Regression and correlation practice problems

Recall the two formulas. The slope is $b = r \cdot \frac{s_y}{s_x}$, where $s_x$ and $s_y$ are the standard deviations (typical spread) of x and y and $r$ is the correlation. The intercept is $a = \bar{y} - b\bar{x}$, where $\bar{x}$ ("x-bar") and $\bar{y}$ ("y-bar") are the means of x and y.

Compute the slope. $b = r \cdot \frac{s_y}{s_x} = 0.75 \cdot \frac{8}{2} = 0.75 \cdot 4 = 3$ kg per gram.

Compute the intercept using the slope from the previous step. $a = \bar{y} - b\bar{x} = 50 - 3 \cdot 10 = 50 - 30 = 20$ kg.

Write the line. $\hat{y} = 20 + 3x$.

Interpret the slope in context. The slope is $3$ kg per gram, so for each additional gram of weekly fertilizer, the model predicts the yield increases by 3 kg.

Check that the line passes through $(\bar{x}, \bar{y}) = (10, 50)$. At $x = 10$, $\hat{y} = 20 + 3 \cdot 10 = 50$, which matches $\bar{y}$.

Find the means. $\bar{x} = \frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5$ printers and $\bar{y} = \frac{5 + 9 + 11 + 15}{4} = \frac{40}{4} = 10$ kWh.

Find the deviations from each mean. For $x$: $2 - 5 = -3$, $4 - 5 = -1$, $6 - 5 = 1$, $8 - 5 = 3$. For $y$: $5 - 10 = -5$, $9 - 10 = -1$, $11 - 10 = 1$, $15 - 10 = 5$.

Sum the products of the paired deviations for the top of the slope. $\sum (x - \bar{x})(y - \bar{y}) = (-3)(-5) + (-1)(-1) + (1)(1) + (3)(5) = 15 + 1 + 1 + 15 = 32$.

Sum the squared x-deviations for the bottom of the slope. $\sum (x - \bar{x})^2 = (-3)^2 + (-1)^2 + 1^2 + 3^2 = 9 + 1 + 1 + 9 = 20$.

Compute the slope. $b = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2} = \frac{32}{20} = 1.6$ kWh per printer.

Compute the intercept. $a = \bar{y} - b\bar{x} = 10 - 1.6 \cdot 5 = 10 - 8 = 2$ kWh.

Write the line. $\hat{y} = 2 + 1.6x$.

Part (a): find the predicted burn time at $x = 8$. $\hat{y} = 2 + 1.5 \cdot 8 = 2 + 12 = 14$ hours.

Compute the residual, observed minus predicted. $\text{residual} = y - \hat{y} = 15 - 14 = +1$ hour.

Read the sign. The residual is positive, so the observed value sits above the line and the model underpredicted this candle's burn time by 1 hour.

Part (b): find the predicted burn time at $x = 10$. $\hat{y} = 2 + 1.5 \cdot 10 = 2 + 15 = 17$ hours.

Rearrange the residual formula to observed equals predicted plus residual. $y = \hat{y} + \text{residual} = 17 + (-2) = 15$ hours.

Read the sign. The negative residual means the model overpredicted: this 10-ounce candle burned 2 hours less than the line predicted.

Interpret the slope. Because $x$ is measured in hundreds of kilometers, the slope $-50$ means that for each additional 100 km on the odometer, the model predicts the resale price drops by 50 dollars.

Interpret the intercept. The intercept $1500$ is the predicted price at $x = 0$ (0 km). Since the data start at 500 km, $x = 0$ is outside the observed range, so this is an extrapolation, not a value read directly from the data.

Interpret the correlation. $r = -0.9$ is close to $-1$, so there is a strong negative linear association: e-bikes with more kilometers tend to sell for less.

Compute the coefficient of determination. $r^2 = (-0.9)^2 = 0.81$.

Interpret $r^2$. About 81 percent of the variation in resale price is explained by the linear relationship with odometer reading.

Part (a): compute the slope. $b = r \cdot \frac{s_y}{s_x} = 0.8 \cdot \frac{15}{5} = 0.8 \cdot 3 = 2.4$ calls per agent.

Compute the intercept. $a = \bar{y} - b\bar{x} = 60 - 2.4 \cdot 20 = 60 - 48 = 12$ calls.

Write the line. $\hat{y} = 12 + 2.4x$.

Part (b): find the predicted number of calls at $x = 25$. $\hat{y} = 12 + 2.4 \cdot 25 = 12 + 60 = 72$ calls.

Compute the residual, observed minus predicted. $\text{residual} = y - \hat{y} = 80 - 72 = +8$ calls.

Interpret the residual. It is positive, so the model underpredicted: this shift resolved 8 more calls than the line predicted for 25 agents.

Part (a): interpret the correlation. $r = 0.7$ is positive and moderately strong, so neighborhoods with more coffee shops tend to have higher average rent, and the association is roughly linear.

Part (b): compute $r^2$. $r^2 = (0.7)^2 = 0.49$.

Interpret $r^2$. About 49 percent of the variation in average rent is explained by the linear relationship with the number of coffee shops; the other 51 percent is not.

Part (c): weigh causation. The data are observational, not from an experiment, so a correlation on its own cannot establish cause.

Name a lurking variable. Something like neighborhood population density or overall desirability could raise both the number of coffee shops and the rent, producing the association without one causing the other.

Part (a): compute the slope, then the intercept. $b = r \cdot \frac{s_y}{s_x} = 0.8 \cdot \frac{20}{10} = 0.8 \cdot 2 = 1.6$ kg per cm, and $a = \bar{y} - b\bar{x} = 30 - 1.6 \cdot 40 = 30 - 64 = -34$ kg. The line is $\hat{y} = -34 + 1.6x$.

Part (b): interpret the slope. The slope $1.6$ means that for each additional centimeter of shell length, the model predicts the weight increases by 1.6 kg.

Part (c): interpret the intercept. The intercept $-34$ is the predicted weight at a shell length of 0 cm, which is a negative weight and therefore impossible. It also lies far outside the 30 to 55 cm range of the data, so it is an extrapolation with no logical meaning here.

Part (d): compute $r^2$. $r^2 = (0.8)^2 = 0.64$, so about 64 percent of the variation in turtle weight is explained by the linear relationship with shell length.

Part (e): predict at $x = 50$. $\hat{y} = -34 + 1.6 \cdot 50 = -34 + 80 = 46$ kg.

Classify the prediction. Because 50 cm lies between the smallest and largest measured shells, 30 and 55 cm, this is interpolation, so the prediction is reasonably reliable.

Part (a): interpret the slope. The slope $-0.25$ means that for each additional degree Celsius, the model predicts the dissolved oxygen falls by 0.25 mg/L.

Part (b): find the predicted value at $x = 20$. $\hat{y} = 14 - 0.25 \cdot 20 = 14 - 5 = 9$ mg/L.

Compute the residual, observed minus predicted. $\text{residual} = y - \hat{y} = 8.5 - 9 = -0.5$ mg/L.

Interpret the residual. It is negative, so the observed value sits below the line and the model overpredicted the dissolved oxygen by 0.5 mg/L at 20 degrees.

Part (c): compute $r^2$. $r^2 = (-0.95)^2 = 0.9025$, so about 90.3 percent of the variation in dissolved oxygen is explained by the linear relationship with temperature.

Part (d): read the residual plot. A U-shaped pattern is curvature, which is the direct sign that a line is not the most appropriate model, even though $r^2$ is high. A large $r^2$ does not confirm linearity; only apparent randomness in the residual plot does, so the true relationship here bends and the line misses in a systematic pattern.