Computer regression output practice problems

(a) Find the column that holds the equation. Only the Coef column feeds the line. SE Coef, T, and P describe how precisely each coefficient was estimated and never appear in the equation itself.

(a) Read the two coefficients. The Constant row gives the y-intercept $a = 12.50$ dollars, and the Hours row gives the slope $b = 6.40$ dollars per hour.

(a) Write the line with the variables named. Let $x$ be the length of a firing in hours and $\hat{y}$ (read "y-hat") the predicted electricity cost in dollars. Then $\hat{y} = 12.50 + 6.40x$.

(a) Sanity-check the line at a firing length inside the data. At $x = 8$ hours, $\hat{y} = 12.50 + 6.40 \times 8 = 12.50 + 51.20 = 63.70$ dollars, a plausible electricity cost for one firing.

(b) Name the cells. The intercept is the Coef entry on the Constant row, and the slope is the Coef entry on the row labeled with the explanatory variable, here Hours. Software labels that row with the variable's name, not with the word slope, which is why the row heading matters.

(c) Locate $x = 0$ relative to the data. The firings ran from 4 to 12 hours, so $x = 0$ sits 4 hours below the shortest firing in the data, making any statement about it an extrapolation.

(c) Add the context problem. A firing of zero hours is not a firing, so there is no real quantity for 12.50 dollars to measure. The intercept is the number that positions the line over the 4 to 12 hour range, not a startup cost the studio pays.

(a) Pull the slope out of the output. It is the Coef entry on the Dogs row, $b = 3.75$, and its units are the units of $y$ over the units of $x$, so gallons per dog.

(a) Assemble the four required pieces: the word predicted, a one-unit increase in the explanatory variable, the phrase on average, and the units of the response in context.

(a) Write the sentence. For each additional dog boarded, the predicted daily water use increases by 3.75 gallons, on average.

(b) Name the first omission. Dropping predicted turns a statement about the fitted line into a claim about what any particular dog actually does. The line describes an average pattern across 27 days, and no individual dog was measured drinking or being washed with 3.75 gallons.

(b) Point at the output for the evidence. S = 18.41 gallons is the standard deviation of the residuals, so real days miss the line by about 18 gallons in either direction. A shortcut that promises 3.75 gallons per dog hides that spread.

(c) Scale the slope. The slope is the predicted change for a one-unit increase, so two more dogs predict $2 \times 3.75 = 7.50$ more gallons.

(c) Note what you did not do. You never needed the intercept 41.00, because a change in the prediction depends only on the slope and the change in $x$.

(a) Build the equation from the Coef column. The Constant row gives $a = 15.50$ and the Days row gives $b = 0.28$, so $\hat{y} = 15.50 + 0.28x$, with $x$ the days since veraison and $\hat{y}$ the predicted sugar content in degrees Brix.

(a) Substitute $x = 30$. First the slope term: $0.28 \times 30 = 8.40$.

(a) Add the intercept. $\hat{y} = 15.50 + 8.40 = 23.90$ degrees Brix.

(b) Compare 30 with the range of the data. The sampled dates ran from 10 to 40 days, and $10 < 30 < 40$, so predicting at 30 days is interpolation and the prediction is reasonably trustworthy.

(c) Run the arithmetic anyway. $0.28 \times 60 = 16.80$, so $\hat{y} = 15.50 + 16.80 = 32.30$ degrees Brix.

(c) Locate 60 relative to the data. It sits 20 days beyond the largest $x$ in the sample, so this is extrapolation, and the further you go past the data the less reliable the predicted value becomes.

(c) Add the context check that makes the point concrete. The line keeps adding 0.28 Brix every day forever, but real grapes stop accumulating sugar and eventually raisin on the vine, so 32.30 Brix is the line's arithmetic rather than a defensible prediction. The regression carries no information about days the vineyard never sampled.

(a) Write the line from the Coef column. $\hat{y} = 18.00 + 1.50x$, with $x$ the load weight in pounds and $\hat{y}$ the predicted drying time in minutes.

(a) Predict at the observed weight. $1.50 \times 40 = 60.00$, so $\hat{y} = 18.00 + 60.00 = 78.00$ minutes.

(a) Apply the residual definition, observed minus predicted. $\text{residual} = y - \hat{y} = 70 - 78.00 = -8.00$ minutes.

(b) Read the sign. The residual is negative, so the observed drying time sits below the line: the model overpredicted this load by 8 minutes, and the load dried faster than the line said it would.

(c) Find the number that sets the scale for a residual. S = 8.176 minutes is the standard deviation of the residuals, the typical distance between an observed drying time and the predicted one.

(c) Compare. The residual $-8.00$ minutes is slightly smaller in size than S = 8.176 minutes, so this load misses by about one typical amount. Nothing marks it as an outlier; a model with this much scatter produces misses of roughly 8 minutes routinely.

(a) Recognize what R-Sq is. R-Sq is the coefficient of determination $r^2$ (r-squared) written as a percent, so $r^2 = 50.4\% = 0.504$.

(a) Take the square root to get the size of the correlation. $\sqrt{0.504} = 0.7099$, so the correlation has size 0.71 to two decimals. Check it back: $0.71^2 = 0.5041$, which is 50.41 percent and prints as 50.4 percent, matching the output.

(a) Attach the sign from the slope. The Coef entry on the Stops row is $-2.40$, which is negative, so $r = -0.71$.

(b) Say why the slope decides. The slope and the correlation are linked by $b = r \cdot \frac{s_y}{s_x}$, where $s_y$ and $s_x$ are the standard deviations of the response and the explanatory variable. Standard deviations are never negative, so $b$ and $r$ always carry the same sign.

(b) Say why R-Sq cannot decide. Squaring throws the sign away: both $0.71$ and $-0.71$ square to $0.5041$. R-Sq is therefore consistent with a strong positive or a strong negative association, and you need the slope, the T value, or the scatterplot to break the tie.

(c) Check the analyst's claim. SE Coef is a standard error, which is a measure of spread and so is always positive, and $T = \frac{\text{Coef}}{\text{SE Coef}}$. Dividing by a positive number cannot change a sign, so T always matches the slope: $\frac{-2.40}{0.4346} = -5.52$. The analyst is right, and the slope, the T value, and $r$ all share one sign.

Finish in context. $r = -0.71$ is a moderately strong negative linear association: routes with more stops per mile tend to run at lower average speeds.

(a) Identify the statistic. R-Sq = 88.4% is the coefficient of determination $r^2$ (r-squared), printed as a percent, and it is a proportion of variation, not of anything else.

(a) Name the two variables in the right roles. The response is the monthly water use and the explanatory variable is the number of residents.

(a) Write the interpretation. About 88.4 percent of the variation in monthly water use among these 18 dorm buildings is explained by the linear relationship with the number of residents.

(b) Subtract from 100. $100 - 88.4 = 11.6$ percent of the variation in water use is left unexplained by this line, and it comes from everything else that differs between buildings: fixture age, whether there is a laundry room, irrigation, and how much any group of students happens to use.

(c) Correct the first student. R-Sq counts variation, not points. It says nothing about how many buildings sit on the line, and in fact a data set can have R-Sq of 88.4 percent with no building landing exactly on it. The statistic that measures how close the buildings sit is S = 11.13 thousand gallons.

(c) Correct the second student. The missing words are the variation in. The line does not explain 88.4 percent of a building's water use; a 400-resident dorm would use a great deal of water under any model. What the line explains is 88.4 percent of the differences in water use from building to building.

(a) Read the line under the table. S = 4.765 is the standard deviation of the residuals, written $s$ in most textbooks. It is computed from the residuals, so it carries the units of the response variable: minutes.

(b) Interpret it as a typical miss. Actual taxi-out times sit about 4.765 minutes away from the times this line predicts, on average, so roughly 4.8 minutes is the size of a typical prediction error for this model.

(c) Identify 0.3052 correctly. It is the SE Coef entry on the Queue row, the standard error of the estimated slope. It measures how much the estimated slope of 2.20 would move from one sample of 40 departures to another, and its units are minutes per aircraft.

(c) Separate the two ideas. SE Coef describes the uncertainty in a coefficient, while S describes how far individual departures fall from the fitted line. They answer different questions, and 0.3052 is far too small to be a plausible error for a taxi-out time measured in tens of minutes.

(d) Build the line and predict. $\hat{y} = 9.50 + 2.20x$, so at $x = 8$, $2.20 \times 8 = 17.60$ and $\hat{y} = 9.50 + 17.60 = 27.10$ minutes.

(d) Attach the typical error. Departures with 8 aircraft ahead are predicted to taxi out for 27.10 minutes, and their actual times typically land within about 4.8 minutes of that, so times in the low 20s or low 30s are ordinary rather than surprising.

(a) Read the Coef column. The Constant row gives $a = 82.00$ and the Wind row gives $b = -3.50$. Let $x$ be the midday wind speed in mph and $\hat{y}$ the predicted number of kayak rentals, so $\hat{y} = 82.00 - 3.50x$.

(b) Substitute $x = 14$. $3.50 \times 14 = 49.00$, so $\hat{y} = 82.00 - 49.00 = 33.00$ rentals. Note that 14 mph sits inside the observed 3 to 22 mph range, so this is interpolation.

(c) Take observed minus predicted. $\text{residual} = y - \hat{y} = 41 - 33.00 = +8.00$ rentals.

(c) Read the sign. The residual is positive, so that day sits above the line: the outfitter rented 8 more kayaks than the model predicted for 14 mph wind, and the model underpredicted the day.

(d) Convert R-Sq to $r$, size first. $r^2 = 75.7\% = 0.757$ and $\sqrt{0.757} = 0.8701$, so the correlation has size 0.87. Check it back: $0.87^2 = 0.7569$, which prints as 75.7 percent.

(d) Fix the sign from the slope. The slope $-3.50$ is negative, so $r = -0.87$: a strong negative linear association, with windier days renting fewer kayaks.

(e) Interpret R-Sq with the word variation. About 75.7 percent of the variation in daily kayak rentals across these 35 days is explained by the linear relationship with midday wind speed, leaving 24.3 percent to weather other than wind, day of the week, holidays, and everything else the model ignores.

(f) Compare the residual with S. S = 8.053 rentals is the standard deviation of the residuals, so a typical day misses the line by about 8 rentals. The residual of $+8.00$ is right at that size, so this day is an ordinary miss rather than an unusual one.