Unit 5 test

AP Statistics Unit 5 Test

15 multiple-choice questions and 1 free-response question, every one from Unit 5: Regression Analysis, written in the format of the AP Statistics Exam.

Take the multiple choice on its own, timed at the exam's pace or in practice mode with each answer checked as you give it. Or sit the whole test, free response included, with each section on its own clock; the free-response scoring guide opens when that section ends.

AP Statistics Unit 5 Test

15 multiple-choice and 1 free-response question, a shorter set laid out like the AP Statistics Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time
Section I: Multiple Choice	15 of the exam's 42	32 minutes 9 seconds
Section II: Free Response	1 of the exam's 4	22 minutes 30 seconds

A graphing calculator with statistical capabilities is expected on both sections.

One section at a time. When a section's clock reaches zero, the section ends and is scored.
Multiple choice is scored when its section ends, with an explanation for every choice.
Free response is never scored by the page. When the section ends, you see the scoring guide and a model response for each part and tick the points your answer earns.
At the end you get your score and the time you used for each section.
Want the answers as you go? Pick Study mode below. Each multiple-choice answer is checked as soon as you give it, with its explanation, and each free-response question has a button that opens its scoring guide.
Your answers are kept in this browser tab if the page reloads, and nowhere else.

Time for each section Standard time Time and a half Double time Untimed Study mode

Choose before Section I starts. Time and a half and double time match common AP testing accommodations; untimed shows the time you use without ending a section. Study mode is untimed and shows each answer as you go.

Section I: Multiple Choice

15 questions · 32 minutes 9 seconds

A school district's technology office wants to use the number of devices connected to a school's Wi-Fi network to predict the network's average download speed. At 40 randomly chosen moments during the school day, staff recorded both values. Which describes the graph the office should make?
Answer: C.
1. A
  This puts the variables on the wrong axes. The explanatory variable, the one used to predict, goes on the horizontal axis. Here that is the number of devices connected, and download speed is the response.
2. B
  Two histograms show each variable on its own, so the pairing is lost: you cannot see which speed went with which number of devices. A scatterplot keeps each moment's two values together as one point.
3. C
  Correct. The number of devices connected is the explanatory variable, so it goes on the horizontal axis, and download speed, the response, goes on the vertical axis. Each of the 40 moments gives one ordered pair and one point.
4. D
  Sorting the two lists separately breaks the pairs. Each point has to show the two values recorded at the same moment, and after sorting, a device count would be matched with a speed from a different moment.
An engineer recorded the wind speed, in meters per second, and the power output, in kilowatts, of a small wind turbine at 10 moments.
$\begin{matrix} Wind speed & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ Power output & 0.4 & 0.9 & 1.9 & 3.1 & 5.0 & 7.4 & 10.4 & 14.3 & 18.8 & 24.6 \end{matrix}$
Which best describes a scatterplot of power output against wind speed?
Answer: A.
1. A
  Correct. The points follow a smooth pattern closely, so the association is strong, and power output increases as wind speed increases, so it is positive. The pattern bends: from 3 to 4 meters per second the output rises 0.5 kilowatt, and from 11 to 12 it rises 5.8 kilowatts, so the form is curved and gets steeper.
2. B
  Rising every time is a statement about direction, not form. A linear form needs roughly equal rises for equal steps in wind speed, and here the rise per meter per second grows from 0.5 to 5.8 kilowatts.
3. C
  Strength is how closely the points follow the pattern, whatever its shape. These points follow a smooth curve with almost no scatter, so the association is strong even though it is not linear.
4. D
  The curve bends the other way. Leveling off would mean smaller and smaller rises, but the rises grow, from 0.5 kilowatt at the low end to 5.8 kilowatts at the high end.
An electric utility recorded the electricity used in January 2025 and in January 2026, in kilowatt-hours, by 12 households.
$\begin{matrix} 2025 & 610 & 742 & 815 & 880 & 934 & 1,012 & 1,086 & 1,150 & 1,233 & 1,305 & 1,390 & 1,478 \\ 2026 & 588 & 760 & 781 & 851 & 960 & 975 & 1,049 & 1,182 & 1,190 & 1,271 & 1,421 & 1,436 \end{matrix}$
A scatterplot shows the 2025 use on the horizontal axis and the 2026 use on the vertical axis, with the line $y = x$ drawn on it. The utility claims that most of these 12 households used less electricity in January 2026 than in January 2025. Which statement is best supported by the scatterplot?
Answer: D.
1. A
  A positive association says that households using more in 2025 tended to use more in 2026. It says nothing about whether use went up or down from one year to the next. That is shown by where each point sits relative to the line $y = x$ .
2. B
  With 2026 on the vertical axis, a point above the line $y = x$ is a household that used more in 2026. Only 4 households did that. The other 8 points lie below the line.
3. C
  The line $y = x$ compares each household with itself, which the scatterplot is set up to do. The means (about 1,053 kilowatt-hours in 2025 and 1,039 in 2026) could drop because of a few households, and they cannot say how many households used less.
4. D
  Correct. A point below the line $y = x$ has a 2026 value smaller than its 2025 value. Eight of the 12 households are below the line, so most of them used less electricity in January 2026, which supports the claim.
For 50 counties in one state, the correlation between the number of public charging stations for electric cars per 10,000 residents and the county's median household income is $r = 0.74$ . A county commissioner says, "Building more charging stations will raise incomes in our county." Which response is best?
Answer: B.
1. A
  No correlation, however strong, shows cause by itself. These counties were observed, not assigned numbers of stations, so the association could come from something else that goes with both variables.
2. B
  Correct. This is an observational study. Counties that are more urban, for example, tend to have higher incomes and also more electric cars and more charging stations. A correlation does not show that changing the number of stations would change incomes.
3. C
  This reads $r$ as if it were $r^{2}$ and then adds cause. Even $r^{2} = {0.74}^{2} \approx 0.55$ would describe variation explained by a linear relationship, not variation caused by the stations.
4. D
  Reversing the direction of cause is still a causal claim, and these data cannot support it either. Higher incomes might lead to more stations, but an observational correlation cannot show that, or rule out other explanations.
For 30 tablet models, the correlation between screen size, in inches, and weight, in grams, is $r = 0.84$ . A reviewer converts every screen size to centimeters (1 inch is 2.54 centimeters) and every weight to ounces (1 ounce is 28.35 grams). Then the reviewer makes a new scatterplot with weight on the horizontal axis and screen size on the vertical axis. What is the correlation for the new scatterplot?
Answer: D.
1. A
  Switching which variable goes on which axis does not change the direction of the association. Heavier tablets still tend to have larger screens, so $r$ stays positive.
2. B
  This applies both conversion factors to $r$ : $0.84 \times \frac{2.54}{28.35} \approx 0.08$ . The correlation has no units, so changing the units of either variable leaves it alone.
3. C
  This divides $r$ by the inch-to-centimeter factor: $\frac{0.84}{2.54} \approx 0.33$ . Correlation is unit-free, and multiplying every value of a variable by a positive number does not change it.
4. D
  Correct. The correlation has no units, so converting inches to centimeters and grams to ounces does not change it. It also measures the association between the two variables the same way whichever one is on the horizontal axis, so $r$ is still 0.84.
A rowing club charges each member a season fee of 40 dollars plus 6 dollars for every practice session the member attends. Last season its 25 members attended from 4 to 25 sessions each. The treasurer made a scatterplot of the total amount each member paid against the number of sessions the member attended. Which statement about the correlation for these data is correct?
Answer: A.
1. A
  Correct. Every member's total is exactly $40 + 6 x$ dollars for $x$ sessions, so all 25 points lie on one line that rises. A perfect positive linear association gives $r = 1$ .
2. B
  A linear relationship does not have to pass through the origin. The fee shifts the line up by 40 dollars, but every point is still on the line, so the correlation is still 1.
3. C
  The fee is the same for everyone, so it does not change which members paid more. The amount paid rises by 6 dollars with every extra session, so the association is perfect, not absent.
4. D
  The correlation is unit-free and can be found for any two quantitative variables, whatever their units. Here it is exactly 1.
A sports nutritionist measured 40 adults whose lean body mass ranged from 38 to 72 kilograms. The least-squares line for predicting resting metabolic rate, in calories per day, from lean body mass $x$ , in kilograms, is $\hat{y} = 370 + 21.6 x$ . What is the predicted resting metabolic rate for an adult with a lean body mass of 55 kilograms?
Answer: B.
1. A
  $21.6 (55) = 1,188$ leaves out the intercept. The prediction starts at 370 and adds the slope times $x$ .
2. B
  Correct. Substitute 55 for $x$ : $\hat{y} = 370 + 21.6 (55) = 370 + 1,188 = 1,558$ calories per day. Since 55 kilograms is inside the 38 to 72 kilograms in the data, this is an interpolation.
3. C
  This swaps the roles of the two numbers, as if the line were $\hat{y} = 21.6 + 370 x$ . In $\hat{y} = a + b x$ the number multiplied by $x$ is the slope, 21.6.
4. D
  This adds 370 and 21.6 before multiplying: $(370 + 21.6) (55) = 21,538$ . Only the slope is multiplied by $x$ .
A cloud storage company timed 60 uploads of files from 0.2 to 6.5 gigabytes on a standard home connection. The least-squares line for predicting upload time, in minutes, from file size $x$ , in gigabytes, is $\hat{y} = 0.8 + 2.35 x$ . One gigabyte is 1,000 megabytes. What is the predicted upload time for an 850-megabyte file?
Answer: B.
1. A
  $2.35 (0.85) \approx 2.00$ leaves out the intercept. The line predicts $0.8 + 2.00 = 2.80$ minutes.
2. B
  Correct. The line uses gigabytes, so 850 megabytes is $\frac{850}{1,000} = 0.85$ gigabyte. Then $\hat{y} = 0.8 + 2.35 (0.85) \approx 2.80$ minutes.
3. C
  This converts 850 megabytes to 8.5 gigabytes by dividing by 100. One gigabyte is 1,000 megabytes, so the file is 0.85 gigabyte, and 8.5 would also be beyond the 6.5 gigabytes in the data.
4. D
  This puts 850 into the equation as if $x$ were in megabytes. The line was fit with $x$ in gigabytes, so $x = 0.85$ .
Questions 9 to 11 refer to the following.
Hydrologists recorded, for each of 22 years, the snowpack at a mountain site on April 1, measured in inches of water, and the mean flow of the river below the site in June, in cubic meters per second. The snowpack in those years ranged from 8.4 to 31.6 inches. Computer output for the least-squares regression of mean June flow on April 1 snowpack is shown.
$\begin{matrix} Predictor & Coef \\ Constant & 5.55 \\ Snowpack & 1.758 \\ R-Sq = 93.3 % \end{matrix}$

This year only 4.5 inches of snowpack had built up at the site by April 1, the lowest amount ever recorded there. The hydrologists want a forecast of this June's mean flow from the line. What should they report?
Answer: C.
1. A
  $1.758 (4.5) \approx 7.9$ leaves out the intercept. The prediction is $5.55 + 1.758 (4.5) \approx 13.5$ .
2. B
  Interpolation means the value of $x$ lies inside the interval of $x$ -values used to fit the line, which here is 8.4 to 31.6 inches. That interval does not start at 0, so 4.5 inches is outside it.
3. C
  Correct. $\hat{y} = 5.55 + 1.758 (4.5) \approx 13.5$ cubic meters per second. The snowpacks in the data ran from 8.4 to 31.6 inches, so 4.5 inches is beyond them, the prediction is an extrapolation, and it is less reliable.
4. D
  R-Sq describes how well the line fits the years that were observed. It cannot say whether the straight-line pattern continues below 8.4 inches, where there are no data.
Uses the material above question 9.

In one of the 22 years, the snowpack was 20.4 inches and the residual for that year was $- 4.7$ cubic meters per second. What does this residual tell you about that year?
Answer: A.
1. A
  Correct. A residual is observed minus predicted. A residual of $- 4.7$ means the observed flow, 36.7, was 4.7 below the predicted flow of about 41.4 cubic meters per second, so the line overestimated that year's flow.
2. B
  This reverses the sign. A negative residual means the observed value is below the prediction, so the flow was less than predicted, not more.
3. C
  This swaps the variables. The line predicts June flow from snowpack, so the residual is measured in the units of flow, the response, and compares flows, not snowpacks.
4. D
  A residual compares an observation with the line's prediction for its own $x$ -value, not with the mean of all the observations. That year's flow of 36.7 was 3.4 below the mean of about 40.1, which is a different number.
Uses the material above question 9.

The hydrologists need to tell the town council, in plain terms, what the number 1.758 in the output means. What should they say?
Answer: D.
1. A
  The slope is in units of the response per unit of the explanatory variable: cubic meters per second of flow for each inch of snowpack. It is an amount added to the predicted flow, not a percent of it.
2. B
  This reverses the variables. The output is for the regression of flow on snowpack, so snowpack is the explanatory variable and the slope gives a change in predicted flow.
3. C
  That describes the intercept, the Constant row, 5.55. The number 1.758 sits in the Snowpack row, so it is the slope.
4. D
  Correct. The slope is the predicted change in the response for a one-unit increase in the explanatory variable: for each additional inch of April 1 snowpack, the predicted mean June flow is about 1.758 cubic meters per second higher.
A monitoring station reports the yearly mean concentration of carbon dioxide in the air, in parts per million (ppm). For the years 1980, 1985, and so on through 2020, the least-squares line for predicting concentration from year has $r = 0.995$ . The residuals, in ppm, are shown in order of year.
$\begin{matrix} Year & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 & 2010 & 2015 & 2020 \\ Residual & 3.5 & 0.9 & - 0.3 & - 2.4 & - 3.1 & - 2.2 & - 1.3 & 0.7 & 4.1 \end{matrix}$
Which conclusion about the linear model is best supported?
Answer: D.
1. A
  A correlation close to 1 does not by itself mean a linear model is appropriate. A curved pattern can still give $r$ near 1, and only the residuals show whether the line misses in a systematic way.
2. B
  The residuals of a least-squares line always add to 0, whatever the shape of the data, so that cannot show the model is appropriate. What matters is whether they form a pattern.
3. C
  The size of the residuals is not what decides the form. A linear model can be appropriate with large residuals if they are scattered with no pattern. The problem here is the curved pattern, not the size.
4. D
  Correct. The residuals run from positive in 1980 and 1985, to negative from 1990 to 2010, to positive again in 2015 and 2020. That is curvature in the residual plot: the concentration rises faster over time than a straight line allows, so a linear model is not the most appropriate model.
A heating company used 30 jobs in houses with 2 to 9 rooms to fit the least-squares line $\hat{y} = 1.9 + 2.65 x$ , which predicts the time, in hours, to install a heat pump from the number of rooms $x$ it has to serve. One installation in a house with 4 rooms took 10.8 hours. What is the residual for this installation?
Answer: A.
1. A
  Correct. The predicted time is $\hat{y} = 1.9 + 2.65 (4) = 12.5$ hours. The residual is observed minus predicted: $10.8 - 12.5 = - 1.7$ hours. The installation took less time than the line predicts.
2. B
  This subtracts only the slope term, $10.8 - 2.65 (4) = 0.2$ . The prediction also includes the intercept, 1.9.
3. C
  This is predicted minus observed, $12.5 - 10.8$ . A residual is observed minus predicted, so the sign is reversed.
4. D
  This subtracts only the intercept, $10.8 - 1.9 = 8.9$ . The predicted time for a house with 4 rooms is $1.9 + 2.65 (4) = 12.5$ .
An energy analyst used 80 office buildings to fit a least-squares line that predicts a building's yearly electricity use, in megawatt-hours, from its number of floors. What makes this line the least-squares line?
Answer: C.
1. A
  The residuals of the least-squares line do add to 0, but so do the residuals of every line through the point $(\overline{x}, \overline{y})$ , whatever its slope. Adding to 0 does not single out one line.
2. B
  A good fit is not about how many points the line hits. The least-squares line may pass through none of the 80 points.
3. C
  Correct. The least-squares line is the one line that makes the sum of the squares of the residuals as small as possible. Each residual is a vertical distance from a point to the line, measured in megawatt-hours of electricity.
4. D
  The line predicts electricity use, so its misses are measured vertically, in megawatt-hours. Horizontal distances would treat the number of floors as the response, which is the regression the other way around.
For 30 lakes in one region, the correlation between the percent of land around the lake used for farming and the lake's water clarity, in meters, is $r = - 0.70$ . What percent of the variation in water clarity among these lakes is explained by the linear relationship with the percent of farmland?
Answer: B.
1. A
  The coefficient of determination is $r^{2}$ , and a square is never negative. The negative sign of $r$ gives the direction of the association, not a negative amount of variation.
2. B
  Correct. The proportion of variation explained is $r^{2} = (- 0.70)^{2} = 0.49$ , so about 49% of the variation in water clarity is explained by the linear relationship with the percent of farmland.
3. C
  $1 - r^{2} = 0.51$ is the share of the variation that the linear relationship does not explain.
4. D
  This reads $| r |$ as the share explained. The share of variation explained is $r^{2} = 0.49$ , not $r$ .

Section II: Free Response

1 question · 22 minutes 30 seconds

Question 2: Multi-Focus on Practices 3 and 4 10 points

An energy auditor measures how leaky a house is with a blower-door test. A fan pulls air out of the house, and the test reports how many times per hour the air inside would be replaced at a fixed pressure, called air changes per hour (ACH). A higher ACH means a leakier house. The auditor selected a random sample of 14 single-family houses in one town and recorded each house's age, in years, and its ACH. The data are shown in the table. Computer output for the least-squares regression of ACH on age is shown below.
$\begin{matrix} Predictor & Coef \\ Constant & 2.154 \\ Age & 0.0989 \\ R-Sq = 96.8 % \end{matrix}$

Age and air leakage of 14 randomly selected houses
House Age (years) Air changes per hour (ACH)

A 4 2.9
B 9 2.3
C 13 3.8
D 18 4.2
E 22 4.2
F 27 4.3
G 31 5.6
H 36 6.0
J 42 6.1
K 47 6.3
L 53 7.8
M 61 8.6
N 68 8.5
P 78 9.9

Use the given information to respond to parts A, B, C, D, and E. Label any subparts (e.g., i and ii) that may be present.
1. A.
  
  Describe the relationship between the age of a house and its ACH for these 14 houses.
  Question 2, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Describes the direction (positive), the form (linear) and the strength (strong) of the association, in the context of house age and ACH. All three features and the context are required. The strength may be supported by R-Sq = 96.8% or by $r \approx 0.984$ .Earns it: "There is a strong, positive, linear association between a house's age and its ACH: older houses in the sample tend to be leakier."; "Strong positive linear relationship between age and air changes per hour, with r about 0.984."Does not: "Strong, positive and linear." (no context); "Older houses leak more because they are older." (a causal claim, with no form or strength); "As age goes up 1 year, ACH goes up 0.0989." (gives the slope, not the form and strength)
  Model response
  There is a strong, positive, linear association between the age of a house and its air leakage. Older houses in the sample tend to have a higher ACH, and the points lie close to a straight line (R-Sq = 96.8%, so $r \approx 0.984$ ). No house stands apart from the pattern.
2. B.
  
  i. Interpret the slope of the least-squares regression line in context.
  
  ii. Interpret the value of R-Sq in context.
  Question 2, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Interprets the slope: for each additional year of age, the predicted ACH increases by about 0.0989. The interpretation must say predicted (or on average) and be in context.Earns it: "For each additional year of age, the predicted ACH increases by about 0.0989."; "Each extra year goes with about 0.099 more air changes per hour, on average."Does not: "For each additional year, the ACH increases by 0.0989." (a deterministic statement; the slope describes the predicted change); "Each year a house becomes about 0.0989 percent leakier." (reads the slope as a percent); "For each additional ACH, the age increases by 0.0989 years." (reverses the variables)
  - 1 point Interprets R-Sq: about 96.8% of the variation in ACH among the houses is explained by the linear relationship with age.Earns it: "About 96.8% of the variation in the houses' ACH is explained by the linear relationship between ACH and age."Does not: "Age causes 96.8% of a house's air leakage." (R-Sq is about variation explained by the line, not about cause); "The line predicts 96.8% of the houses correctly." (not what R-Sq measures); "The correlation is 0.968." (R-Sq is $r^{2}$ , not $r$ )
  Model response
  i. For each additional year of age, the predicted ACH of a house increases by about 0.0989 air changes per hour.
  
  ii. About 96.8% of the variation in ACH among these houses is explained by the linear relationship between ACH and the age of the house.
3. C.
  
  House K is 47 years old and has an ACH of 6.3.
  
  i. Calculate the residual for House K. Show your work.
  
  ii. Interpret the residual in context.
  Question 2, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Calculates the predicted ACH, $2.154 + 0.0989 (47) \approx 6.80$ , and the residual, $6.3 - 6.80 = - 0.50$ . Work must be shown. Minor rounding differences are accepted.Earns it: "Predicted ACH $= 2.154 + 0.0989 (47) = 6.80$ , so the residual is $6.3 - 6.80 = - 0.50$ ."Does not: " $6.80 - 6.3 = 0.50$ " (predicted minus observed; the sign is reversed); " $6.3 - 2.154 = 4.15$ " (leaves out the slope term); " $- 0.50$ " with no work shown
  - 1 point Interprets the residual: House K's ACH is about 0.50 lower than the line predicts for a house 47 years old, so the line overestimates its air leakage. The interpretation must be consistent with the residual calculated in (i).Earns it: "House K's ACH is about 0.50 below what the line predicts for a 47-year-old house. The line overestimates how leaky it is."Does not: "House K is 0.50 years younger than predicted." (the residual is in the units of the response, ACH); "The line underestimates House K." (a negative residual means the prediction is too high)
  Model response
  i. Predicted ACH $= 2.154 + 0.0989 (47) \approx 6.80$ . Residual $= 6.3 - 6.80 = - 0.50$ .
  
  ii. House K's ACH is about 0.50 air changes per hour lower than the regression line predicts for a 47-year-old house. The line overestimates its air leakage.
4. D.
  
  A residual plot for the regression shows the 14 residuals, which range from $- 0.74$ to $0.41$ , scattered above and below 0 with no curved or other pattern as age increases.
  
  i. Based on the residual plot, is a linear model appropriate for these data? Justify your answer.
  
  ii. A classmate says, "The correlation is close to 1, so we already knew a linear model was appropriate and did not need a residual plot." Explain why the classmate's reasoning is incorrect.
  Question 2, part D: your response. It stays in this browser tab and is never sent anywhere.
  Score part D
  - 1 point States that a linear model is appropriate, because the residual plot shows no pattern: the residuals are scattered above and below 0 with no curvature.Earns it: "Yes. The residuals are scattered randomly above and below 0 with no curve, so a line fits the form of the data."Does not: "Yes, because the residuals are small." (judges the size of the residuals, not their pattern); "Yes, because $r$ is close to 1." (does not use the residual plot); "No, because the residuals are not all 0." (no model fits every point exactly)
  - 1 point Explains that a correlation close to 1 (or $- 1$ ) does not by itself show that a linear model is appropriate, because data with a curved pattern can also have a correlation close to 1. The residual plot is what shows whether the form is linear.Earns it: "A curved pattern can still have r near 1, so r cannot tell you the form is linear. You need the residual plot to check for curvature."Does not: "Correlation does not imply causation." (true, but it answers a different question); "r only measures strength." (does not explain why strength cannot settle the form)
  Model response
  i. Yes. The residuals are scattered above and below 0 with no curved or other pattern, so a linear model is appropriate for the relationship between age and ACH.
  
  ii. A correlation close to 1 shows a strong linear association, but it does not show that a line is the right form. Data that follow a curve can also have a correlation close to 1, and then the line would miss in a systematic way. Only the residual plot shows whether such a pattern is there.
5. E.
  
  i. Use the regression line to predict the ACH of a house in the town that is 110 years old. Show your work, and explain whether this prediction should be trusted.
  
  ii. The auditor's report says, "These data show that the aging of a house causes it to become leakier." Explain whether the data support this statement.
  Question 2, part E: your response. It stays in this browser tab and is never sent anywhere.
  Score part E
  - 1 point Calculates the predicted ACH, $2.154 + 0.0989 (110) \approx 13.03$ , with work shown.Earns it: " $2.154 + 0.0989 (110) = 13.03$ ACH"Does not: " $0.0989 (110) = 10.88$ " (leaves out the intercept)
  - 1 point Explains that the prediction should not be trusted much, because 110 years is beyond the ages of the houses in the data (4 to 78 years), so it is an extrapolation and the linear pattern may not continue that far.Earns it: "It is an extrapolation: the oldest house in the sample is 78 years old, and we do not know that the pattern holds out to 110 years."Does not: "It can be trusted because R-Sq is 96.8%." (R-Sq describes the fit to the observed ages only); "It is an extrapolation because 110 is a big number." (does not compare with the ages in the data)
  - 1 point Explains that the data do not support the causal statement, because this is an observational study: the houses were sampled, not assigned ages, so older houses can differ in other ways that affect leakage, such as the building methods and standards used when they were built.Earns it: "No. The auditor only observed houses. Older houses were also built with older methods and standards, which could explain the extra leakage, so the data cannot show that aging itself is the cause."Does not: "Correlation is not causation." (no reason tied to how these data were collected); "Yes, because the correlation is very strong." (strength does not establish cause)
  Model response
  i. Predicted ACH $= 2.154 + 0.0989 (110) \approx 13.03$ air changes per hour. This prediction should not be trusted much. The houses in the sample are 4 to 78 years old, so 110 years is an extrapolation, and the straight-line pattern may not continue that far.
  
  ii. The data do not support the statement. This is an observational study: the auditor recorded the ages and ACH of houses as they are and did not assign ages to houses. Older houses were built with different methods and under different standards than newer ones, and that, rather than aging itself, could account for some or all of the extra leakage.
Your score: 0 of 10, from the boxes you ticked.

House	Age (years)	Air changes per hour (ACH)
A	4	2.9
B	9	2.3
C	13	3.8
D	18	4.2
E	22	4.2
F	27	4.3
G	31	5.6
H	36	6.0
J	42	6.1
K	47	6.3
L	53	7.8
M	61	8.6
N	68	8.5
P	78	9.9

House	Age (years)	Air changes per hour (ACH)
A	4	2.9
B	9	2.3
C	13	3.8
D	18	4.2
E	22	4.2
F	27	4.3
G	31	5.6
H	36	6.0
J	42	6.1
K	47	6.3
L	53	7.8
M	61	8.6
N	68	8.5
P	78	9.9

AP Statistics Unit 5 Test

Section I: Multiple Choice

Section II: Free Response

Question 2: Multi-Focus on Practices 3 and 4 10 points

Your results

House	Age (years)	Air changes per hour (ACH)
A	4	2.9
B	9	2.3
C	13	3.8
D	18	4.2
E	22	4.2
F	27	4.3
G	31	5.6
H	36	6.0
J	42	6.1
K	47	6.3
L	53	7.8
M	61	8.6
N	68	8.5
P	78	9.9