Regression Word Problems

Show your work for each problem.

1.
The model $\hat{y} = 6x + 48$ predicts a test score $\hat{y}$ (points) from weekly study hours $x$ . What does the slope $6$ mean?
1. a)A student who studies 0 hours scores 6 points
2. b)Each extra hour of study per week goes with about 6 more points, on average
3. c)6 students took part in the study
4. d)The score is always 6 times the number of hours
Answer: Each extra hour of study per week goes with about 6 more points, on average
The slope is a rate: for each extra hour of study per week, the predicted score rises by about $6$ points, on average. The number $48$ is the intercept, not the slope.
2.
Using $\hat{y} = 6x + 48$ (test score from weekly study hours $x$ ), predict the score of a student who studies $4$ hours per week.
Answer: 72
$\hat{y} = 6 \times 4 + 48 = 24 + 48 = 72$ points.
3.
A regression line was fitted to data with $x$ values from $2$ to $10$ . For which value of $x$ is a prediction an extrapolation?
1. a)5
2. b)15
3. c)7.5
4. d)9
Answer: 15
Only $15$ lies outside the interval from $2$ to $10$ , so predicting there is extrapolation. $5$ , $7.5$ and $9$ are inside the data (interpolation).
4.
A data point has actual value $y = 80$ , and the regression line predicts $\hat{y} = 74$ for it. What is its residual?
Answer: 6
$e = y - \hat{y} = 80 - 74 = 6$ . The residual is positive, so the point lies $6$ units above the line.
5.
A regression of exam score on study time has $R^2 = 0.64$ . Which statement is correct?
1. a)64% of the data points lie exactly on the line
2. b)64% of the variation in exam scores is explained by the linear relationship with study time
3. c)The correlation coefficient is r = 0.64
4. d)Each extra hour of study adds 0.64 points
Answer: 64% of the variation in exam scores is explained by the linear relationship with study time
$R^2 = 0.64$ means $64\%$ of the variation in exam scores is explained by the linear relationship with study time. It says nothing about how many points lie on the line, and $r = \pm 0.8$ , not $0.64$ .
6.
The correlation between daily steps and calories burned is $r = 0.7$ . What is $R^2$ ? Give it as a decimal.
Answer: 0.49
$R^2 = r^2 = 0.7^2 = 0.49$ : about $49\%$ of the variation in calories burned is explained by the number of steps.
7.
A dealer's model for used cars is $\hat{y} = 18000 - 1500x$ , where $\hat{y}$ is the price in dollars and $x$ the age in years (data for cars from 0 to 8 years old). What does the intercept $18000$ mean?
1. a)A new car (age 0) is predicted to cost about 18000 dollars
2. b)A new car is worth 1500 dollars
3. c)After 18000 years the car is worth nothing
4. d)The car loses 18000 dollars each year
Answer: A new car (age 0) is predicted to cost about 18000 dollars
At $x = 0$ (a new car) the model predicts $\hat{y} = 18000$ dollars. Age $0$ is part of the data, so this interpretation makes sense. The $1500$ is the slope: the predicted value drops by about $1500$ dollars per year.
8.
A café uses $\hat{y} = 110 - 3x$ to predict the number of hot drinks sold from the morning temperature $x$ in °C (data from $0$ to $20$ °C). How many hot drinks does it predict for $12$ °C?
Answer: 74
$\hat{y} = 110 - 3 \times 12 = 110 - 36 = 74$ hot drinks. Since $12$ °C is inside the data range, this is an interpolation.
9.
In a model of delivery times, one delivery has a residual of $-5$ minutes. What does this mean?
1. a)The delivery took 5 minutes longer than predicted
2. b)The model predicted a time of 5 minutes
3. c)The slope of the model is -5
4. d)The delivery took 5 minutes less than predicted
Answer: The delivery took 5 minutes less than predicted
$e = y - \hat{y} = -5$ means $y = \hat{y} - 5$ : the delivery took $5$ minutes less than predicted, so the point lies below the line.
10.
A plant's height is modelled by $\hat{y} = 2.5x + 4$ (cm, $x$ in weeks). By how many cm does the predicted height increase over $4$ more weeks?
Answer: 10
Change in $\hat{y}$ = slope $\times$ change in $x$ = $2.5 \times 4 = 10$ cm.
11.
The café model is $\hat{y} = 110 - 3x$ (hot drinks, $x$ in °C). On a $15$ °C morning the café sold $75$ hot drinks. Find the residual of this morning.
Answer: 10
1. What is $3 \times 15$ ? 45
2. What is the predicted number of hot drinks $\hat{y}$ ? 65
3. What is the residual $y - \hat{y}$ ? 10
12.
For the study-hours data, the sum of squared residuals is $SS_{\text{res}} = 40$ and the total variation of the scores is $SS_{\text{tot}} = 400$ . Find $R^2$ and say what percentage of the variation in scores the line explains.
Answer: 0.9
1. What is $\frac{SS_{\text{res}}}{SS_{\text{tot}}} = \frac{40}{400}$ ? 0.1
2. What is $R^2 = 1 - \frac{SS_{\text{res}}}{SS_{\text{tot}}}$ ? 0.9
3. What percentage of the variation does the line explain (number only)? 90
13.
The model $\hat{y} = 6x + 48$ was fitted to students who study from $1$ to $5$ hours per week (tests out of $100$ points). Predict the score for $12$ hours and decide what kind of prediction this is.
Answer: 120
1. What is $6 \times 12$ ? 72
2. What score does the model predict for 12 hours? 120
3. Is this prediction an interpolation or an extrapolation? extrapolation
14.
The dealer's model is $\hat{y} = 18000 - 1500x$ (price in dollars, $x$ = age in years, data from 0 to 8 years). A $4$ -year-old car was sold for 13000 dollars. Find the predicted price and the residual. Was the car sold above or below the predicted price?
Answer: 1000
1. What is $1500 \times 4$ ? 6000
2. What is the predicted price $\hat{y}$ (number only)? 12000
3. What is the residual (number only)? 1000
4. Was the car sold above or below the predicted price? above
15.
A student counts the push-ups $y$ she can do after $x$ weeks of training:
$x$ 1 2 3 4
$y$ 3 5 6 10
Find the least-squares regression line $\hat{y} = mx + b$ .
Answer: y = 2.2x + 0.5
1. What is the mean $\bar{x}$ ? 2.5
2. What is the mean $\bar{y}$ ? 6
3. What is $\sum(x-\bar{x})(y-\bar{y})$ ? 11
4. What is $\sum(x-\bar{x})^2$ ? 5
5. What is the slope $m$ ? 2.2
6. What is the intercept $b = \bar{y} - m\bar{x}$ ? 0.5
16.
The café model is $\hat{y} = 110 - 3x$ (hot drinks, $x$ in °C, data from $0$ to $20$ °C), with $R^2 = 0.9$ . On an $8$ °C morning the café sold $90$ hot drinks. Find the prediction, the residual, the correlation coefficient $r$ and the percentage of variation explained.
Answer: -0.95
1. What does the model predict for $8$ °C? 86
2. What is the residual of this morning? 4
3. What is $r$ , rounded to two decimals? -0.95
4. What percentage of the variation in hot drinks sold is explained (number only)? 90
17.
An ice-cream stand collected data for temperatures from $10$ to $30$ °C. The regression line of ice creams sold per hour on temperature has slope $1.5$ , and the means are $\bar{x} = 20$ °C and $\bar{y} = 50$ ice creams. Find the intercept, predict sales at $30$ °C, and find the residual of an hour at $30$ °C with $61$ ice creams sold.
Answer: -4
1. What is $1.5 \times 20$ ? 30
2. What is the intercept $b = \bar{y} - m\bar{x}$ ? 20
3. What does the line predict at $30$ °C? 65
4. What is the residual of the hour with $61$ ice creams? -4
18.
With $\hat{y} = 6x + 48$ (test score from weekly study hours, data from $1$ to $5$ hours), how many hours of study does the model say are needed for a predicted score of $84$ ? Can this answer be trusted?
Answer: 6
1. Subtract the intercept: what is $84 - 48$ ? 36
2. Divide by the slope: what is $x$ ? 6
3. Is $x = 6$ an interpolation or an extrapolation? extrapolation