Regression Analysis: Unit Diagnostic (Statistics Unit 5)

Use this diagnostic to find which Unit 5 topics to revisit: two short questions for each of Topics 5.1 to 5.5.

These are original Marlbridge practice questions, not past exam questions, with fictional data. The diagnostic is not calibrated and gives no predicted score.

How to take it. Allow 30 minutes; answer everything before opening any answer. Graphing calculator allowed. Every model ŷ = a + bx gives a predicted response; residual = observed y − predicted y. The interval of x-values in the data includes its end points.

Question 1 (multiple choice · 5.1)

Which of these is a bivariate quantitative data set?

(A) The heights of 25 sunflowers in one bed and the heights of 25 different sunflowers in another bed
(B) For each of 25 sunflowers, the hours of sunshine it received and the height it reached
(C) For each of 25 sunflowers, its flower colour (yellow, red or orange) and its height
(D) The heights of 25 sunflowers, measured on one day

Answer and explanation

Answer: (B). Two quantitative values are recorded on the same individual, so each sunflower gives one ordered pair (x, y).

(A) has two quantitative lists, but on different plants, so the values cannot be paired.
(C) pairs a categorical variable with a quantitative one.
(D) is one quantitative variable.

If you missed this: Topic 5.1 study guide.

Question 2 (short answer · 5.1)

A fictional cycling club recorded the average gradient (%) and the average speed (km/h) of ten rides.

Gradient (%)	1	2	2.5	3	4	4.5	5	6	6.5	7
Speed (km/h)	29.5	28.0	17.0	26.1	24.0	23.4	21.8	20.5	19.6	18.4

(a) Which variable belongs on the x-axis of a scatterplot? Why? (b) Describe the association in context. (c) A rider claims: “Every ride with a gradient below 3% averaged more than 25 km/h.” Is the claim supported?

Answer and explanation

(a) Gradient. The club would use the route’s gradient to predict its speed, so gradient is explanatory and speed is the response (y-axis).

(b) Apart from one ride, there is a strong, negative, linear association between gradient and average speed: rides on steeper routes tended to be slower. One ride is unusual: at 2.5% the club averaged only 17.0 km/h, far below the 26 to 28 km/h of rides with similar gradients. There are no clusters.

(c) Not supported. Three rides had gradients below 3%. Two averaged more than 25 km/h, but the 2.5% ride averaged 17.0 km/h. One counterexample is enough.

If you missed this: Topic 5.1 study guide, “Describing a scatterplot: four features”.

Question 3 (multiple choice · 5.2)

For 120 students at a fictional college, the correlation between the number of classes missed and the final mark is r = −0.91. The scatterplot is linear. Which statement is correct?

(A) Each missed class causes a student’s final mark to fall.
(B) The association is weak, because r is negative.
(C) There is a strong, negative, linear association: students who missed more classes tended to have lower final marks.
(D) 91% of students who missed many classes got low marks.

Answer and explanation

Answer: (C). The sign gives the direction; |−0.91| is close to 1, so the linear association is strong.

(A) claims cause from an observational study. Motivation, for example, could affect both attendance and marks.
(B) confuses direction with strength.
(D) reads r as a percentage of students. r is not a proportion of anything.

If you missed this: Topic 5.2 study guide.

Question 4 (multiple choice · 5.2)

Eight points lie almost exactly on a rising straight line, with r = 0.9997. A ninth point is added. Its x-value is in the middle of the others, but its y-value is far below the pattern. What happens to r?

(A) r increases, because there are more points.
(B) r stays the same, because r has no units.
(C) r becomes negative, because the new point lies below the line.
(D) r moves closer to 0, because the new point is far from the linear pattern.

Answer and explanation

Answer: (D). r is not resistant. In one fictional example of this kind, r falls from 0.9997 to 0.80.

(A) is wrong: extra points only strengthen r if they fit the pattern.
(B) mixes up two facts: r has no units, but it does depend on the data.
(C) is too strong: the other eight points still show a positive association.

If you missed this: Topic 5.2 study guide, “When r misleads (2)”.

Question 5 (multiple choice · 5.3)

A fictional park uses ŷ = 3.2 + 0.65x to predict the height (m) of a tree from its age x (years). The trees in the data were 2 to 15 years old. Which statement is correct?

(A) For a 20-year-old tree, the predicted height is 16.2 m, an interpolation.
(B) For an 8-year-old tree, the predicted height is 8.4 m, an interpolation.
(C) For a 15-year-old tree, the prediction of 12.95 m is an extrapolation, because 15 is the largest age.
(D) An 8-year-old tree will be exactly 8.4 m tall.

Answer and explanation

Answer: (B). ŷ = 3.2 + 0.65(8) = 8.4 m, and 8 lies inside 2 to 15.

(A) has the right arithmetic, but 20 is outside the data: an extrapolation.
(C) forgets that the interval includes its end points.
(D) treats ŷ as an exact value; real trees vary around the line.

If you missed this: Topic 5.3 study guide.

Question 6 (multiple choice · 5.3)

A fictional town’s population is modelled by ŷ = 8.2 + 0.31x, where x is years since 2015 and ŷ is the predicted population in thousands. The data cover 2016 to 2025. What does the model predict for 2030?

(A) 12.85 thousand; an interpolation
(B) 637.5 thousand; an extrapolation
(C) 12.85 thousand; an extrapolation, 5 years beyond the data
(D) 4.65 thousand; an extrapolation

Answer and explanation

Answer: (C). For 2030, x = 15: ŷ = 8.2 + 0.31(15) = 12.85 thousand. The data stop at x = 10, so this is an extrapolation and less reliable.

(A) misclassifies a year outside 2016 to 2025.
(B) substitutes x = 2030 instead of x = 15.
(D) leaves out the intercept 8.2.

If you missed this: Topic 5.3 study guide, “Interpolation and extrapolation”.

Question 7 (multiple choice · 5.4)

A fictional nursery predicts the price ($) of a bonsai tree from its age x (years) with ŷ = 14 + 6.2x. A 9-year-old bonsai sold for $62. Which is correct?

(A) The residual is +$7.80; the model underpredicted its price.
(B) The residual is −$7.80; the model underpredicted its price.
(C) The residual is $69.80.
(D) The residual is −$7.80; the model overpredicted its price.

Answer and explanation

Answer: (D). ŷ = 14 + 6.2(9) = $69.80. Residual = 62 − 69.80 = −$7.80. The actual price was below the prediction, so the model predicted too high.

(A) calculates predicted − observed.
(B) has the right number but the wrong word: negative means overpredicted.
(C) is the predicted price, not the residual.

If you missed this: Topic 5.4 study guide.

Question 8 (short answer · 5.4)

A fictional bakery recorded its oven temperature (°C) every 2 minutes after switching it on.

Time (min)	0	2	4	6	8	10	12
Temperature (°C)	20	75	120	152	175	190	197

Technology gives ŷ = 45.29 + 14.57x and r = 0.96.

(a) Find the residuals at 0, 6 and 12 minutes. (b) Interpret the residual at 6 minutes. (c) The other residuals are 0.57 (2 min), 16.43 (4 min), 13.15 (8 min) and −0.99 (10 min). Describe the residual plot. Is a linear model appropriate?

Answer and explanation

(a) 0 min: 20 − 45.29 = −25.29 °C. 6 min: ŷ = 132.71, so 152 − 132.71 = 19.29 °C. 12 min: ŷ = 220.13, so 197 − 220.13 = −23.13 °C.

(b) At 6 minutes the oven was 19.29 °C hotter than the model predicted, so the model underpredicted its temperature.

(c) In time order the residuals are negative, then positive from 2 to 8 minutes, then negative: an arch, not random scatter. The oven heats fast at first (55 °C in the first 2 minutes) and slowly later (7 °C in the last 2). A linear model is not the most appropriate, even though r = 0.96.

If you missed this: Topic 5.4 study guide, “Residual plots”.

Question 9 (multiple choice · 5.5)

At a fictional ski resort, the least-squares line for predicting daily lift-pass sales (hundreds) from the maximum temperature (°C) has slope −1.6 and r² = 0.81. Which is correct?

(A) r = −0.90; about 81% of the variation in daily lift-pass sales is explained by the linear relationship with maximum temperature.
(B) r = 0.90; about 81% of the variation in daily lift-pass sales is explained by the linear relationship with maximum temperature.
(C) r = −0.81; about 90% of the variation in daily lift-pass sales is explained by the linear relationship with maximum temperature.
(D) r = −0.90; the line predicts sales correctly on about 81% of days.

Answer and explanation

Answer: (A). √0.81 = 0.90, and r takes the sign of the slope.

(B) forgets the sign.
(C) swaps r and r².
(D) misreads r² as a success rate.

If you missed this: Topic 5.5 study guide.

Question 10 (short answer · 5.5)

A fictional fruit stall recorded the price of cherries ($ per kg) and the mass sold (kg) on seven days.

Price ($/kg)	4	5	5.5	6	7	8	9
Sold (kg)	48	41	40	35	30	22	19

(a) Use technology to find the least-squares regression line, r and r². (b) Interpret the slope in context. (c) Should the y-intercept be interpreted? Explain. (d) Interpret r² in context.

Answer and explanation

(a) predicted mass sold = 71.64 − 5.99 × (price), with r = −0.9944 ≈ −0.99 and r² = 0.9888.

(b) For each $1 per kg increase in price, the predicted mass of cherries sold decreases by about 5.99 kg.

(c) No. It predicts 71.64 kg sold at a price of $0. Prices in the data run from $4 to $9, so $0 is a far extrapolation, and the stall does not give cherries away.

(d) About 98.9% of the variation in the mass of cherries sold is explained by the linear relationship with price.

If you missed this: Topic 5.5 study guide, “Interpreting the slope and the y-intercept”.

Your next step

Topic	Question(s)	If you missed it, read
5.1	1, 2	Scatterplots
5.2	3, 4	Correlation
5.3	5, 6	Linear regression models
5.4	7, 8	Residuals
5.5	9, 10	Least-squares regression

How to use your result

Mark each question right, partly right or wrong. A description or interpretation with no context is only partly right.
Fix weak topics in order: residuals (5.4) need predictions from 5.3, and r² (5.5) builds on r from 5.2.
Read the study guide, then do its practice set.
Then try the Unit 5 mixed review, which combines topics.