correct classification and a brief reliability comment for each.

Substituting $x = 2080$ into the line gives a numerical prediction (the question's specific equation would supply the number).

This is an extreme extrapolation, $60$ years beyond the latest data point. Possible reasons the prediction may be wrong: changes in regional employment, climate impact on agriculture, policy changes, immigration patterns, or saturation of available housing.

Linear growth over $60$ years extrapolated from $30$ years of data is rarely reliable. State the predicted value but caveat strongly that the model assumes the past linear trend continues indefinitely, which is unlikely.

Markers reward identification as extrapolation, the prediction, and at least two specific reasons it may be unreliable.

State the data range

The data covers $x = 5$ to $x = 40$, so any prediction with $x$ inside $[5, 40]$ interpolates and anything outside extrapolates.

Locate the prediction

Here $x = 22$, and $5 \le 22 \le 40$, so this value lies inside the observed range.

Answer

$x = 22$ is inside the data range, so the prediction is interpolation.

State the data range. The data covers $x = 5$ to $x = 40$. Inside $[5, 40]$ is interpolation; outside is extrapolation.

Classify each value.

• $x = 55$: this is greater than $40$, so it lies above the range. This is extrapolation.
• $x = 2$: this is less than $5$, so it lies below the range. This is extrapolation.

Answer: both predictions are extrapolation, because $x = 55$ and $x = 2$ each lie outside the data range $[5, 40]$.

Substitute to get the prediction. Put $x = 9$ into the line:

State the data range and classify. The data covers $x = 4$ to $x = 16$, and $4 \le 9 \le 16$, so $x = 9$ is inside the range.

Answer: the prediction is $y = 37$, and because $x = 9$ is inside the data range it is interpolation.

State the data range. The data covers ages $2$ to $10$, so ages inside $[2, 10]$ interpolate and ages outside extrapolate.

Classify each age.

• Age $6$: since $2 \le 6 \le 10$, this is inside the range, so interpolation.
• Age $14$: since $14 > 10$, this is outside the range, so extrapolation.

Answer: age $6$ is interpolation (inside the data range) and age $14$ is extrapolation (outside the data range).

Convert the years and note the data range. With $t$ measured from $2000$, the data covers $t = 0$ to $t = 15$ (the years $2000$ to $2015$). The year $2010$ is $t = 10$ and $2035$ is $t = 35$.

Predict and classify $2010$. Substitute $t = 10$:

Since $0 \le 10 \le 15$, this is interpolation.

Predict and classify $2035$. Substitute $t = 35$:

Since $35 > 15$, this is extrapolation.

Answer: the $2010$ prediction of $5400$ is interpolation; the $2035$ prediction of $8400$ is extrapolation, $20$ years beyond the data.

State the data range. The data covers months $1$ to $18$, so a month inside $[1, 18]$ interpolates and a month outside extrapolates.

Predict and classify month $12$. Substitute $m = 12$:

Since $1 \le 12 \le 18$, this is interpolation, so the predicted $$14600$ is reliable.

Predict month $48$ and assess reliability. Substitute $m = 48$:

Since $48 > 18$, this is extrapolation. The line assumes revenue keeps climbing by $$800$ every month forever, but the cafe will eventually saturate its local market and growth will flatten, so the predicted $$43400$ is likely to overstate revenue.

Answer: month $12$ gives $$14600$ (interpolation, reliable); month $48$ gives $$43400$ (extrapolation, unreliable because of market saturation).

State the data range. The data covers ages $3$ to $11$, so ages inside $[3, 11]$ interpolate.

Predict and classify age $7$. Substitute $a = 7$:

Since $3 \le 7 \le 11$, this is interpolation, so $118$ cm is reliable.

Predict age $30$ and assess. Substitute $a = 30$:

Since $30 > 11$, this is substantial extrapolation. The model keeps adding $6$ cm every year, but human growth stops in the late teens (a biological limit), so a predicted height of $256$ cm is impossible.

Answer: age $7$ gives $118$ cm (interpolation, reliable); age $30$ gives $256$ cm, an unreliable extrapolation because growth stops in adolescence.

State the data range. The rainfall data covers $20$ to $60$ cm, so a value inside $[20, 60]$ interpolates.

Predict and classify $45$ cm. Substitute $w = 45$:

Since $20 \le 45 \le 60$, this is interpolation.

Predict and classify $15$ cm. Substitute $w = 15$:

Since $15 < 20$, this is extrapolation.

Compare reliability. The $45$ cm prediction is more reliable because it interpolates. The $15$ cm prediction extrapolates below the data, where a drought-level rainfall could stunt the crop far more than the line suggests, so $16$ tonnes per hectare may be too high.

Answer: $45$ cm gives $43$ t/ha (interpolation, more reliable); $15$ cm gives $16$ t/ha (extrapolation, less reliable because severe water shortage breaks the linear trend).

Note the data range. With $t$ measured from $1995$, the data covers $t = 0$ to $t = 25$. The year $2015$ is $t = 20$ and $2045$ is $t = 50$.

Predict and classify $2015$. Substitute $t = 20$:

Since $0 \le 20 \le 25$, this is interpolation.

Predict $2045$. Substitute $t = 50$:

Since $50 > 25$, this is extrapolation, $25$ years past the last data point.

Compare reliability. The $2015$ figure is more reliable because it interpolates. The $2045$ figure of $20500$ assumes the past linear trend simply continues; in reality the town could stall if a major employer closes, or stop growing once available housing land runs out.

Answer: $2015$ gives $13000$ (interpolation, reliable); $2045$ gives $20500$ (substantial extrapolation, unreliable because of possible employer loss and housing saturation).

State the data range. The data covers weeks $2$ to $26$, so a week inside $[2, 26]$ interpolates.

Predict and classify week $20$. Substitute $w = 20$:

Since $2 \le 20 \le 26$, this is interpolation, so $$8800$ is reliable.

Predict and classify week $52$. Substitute $w = 52$:

Since $52 > 26$, this is extrapolation.

Compare mild and substantial extrapolation. Week $28$ is only $2$ weeks past the data, so mild extrapolation; it is defensible because the trend has little room to drift. Week $52$ is $26$ weeks past, doubling the time span, so the prediction relies far more heavily on the unverified assumption that linear growth continues, and the local market is likely to saturate first.

Answer: week $20$ gives $$8800$ (interpolation); week $52$ gives $$20000$ (substantial extrapolation, far less reliable than a mild extrapolation to week $28$).

State the data range. The data covers ages $4$ to $12$, so ages inside $[4, 12]$ interpolate.

Predict and classify age $9.5$. Substitute $a = 9.5$:

Since $4 \le 9.5 \le 12$, this is interpolation, so $134.25$ cm is reliable.

Classify age $13$. Age $13$ is just $1$ year past the range, so mild extrapolation. It could be defensible, since a $13$ year old is still growing and the trend is unlikely to change sharply in one year. (For reference, the line gives $5.5 \times 13 + 82 = 153.5$ cm.)

Predict age $40$ and assess. Substitute $a = 40$:

Since $40$ is far above $12$, this is substantial extrapolation. Growth stops in the late teens (a biological limit), yet the line keeps adding $5.5$ cm per year, so the predicted $302$ cm is impossible.

Answer: age $9.5$ gives $134.25$ cm (interpolation); age $13$ is mild extrapolation, possibly defensible; age $40$ gives $302$ cm, an unreliable extrapolation because human growth ceases in adolescence.

State the data range. The temperature data covers $18$ to $38$ degrees, so a value inside $[18, 38]$ interpolates.

Predict and classify $30$ degrees. Substitute $T = 30$:

Since $18 \le 30 \le 38$, this is interpolation, so $60$ MW is reliable.

Predict and classify $44$ degrees. Substitute $T = 44$:

Since $44 > 38$, this is extrapolation, $6$ degrees beyond the data.

Assess the $44$ degree prediction. On extreme heat days demand can rise faster than the line predicts, because almost every air conditioner runs at once, so the linear rate of $1.4$ MW per degree may understate true demand. At only $6$ degrees beyond a $20$ degree data span this is mild extrapolation, so the figure is a rough guide rather than worthless.

Answer: $30$ degrees gives $60$ MW (interpolation, reliable); $44$ degrees gives $79.6$ MW (mild extrapolation, possibly an underestimate because demand surges non-linearly in extreme heat).