TEKS Support 
This unit contains activities that support the following knowledge and skills elements of the TEKS.
(1) (A) 
X 
(4) (A) 

(1) (B) 
X 
(4) (B) 
X 
(1) (C) 
X 

(2) (A) 
X 
(8) (A) 

(2) (B) 
X 
(8) (B) 

(2) (C) 
(8) (C) 
X 

(2) (D) 
X 

(3) (A) 
X 
(9) (A) 

(3) (B) 
(9) (B) 

(3) (C) 
The mathematical prerequisites for this unit are
The mathematical topics included or taught in this unit are
The equipment list for this unit is
Context Overview 
This unit investigates statistical concepts involved with prediction. Archaeologists, criminologists, and doctors all have an interest in predicting people’s heights from knowledge of another variable, such as the length of bones in the body or the distance between footsteps. In this unit, students collect their own data on height, forearm length, and stride length. Based on their data, they determine models to predict height from forearm or stride length. For the final project, students analyze skeletal data from the Forensic Anthropological Data Bank and determine models to predict height from the length of long bones in the arms and legs.
In addition to predicting height, students investigate an ecological problem. The manatee is an endangered species with a rising death rate. About onethird of all manatee deaths are attributable to human causes, and among these the leading cause is contact with powerboats. Based on data collected by the Florida Department of Environmental Protection, students decide how increases in powerboat registrations are affecting the life of the manatee.
Mathematical Development 
The two major mathematical goals of the unit are to explore bivariate data analysis and to further understanding of linear relationships.
This unit expands students’ acquaintance with data analysis, emphasizing not simply describing data but also using data to make predictions. The unit begins by examining a proportional relationship between people’s head length and their height. Based on data collected from their classmates, students determine the best multiplier for this relationship. In addition, they interpret the meaning of slope in this context.
Later in the unit, students display data using dot plots (onevariable data) and scatter plots (twovariable data) and then use their displays to make predictions. They assess the precision of their predictions based on the variability in the data. Midway through the unit, students encounter a major problem: Given a scatter plot, how do you select the "best" line to describe the data? After looking at several methods and comparing the resulting models, students are introduced to the leastsquares line. They use residual plots to judge the adequacy of the linear regression model to describe the pattern in a scatter plot. In addition, they examine the effect that outliers have on the leastsquares line and learn to select the best variable for making a prediction.
Teacher Notes 
Preparation Reading—Let the Bones Speak! 
This reading introduces the major contextual theme of the unit and leaves students with a question: What can you predict about a person from a few bones?
Make sure students begin this unit by reading this preparation reading. Information from this reading is used throughout the unit.
Activity 1—Using Your Head 

Materials Needed 
Meter sticks (at least 2) Rulers 
Mathematically, relationships between height and head length, examined from the perspective of artists, focus attention on models of the form y = mx and the meaning of slope. At the end of the activity, students use one such model to predict a person’s height based on the length of his or her skull bone.
Students should work in small groups (3 to 4 students) on this activity. For Item 2(a) each group will have to share its data with another group.
For Item 1 it is not important that students arrive at "correct" answers. What is important is the reasoning that they use to arrive at their answers. For example, some students may decide that the 416mm tibia belongs to the same person as the 413mm femur. They may have arrived at this conclusion based on the data provided in Table 1. Other students may argue that the femur is the longest bone in the body, and thus this tibia belongs to the same person as the 508mm femur. Let students argue this out for themselves. Don’t give them the answer.
If students struggle making a guess in Item 1(e), remind them to use their general knowledge about people’s heights.
At the conclusion of this activity, discuss Item 2(d and e). Students should understand that, when a residual error is positive, the prediction underestimates the actual value; when a residual error is negative, the prediction overestimates the actual value. You would like to choose a model where, in some sense, the positive and negative errors are balanced or tend to cancel each other out.
Discuss Item 4 with your students to continue developing the concepts of slope and rate of change. Here are some suggested approaches: numeric reasoning, algebraic approach, and graphical approach.
Numeric reasoning:
Make a table of values similar to Example 1. Explain to the students that you are using the notation D(Head length) and D(Height) as shorthand for the change in head length and the change in height, respectively. Stress that the direction of the change must be consistent. Positive values indicate increases, negative values indicate decreases.
In this table, start with the preliminary head length, make the indicated change to head length, and note the corresponding change in height. Point out to students that each time the head length increases by 1 cm, the height increases by 7.5 cm. This is true no matter how large the preliminary head lengths are. If the head length increases by 2 cm, then the height increases by 15 or 2 × 7.5 cm.
Headlength of preliminary sketch  D(Head length) from preliminary to final sketch  Height of preliminary sketch (cm)  Height of final sketch (cm)  D(Height) 

8  1  7.5  
9  1  7.5  
10  1  7.5  
10  2  15  
11  2  15 
Example 1. Table of height values
Algebraic approach:
Make sure the students understand the distributivelawbased answer to Item 4.
Graphical approach:
Use Transparency T.1 to illustrate the change in height corresponding to a 1 cm change in head length.
Be sure that students see the connections between the three approaches.
When discussing Item 5, note that groups may have arrived at different predictions if they decided that the length of a deceased person’s skull is smaller than the person’s head length when he or she was living. Using the length of the skull to estimate the length of the person’s head introduces a source of uncertainty (or error) into the prediction process. The second source of error may be the artists’ guidelines. They were meant to be rough guides for drawing figures and not precise methods for predicting height.
Homework 1—Leg Work 

For this assignment, students work with linear models developed by Dr. Mildred Trotter to predict people’s heights based on femur and tibia lengths. At the conclusion of this assignment, they discover that a person’s femur is generally longer than his or her tibia.
This assignment foreshadows results from the final project. Here students are introduced to several of Dr. Trotter’s equations relating height to lengths of leg bones. At the end of this Homework, they should understand that a person’s femur is longer than his or her tibia.
As background, here is some information on Dr. Trotter. Dr. Mildred Trotter had a long and distinguished physical anthropology career that included working as a special consultant to the U. S. government during World War II. Her task during the war involved the identification of skeletal remains of servicemen. At the time, she realized that bone sizes and proportions vary based on age, sex, race, and ethnic background. Forensic scientists and law enforcement agencies are still using Trotter’s formulas for estimating people’s stature based on the lengths of their bones.
Supplemental Activity 1—Under Investigation 

In this activity students investigate graphs of members of the y = mx + b family and learn how changes in m and b affect the graphs. In addition, students discover that the appearance of a graph can be altered by changing the scaling on one or both of the axes.
When discussing Item 2, note that, because the graph’s origin is often not displayed on the calculator screen when plotting data, students should keep in mind the location of the origin in relation to the graph. Give students practice locating the origin with sample calculator windows. Show them the screens and ask them to sketch the origins. Below are some sample windows.
[100,200] × [50,90]
[–10,–5] × [5,15]
[75,100] × [–40,–10]
[–10,–5] × [50,90]
Note: Generally, students will use windows in the first quadrant or select the standard viewing screen.
Activity 2—Measuring Up 

Materials Needed 
Handout 1 (to record class data) Tape measures, rulers, meter sticks Background Reading: Dr. Mildred Trotter’s Study of Military Personnel 
In this activity, students plan how they will measure and collect data on students’ heights and forearm lengths. Then they collect the data from students in their class.
The quotation below sheds light on how the military measured the height of personnel in the 1940s. (Dr. Trotter’s equations were based on military personnel from this time period.) Share this reading with your class; students may be surprised at the level of detail in the regulation. Can they find the one important detail that’s missing? The regulation appears in a paper by Mildred Trotter and Goldine Gleser (1952).
In Mobilization Regulations, War Department, October 15, 1942 (Regulation 10):
"Directions for taking height. Use a board at least 2 inches wide by 80 inches long, placed vertically, and carefully graduated to 1/4 inch between 58 inches from the floor and the top end. Obtain the height by placing vertically in firm contact with the top of the head, against the measuring rod, an accurately square board of about 6 by 6 by 2 inches, best permanently attached to graduated board by a long cord. The individual should stand erect with back to the graduated board, eyes straight to the front."
As detailed as the regulation appears, something was forgotten. In another set of mobilization regulations dated April 19, 1944, the same essential directions were given with the following sentence added:
"The shoes should be removed when the height is taken."
Mobilization Regulations, War Department, April 19, 1944 (Regulation 10). (Trotter and Gleser 1952, 469470).
Homework 2—Follow in My Footsteps 

Determining a standard method for measuring stride length is a bit more complicated than measuring height. For this assignment, students write a set of instructions for measuring a person’s stride length.
At some time prior to Activity 6, you should collect the stride length data from students in the class.
Supplemental Activity 2—Line Up 

This activity reviews determining an equation for a line given its graph. Students use both slopeintercept and pointslope forms to determine equations of lines from their graphs.
This is a review activity for students who are rusty in use of the slopeintercept and pointslope forms for determining the equation of a line. This is optional.
Activity 3—I Predict That 

Materials Needed 
Class data on heights and forearm lengths 
This activity focuses on the idea of variability in data and its relation to the precision of predictions made from the data. Students analyze one variable, student heights. They assess the precision of using the mean as a predictor of height.
The purpose of this activity is for students to see that precision in prediction is linked to variability in the data. Dot plots are introduced as a graphical tool for analysis of onevariable data, and the mean is suggested as a simple predictor for such data.
Discuss Item 3. Notice that, instead of using the minimum and maximum heights for the prediction interval, we narrow the interval by omitting the three smallest and three largest observations. Although this allows for a more precise prediction (the interval is narrower), omitting data also increases the chance that the prediction will be false. Discuss why the increase in precision is probably worth the increased risk of being wrong.
In Item 4, students consider the mean as a predictor. Point out that asking how far off a prediction might be is another way of asking how large the prediction error might be. Point out that the error depends on the spread (variability) of the data.
In Item 5, note that, whenever you see data that are bimodal (appear roughly as two mounds), you should ask if the data contain two subpopulations. If you can identify the subpopulations, which in this case are the boys and the girls, you should analyze each separately and then compare the results.
In Item 6, the girls’ data are less variable (exhibit less spread) than the entire data set. This reduction in variability allows a more precise prediction.
The purpose for Item 8 is to acquaint students with calculator output from onevariable statistics calculations. Check to see that students understand the mathematical notations for sum and mean.
If you run short of time, students can complete Items 10 and 11 on their own. Note that the term outlier is defined in Item 10. In Item 10, students discover the effect that outliers have on the mean and the importance of adjusting predictions when outliers are present. For Item 11, check to see that students are aware of the link between the precision of predictions and the variability of data. This concept will reappear when students analyze precision of predictions that are based on linear models.
Homework 3—Exercising Judgment 

In Item 1, although students will notice that, on average, the mothers who smoked had babies that weighed less than those of mothers who did not smoke, they may not notice that all of the babies who weighed under 6 lb. had mothers who smoked.
In Item 2, each set of data contains an outlier that inflates the mean. Be sure that students recognize this, remove the outliers, and compute the means of the data that remain. For Items 2(f) and (g), check that students understand why a scatter plot is an inappropriate way to display these data. Scatter plots are used when there is an assumption that two quantities obtained as matched pairs are related. There is no such pairing here, and no natural reason to pair particular numbers for the two groups.
This activity reviews determining an equation for a line given its graph. Students use both slopeintercept and pointslope forms to determine equations of lines from their graphs.
This is a review activity for students who are rusty in use of the slopeintercept and pointslope forms for determining the equation of a line. This is optional.
Activity 4—Forearmed Is Forewarned 

Materials Needed 
Graph paper Ruler Spaghetti or toothpicks 
This activity emphasizes the use of scatter plots in identifying and describing relationships. Students face the problem of selecting the "best model" to describe the pattern of a scatter plot. In deciding between two contenders for the best model, students analyze both models’ residuals.
Item 2 is designed to connect analyses of onevariable data (discussed in Activity 3) to twovariable settings, bridging dot plots to scatter plots. By drawing a vertical line to specify a single value of the independent variable, students can interpret the data that fall along that line (or close to it) as a vertical dot plot.
For example, to view the variability in heights for students with forearm length 27 cm, draw the vertical line x = 27. Then look at the range of heights for students with 27cm forearms (or close to 27cm forearms).
After students have calculated a few predicted values and residual errors in Item 4, you may wish to help them use calculator lists to speed their work. See Handout 4 for TI83 calculator instructions.
For Item 5, students may find it helpful to use a tangible object such as uncooked spaghetti (or toothpicks if they’re working on calculator screens) to use as lines. That way, they can easily adjust the line until they are satisfied with how it fits the data.
Homework 4—The Nature of Our Relationship 

This assignment introduces the ideas of direction (positive or negative), form (linear or nonlinear), and strength (strong or weak) of a relationship.
Note: After students have completed this assignment, review some of the new vocabulary words, such as positively and negatively related, linear and nonlinear form, and weak and strong relationships.
Activity 5—Dangerous Waters 

Students fit a leastsquares line to describe the relationship between the number of manatees killed per year and the number of powerboat registrations. They use their model for analysis and prediction. In addition, they learn to use residual plots to assess whether their model is adequate to describe the data.
For Item 2, you will need to teach students to calculate the equation for the leastsquares line on the graphing calculator or computer. Handout 2 contains TI83 instructions for computing the leastsquares line. Handout 3 provides similar instructions for Excel. For other calculators or spreadsheets, check your manual.
Item 4 states two essential criteria related to good fits and defines "residual plot." Stress student understanding of what this plot really means. The randomness of this plot should be the primary criterion for deciding that a model is reasonable. You may refer students to Handout 4 if they need help calculating the residuals on their calculators.
For Item 6, check that students realize that the number of powerboat registrations is in units of 1,000.
You may want to point out that a "good" residual plot looks like a bunch of dots thrown haphazardly at a piece of paper; the dots should appear randomly scattered around the xaxis. If the dots do not look randomly scattered around the xaxis but instead form a clear pattern, you should look for another model to describe your data.
Some calculators give the value of Pearson’s correlation coefficient, r, as part of the output from a linear regression. If this is the case, you may want to tell students that r is a measure of the strength and direction of a linear relationship. However, stress that judging the goodness of a fit should begin with examining the graphs of the original data and the residual plot.
Homework 5—Anscombe’s Data 

Students fit leastsquares lines to four data sets and discover that they get the same equation in all four cases. After examining scatter plots of the data, students learn that the leastsquares equation is an adequate model for describing the pattern in only one of the data sets.
This is a famous data set. You will find it in numerous statistics texts.
Activity 6—The Plot Thickens 

Materials Needed 
Tape measures or meter sticks Partially completed Handout 1 
Students decide which of two independent variables, forearm or stride length, is a better predictor of student height. In addition, students work through an analysis illustrating how forensic data can help solve crimes.
For Item 1, one method for deciding which of two independent variables yields more precise predictions for the same dependent variable is to select the relationship that has the smaller average of squared errors. Taking the average instead of the sum adjusts for situations where the scatter plot of one relationship has more data than the scatter plot of another relationship. For example, this method can be used when comparing the regression equation based on the class heightforearm data to predict height to the regression equation based on the boys’ heightforearm data.
The average of squared errors is one estimate of the variance about the leastsquares line. It is, however, not the one generally used by statisticians. Statisticians generally use the unbiased estimator SSE/(n–2) where n is the number of cases or the sample size, but this is not relevant to student work in this unit.
To speed the completion of Item 1, you may decide to work the item as a whole class activity.
The sample answers to the remainder of the items in this activity are based on the following set of data collected from a set of 9th and 10th graders.
Name 
Gender 
Height 
Stride Length 
Forearm Length 
Scott  Male  166.0  58.25  28.5 
John  Male  178.0  68.5  29.0 
Matt  Male  171.0  58.5  27.2 
Will  Male  165.0  50.125  28.0 
Michael  Male  177.5  58.75  31.3 
Jeffrey  Male  166.0  62.875  28.3 
Even  Male  175.5  59.125  28.6 
Brad  Male  171.0  67.75  31.5 
Lonnie  Male  184.0  68.875  30.5 
William  Male  184.5  66.25  30.8 
Robert  Male  183.5  79.5  30.5 
Karim  Male  172.0  70.5  30.3 
Meredith  Female  164.5  55.875  24.2 
Lee  Female  166.0  52.375  27.3 
Pilar  Female  168.0  55.375  28.0 
Ansley  Female  178.5  59.75  29.1 
Julie  Female  166.0  48.375  27.9 
Becton  Female  159.0  57.125  28.0 
Elizabeth  Female  166.0  64.0  27.4 
Shannon  Female  154.5  57.75  25.8 
Jamie  Female  161.0  63.5  27.0 
Jeris  Female  177.0  69.75  30.1 
Kat  Female  161.0  72.5  26.5 
Blaie  Female  164.0  75.25  28.2 
Frances  Female  174.0  58.5  28.4 
Eliz  Female  164.0  59.75  26.8 
Baily  Female  168.0  55.25  26.4 
For Item 2, if you have not already collected class data on student stride lengths, you should do so. (See Homework 2.) After deciding on a method for collecting the data, each group can be responsible for collecting the data from its members. After groups have collected their data, pool the results. Students should record these results in the last column of Handout 1. If you have already collected the stridelength data, students can read quickly through Items 2 and 3 and begin their work at Item 4.
Homework 6—You Are What You Eat 

In this assignment students discover the drastic effect that outliers can have on a regression line by comparing models computed with and without outliers. Students also learn to seek the interpretation of outliers in particular settings.
After students have completed this assignment, discuss how the presence of outliers affects the values of m and b in the leastsquares equation. The leastsquares equation can be very sensitive to outliers, particularly if they occur at the extremes. In these situations, the leastsquares line does a poor job of describing the pattern of the majority of the data.
In those cases where you can determine that the outliers are "unusual points" that are not representative of the relationship, remove these points and recalculate the equation of the leastsquares line using the remaining data. For example, in Item 2 (the situation with the swimming data) a good argument could be made that the first two times were not "typical" because the swimmer was still learning the butterfly. In this case, it seems reasonable to remove the outliers and refit the model.
Unit Project—Who Am I? 
This project can be adapted for a wide range of student abilities and time constraints. Students can complete their analysis using a spreadsheet or a graphing calculator. Ideally, students work in groups, and each group should present its work in a formal written report. You may want to have groups give oral presentations in addition to or in place of the written report. Work may also be done individually if more time is available.
Below is a brief set of guidelines for reports. You may decide to give more detailed guidelines of your own design.
If possible, let students decide for themselves how they will complete this project. Encourage them to plan what equations they will need to determine and then divide the work among group members. If some groups struggle, you may need to provide additional structure.
The following is a direct method (not necessarily the best method) of addressing the questions in this project.
Students may use several different approaches in developing equations to predict the heights of Bones 1 and Bones 2. First, they should realize that there are two bones that can be used to predict height: the femur and the ulna. So, they should begin predicting height using each of these independent variables.
Note that the data that appear in the student pages of this project are also provided as computer files, as listed:
Column headings are not included in these files. However, the calculator file is a program that stores the data to named lists. See student pages for the heading labels and units of measure.
Handout 1—CLASS DATA RECORDING SHEET 
Female 
Male 

Name 
Forearm 
Height 
Strident 
Name 
Forearm 
Height 
Stride 

Handout 2—TI83 INSTRUCTIONS FOR FINDING THE LEASTSQUARES LINE 
Here’s how to use the TI83 to calculate the leastsqures line and its residuals.
Press ENTER again and the residuals will appear. How many are positive and how many negative? What is the absolute value of the largest residual? How can you use the features on your calculator to find the sum of the residuals?
Handout 3—EXCEL 4.0 GUIDANCE FOR FINDING THE LEASTSQUARES LINE 
Complete Activity 5 in your text. Use these instructions to assist your work using Excel 4.0.
Handout 4—TI83 INSTRUCTIONS: CALCULATING PREDICTED VALUES AND ERRORS 
Example: Find the errors using the linear model y = 2x + 1.
Recall that prediction errors are defined as Yactual  Ypredicted.
Figure 3. Screen showing formula for L1
Annotated Student Materials 
Preparation Reading—Let the Bones Speak! 
Legend has it that, about 100 years ago, somewhere in Arizona’s Superstition Mountains, a Dutchman by the name of Jacob Walz murdered a group of gold miners in order to claim their mine for himself. Over the years, he would periodically be seen in Phoenix with saddlebags filled with rich ore. Many attempted to follow Walz when he returned to the mine, but he always managed to lose trackers in the rugged wilderness. Walz died in 1891. For over a century, people have searched without success for the Lost Dutchman Mine. Some have lost not only time and money, but their lives. At least two searchers are known to have been murdered during their quest. Others, unable to meet the physical challenges of the rugged area, never returned from their treks and remain missing. 
From time to time, human bones are found in rugged areas such as the Superstition Mountains. Suppose that a contemporary gold digger searching for the Lost Dutchman Mine finds a skull, eight long bones, and numerous bone fragments. He notifies the local authorities who, in turn, send out a team to investigate. After first documenting the exact location and position of the bones at the site, the team records information about the bones, such as their size and general condition. A partial list of information similar to what might be recorded is contained in Table 1.
Bone type 
Number found 
Length (mm) 
Femur  3  413, 414, 508 
Tibia  1  416 
Ulna  2  228, 290 
Radius  1  215 
Humerus  1  357 
Skull  1  230 
Fragments  More than 10  From 30 to 50 mm 
Table 1. Sample record of bones found at site
In such situations, police frequently request help from forensic anthropologists to identify the deceased and determine the cause of death. The bones tell the forensic scientists a story about who the deceased were and frequently how they died. Sometimes the story is an old one, as would be the case if the bones belonged to the gold miners murdered by Walz. Other times, the bones tell a story of more recent crime and provide police with clues that may help them solve a mystery.
Activity 1—Using Your Head 

= 1, 2, 3, 4
In this unit you will be asked to think like a forensic scientist. After studying the data in Table 1 of the preparation reading, you will begin to sort out clues about the deceased from their bones. For the final project at the end of this unit, you will write a report detailing their story.
There are three femurs. Two of the femur bones are very close in length (413 mm and 414 mm) and probably belong to the same person. So, these three bones most likely belong to two people. It is possible that the bones actually belong to more than two people.
Sample answer:
Assume that there are only two people.
Bones 1: femurs–413, 414; tibia–416; ulna–28; and radius–215.
Bones 2: femur–508; ulna–290; humerus–357.
The skull bone could belong to either Bones 1 or Bones 2.
Since the two femurs—413 mm and 414 mm—are close in length, they most likely belong to the same person, Bones 1. From the skeleton in Table 1, it appears reasonable to assume that a person’s tibia should be fairly close in length to her femur, so the 416mm tibia probably belongs to Bones 1. Assuming that there are only two people, the person with shorter legs probably also has shorter arms. That means that the 228mm ulna belongs to Bones 1. The length of a person’s radius should be close to the length of the ulna, and hence the 215mm radius belongs to Bones 1.
Bones 2, the taller of the two people, has a 508mm femur and a 290mm ulna. A person’s humerus should be just a bit longer than his ulna (based on the diagram of the skeleton in the preparation reading). Assuming that there are only two people, the 357 mm humerus belongs to Bones 2. (It’s possible that this bone belongs to a third person.)
Students probably will be most uncertain to whom the skull and humerus belong. Some students may know that the femur is the longest bone in the body and may conclude that the 416mm tibia belongs to Bones 2.
Sample answer:
There is very little evidence to suggest whether the decedents are male or female. Because the femur bones of one of the decedents are considerably shorter than the other, it’s possible that Bones 1 (shorter femur) is female and Bones 2 is male. However, it could just as easily be the case that the decedents are two males, one short and one tall.
Sample answer #1: Most likely the decedents were adults. What would young children be doing out in the Superstition Mountains?
Sample answer #2: When I measured my own forearm to get an estimate of how long an adult’s ulna might be, it measured about 260 mm. (The actual length of my ulna would be a bit less than this measurement.) So, Bones 2 is probably an adult. Bones 1 might be a child.
Sample answer:
Assuming that Bones 1 is female and Bones 2 is male, my guesses are that Bones 1 is 5 feet 5 inches and Bones 2 is 5 feet 10 inches. These guesses were based on my estimates of average heights for adult females and adult males, respectively. While these guesses are fairly rough, they should be within a foot of the actual heights.

Head 
Predicted height 
Actual height 
Residual error: Actual – Predicted 
See sample answer to (d).
See sample answer to (d).
In almost every situation in which predictions are made from data, it is useful to examine the residual errors. Residual errors are defined as the difference between the actual value and the predicted value for each point in your data.
Name 
Head length (cm) 
Predicted height (cm) 
Actual height (cm) 
Residual error 
Jan 
23.0 
161.0 
168.0 
7.0 
Horrace 
25.0 
175.0 
178.0 
3.0 
Betty 
24.0 
168.0 
166.5 
–1.5 
John 
25.0 
175.0 
184.0 
9.0 
Joy 
23.5 
164.5 
170.5 
6.0 
Lora 
22.0 
154.0 
159.0 
5.0 
Positive residual errors indicate that your predictions are too low; negative errors mean predictions are too high; zero residual errors mean the predicted heights match the actual heights.
Sample answer:
Most of the residual errors were positive. So, the rule of thumb seems to frequently underestimate students’ actual heights.
Sample answer based on sample answer to (d):
Use Average ratio of height/head length: (1/6)(168.0/23.0 + 178.0/25.0 + 166.5/24.0 + 184.0/25.0 + 170.0/23.5 + 159.0/22.0) » 7.2.
Using 7.2 as the multiplier produces the following residuals errors.
Name 
Head length (cm) 
Predicted height (cm) 
Actual height (cm) 
Residual error 
Jan 
23.0 
165.6 
168.0 
2.4 
Horrace 
25.0 
180.0 
178.0 
–2.0 
Betty 
24.0 
172.8 
166.5 
–6.3 
John 
25.0 
180.0 
184.0 
4.0 
Joy 
23.5 
169.2 
170.5 
1.3 
Lora 
22.0 
158.4 
159.0 
0.6 
Using this multiplier, two of the six residual errors are negative; the size of errors tends to be smaller. In addition, the errors sum to zero so that the amount overestimated balances the amount underestimated.
H = 7.5L
H = 7.0L
Sample answer: H = 7.2L
Sample answer:
Note that the shaded sections indicate portions of the graphs that represent the realworld relationship between height and head length. (Students may decide to choose a window cropped to capture the shaded sections.)
These three graphs are all lines that pass through the origin. The multiplier controls the amount of inclination: the larger the multiplier the steeper the incline.
Predicted height: H = 7.5(23.0) = 172.5 cm. This estimate is likely to be too high. The graph H = 7.5L lies above the graph H = 7.0L for L > 0. So, for each positive entry for L, the value for H will be larger from the first relationship than from the second.
Each time head length is changed by 1 cm, the height gets changed by 7.5 cm. This is true regardless of the size of the head in the preliminary sketch.
Sample answer:
The decedent was most likely adult. The length of the skull is somewhat smaller than the length of the person’s head.
Sample answer:
Assume that the person was an adult and that head length was 2 cm larger than the skull bone, or 25.0 cm. (The extra 2 cm leaves room for skin and soft tissue and also accounts for shrinkage of the skull due to drying.)
Prediction: 7.5(25.0 cm) = 187.5 cm.
The sample answer of 187.5 cm is about 6 ft 2 in. This is a reasonable height for a tall person.
Sample answer:
First, skull size was used to estimate head length. In addition, the artists’ guidelines are only rough approximations and are not exact for every individual. So the estimate is a very rough one. It may be very far from the person’s actual height.
Sample answer:
It would be helpful to know the relationships between bone lengths and heights. Perhaps data could be collected from the class. Perhaps an artists’ handbook would contain relationships between arm lengths and heights or leg lengths and heights. Perhaps you could get data on bones and peoples’ heights from the Internet and then use these data to determine models that predict height from lengths of bones.
Homework 1—Leg Work 

Dr. Mildred Trotter (18991991), a physical anthropologist, was well known for her work in the area of height prediction based on the length of the long bones in the arms and legs.
Here is one of the relationships proposed by Dr. Trotter.
H = 2.38 F + 61.41 
where H is the person’s height (in cm) and F is the length of the femur (in cm). 
Predicted height for a person with a 38cm femur: 151.85 cm or approximately 152 cm.
Predicted height for a person with a 55cm femur: 192.31 cm or approximately 192 cm.
Figure 3. Axes for height and femur length
Notice that the horizontal axis is scaled from around 35 cm to 60 cm (a slightly wider range than the minimum and maximum femur lengths) with tick marks every 5 units. A zigzag has been added to indicate that there is a break in this scale between 0 and 35.
Refer to answer in (b).
Note: Students may choose a different scaling for the vertical axis.
Jason’s height: (40)(2.38) + 61.41 = 156.61
Jason’s brother’s height: (41)(2.38) + 61.38 = 158.99
Difference: 2.38 cm
Each time you increase the length of the femur by one cm, the height changes by 2.38 cm. This difference has the same value as the multiplier of F, the slope of the linear equation.
Sample answer: 47.5 cm.
F = (H – 61.41)/2.38; femur is the dependent variable and height the independent variable.
(172.7 – 61.41)/2.38 » 46.8 cm
Second Formula:
H = 2.52T + 78.62,
where H and T are measured in cm.
H = (2.52)(41.6) + 78.62 » 183.5 cm. This person is approximately 6 feet tall. This is a reasonable height for a person.
Second formula H = 2.52T + 78.62 for T.
(A doctor might use such an equation to check that the length of a person’s tibia is normal for a person of that height.)
Step 1: subtract 78.62 from both sides of the equation.
H – 78.62 = 2.52T
Step 2: divide both sides by 2.52.
(H – 78.62)/2.52 = T or T = (H – 78.62)/2.52
Approximately 37.3 cm.
H = 1.30(F + T) + 63.29.
(All measurements are in cm.)
H = 1.30(F + T) + 63.29.
The predicted height is 173.8 cm.
The predicted height is (2.38)(42) + 61.41 » 161.4 cm. This is 12.4 cm (or about 5 inches) shorter than the prediction in a).
H = 2.52T + 78.62.
The predicted height is (2.52)(43) + 78.62 » 187 cm. This is 13.2 cm larger than the prediction in (a) and 25.6 cm more than the prediction in (b).
Solving 175 = 2.38F + 61.41 for F gives F » 47.7 cm. Solving 175 = 2.52T + 78.62 for T gives T = 38.2 cm. You would expect the femur to be approximately 9.5 cm longer than the tibia.
Solving 160 = 2.38F + 61.41 for F gives F » 41.4 cm. Solving 160 = 2.52 T + 78.62 for T gives T = 32.3 cm. From these equations, it appears that a person 160 cm tall should have a larger femur than tibia.
Based on the answers to (e) and (f), it appears that a person’s femur should be longer than his or her tibia. When the students measured the bones, they found that the femur of the skeleton was shorter than the tibia. This is the reverse of what Dr. Trotter’s equations indicate. Perhaps the students need to recheck their measurements.
Bones 1: H = 2.38(413.5) + 61.41 » 159.823 cm or approximately 5 ft 3 in. (The average of the two femurs closest in length was used to make this estimation.)
Bones 2: H = 2.38(508) + 61.41 » 182.314 cm or approximately 6 ft.
Using the tibia length from Table 1, H = (2.52)(41.6) + 78.62 » 183.5 cm or approximately 6 ft.
Sample answer:
These calculations do not appear to refute the assumption that there were only two deceased. However, based on these calculations, it appears that the 416mm tibia might belong to Bones 2 (the taller person) rather than Bones 1.
In Activity 1 and in this homework, you examined and interpreted equations established by artists and by a scientist. You used some of Dr. Trotter’s models to estimate the heights of Bones 1 and Bones 2 (described in the preparation reading). Dr. Trotter’s formulas may have challenged some of the assumptions that you made in Item 1(b), Activity 1. However, for the equations given in this homework, she assumed that the deceased were adult white males. If this assumption is not valid, your estimates based on Dr. Trotter’s equations may not be accurate.
Supplemental Activity 1—Under Investigation 

Unlike the artists’ guidelines for drawing figures, Dr. Trotter’s equation,
H = 2.38F + 61.41
(where height, H, and femur length, F, are in cm),
is not a member of the y = mx family, but instead belongs to the larger y = mx + b family. You indicate members of this family by choosing values for m and b. (What were Dr. Trotter’s choices for m and b?)
Recall that Dr. Trotter’s equation
H = 2.38 F + 61.41
was designed to work well for a particular population, adult white males. She later modified her formula by modifying the values of m and b to adjust for age, ethnic background, and gender. To make such adjustments, you will need to know how changes in m and b affect the graph. Complete the following investigation to find out what happens when you make changes to m and b.
Because there are two quantities to change, m and b, it may help to divide the investigation into two parts, as described below.
PART I: KEEP m THE SAME AND CHANGE b.
PART II: KEEP b THE SAME AND CHANGE m.
Repeat Part I, reversing the roles of m and b.
Changing the value of b moves the line up (if b is increased) or down (if b is decreased). In addition, the line will cross the yaxis at b.
The slope, m, determines how steeply the line tilts and (depending if m is positive or negative) whether the line tilts upward or downward as you look along the graph from left to right.
The value of b determines where the line crosses the yaxis. So, yintercept is a descriptive name. The value of m determines the steepness of the line or how much it slopes.
By changing your window settings, you can affect the appearance of a line described by a member of the y = mx + b family without changing the values of m or b. At times, you may want to adjust your window settings to display your graph more effectively. However, you should also be aware that some people, driven by an interest in distorting the truth, will tinker with their window settings until they achieve a graph that satisfies their purpose. Your understanding of how scale change affects the appearance of the line will help you interpret graphs correctly and avoid being misled by their distortions. The next investigation will help you learn the effects on a graph of changing the maximum settings for the horizontal or vertical axis.
Sample answers:
Handdrawn graph (left) and calculator produced graph (right) with the same scale settings.


In each case the value of Ymin = 140. Then Ymax was changed from 200 to 300 and then to 160. The appearance of the graph of Dr. Trotter’s formula changed as indicated below.
The graph appeared flatter when the value of Ymax was increased and became steeper when the value of Ymax was decreased.
Display 1 
Display 2 


The settings for Display 1 were Xmin = 35, Xmax = 60, Ymin = 140, Ymax = 200. For Display 2, Xmax has been changed to 120.
For each Display, the equation of the line is the same and each time the x value changes by 1 unit the yvalue will change by 2.38 units. However, in Display 2, the distance on the horizontal axis representing 1 unit is smaller than in Display 1 since more units must fit on the same display screen. This makes the line appear steeper.
The graph of y = 5.78x + 5 is steeper; 5.78 is larger than 3.48. This means that every time the xvalue is increased by one unit, the yvalue for the first graph will increase by 5.78 units compared to only 3.48 units for the second graph.
The graph of y = 15x + 30. The value for b is 30.
y = (1/2)x + 15 or y = –2x + 5?
The graph of y = –2x + 5.
In this activity you discovered how modifying a member of the y = mx + b family by changing the value of m or b affects its graph. You also discovered that rescaling can change your perception of how steeply a line rises or falls, even though you are graphing the same equation. This understanding will come in handy when you want to select members of the y = mx + b family to describe patterns in data.
Activity 2—Measuring Up 

= 5, 6, 7, 8
Your analysis in Activity 1 and Homework 1 has included parts of a process known as mathematical modeling. The process begins when you identify a problem for which you need an answer or a situation that requires further understanding.
For example, during World War II, the armed services sometimes had problems identifying the remains of dead soldiers. Dr. Mildred Trotter was asked to help. She wondered if there were relationships between the height of a person and the lengths of his long bones.
Having posed this question, Dr. Trotter’s next step was to collect relevant data. She needed measurements of people’s heights and the lengths of their long bones. Her model
H = 2.38 F + 61.41
where femur length is in cm
expresses the relationship she observed between the height measurements and femurlength measurements from her data.
Because it depends on data, the model is only as good as the quality of the data on which it is based. Dr. Trotter took special care to check that her data were collected by people who followed detailed instructions for taking the measurements. In this way, she was able to keep to a minimum the data variability that was due to the measurement process.
Dr. Trotter used lengths of long skeletal bones to predict height. You can’t directly measure the bones in your body. Instead, in this activity, you will design methods for collecting data on students’ heights and the lengths of their forearms. Later you will develop a model to predict classmates’ heights using the lengths of their forearms.
Before you collect your data (height and forearm length from each student in your class), you need to establish a method for taking the measurements. Remember, the worth of your model will depend on the quality of the data that you collect. Everyone who will be doing the measuring must use the same method and then record their data to the same degree of precision (for example, to the nearest eighth of an inch or to the nearest millimeter).
Sample answer:
For height: Stack two meter sticks vertically and tape them to the wall. Have the student to be measured stand in front of the meter sticks. The student should take his or her shoes off and stand up straight. Place a cardboard on top of the student’s head. Hold the cardboard so that it is parallel to the floor. Read the spot where the cardboard touches the meter stick. Record height to the nearest millimeter.
For forearm: Have student hold arm flat on a desk with his or her hand placed palm down. Measure along outside of the arm from the point on the elbow to the top of the knobby bone at the wrist. Measure to the nearest millimeter.
Note: Save your data for use in Activities 3 and 6.
Homework 2—Follow in My Footsteps 

Sometimes all that’s left at a crime scene is a few footprints. However, the length of a person’s stride is also related to the person’s height. Now you will develop a method for measuring a person’s stride. Later you will gather these data and then use your measurements in a model to predict height.
To collect reliable data, you need to carefully plan the method you will use to collect the data. Remember, your model will be only as good as the data on which it is based.
Design a method for measuring the length of a person’s stride.
Here are some items to consider.
How will the person walk? Do you plan to measure from heel to heel or heel to toe? Since step lengths for the same person can vary, does it makes sense to have the person take more than one step and average the results? If so, how many steps should the person take? Determine the measurement instrument (e.g., ruler, tape measure, meter stick) you will use to make the measurement. Specify the precision of the measurement. After you have decided on your method, test your method as you did the methods for measuring height and forearm length. When you are satisfied with your method, describe it with a set of written instructions. Give your instructions to a friend to see if someone else understands what you mean. If necessary, revise your instructions. Save them until your class is ready to collect the stridelength data needed later in this unit. 
Sample answer:
A tape measure with metric reading will be used for measuring. Measurements will be recorded to the nearest tenth of a centimeter.
Setup: Mark a line about 15 feet long with adhesive tape. Mark the starting position with another piece of tape.
Have the person put his or her heels at the edge of the starting position and then tell the the person to take four steps along the marked line. Measure from starting point to back of heel after the person has stopped. Divide by four to get the stride length.
Supplemental Activity 2—Line Up 

This activity gives you an opportunity to practice determining an equation of a line from its graph.
Figure 4. Graphs of four lines
y = 1 when x = 0. This is the yintercept, the value of b.
When Dx = 2, Dy = 1.
Dy/Dx = 1/2; this ratio is the same as the slope.
When Dx = 3, Dy = –6.
Dy/Dx = –2; this ratio is the same as the slope.
Dy = 1 when Dx = 4; slope = Dy/Dx = 1/4; y = (1/4)x.
y = (1/4)x + 3.
Lines A and B have the same steepness and they are parallel; they cross the yaxis at different locations. The slopes for the two lines are equal, m = 1/4; the yintercepts are different, b = 3 for Line A and b = 0 for Line B.
Figure 5. Understory trees in a forest
Species A: approximately 2.6 m; Species B: approximately 1.5 m.
Species A: 3 m; Species B: 6 m.
The two lines in Display 2 are examples of straightline relationships between two variables. In this case the variables are tree height and crown width. The official name for such relationships is linear relationships, and the equations that describe these relationships are called linear equations.
In the section "Let the Bones Speak," you studied linear relationships between bone length and height. Dr. Trotter’s equations are examples of linear equations relating bone length and height variables.
The line for species A. The line goes through the origin. The approximate value for m is 2/3. We found this value for m by starting at the origin. Then we got to another point on the line by moving up 2 meters and across 3 meters.
m » 2/7. We started at the point x = 2.5 and y = 1. We moved 1 unit up and 3.5 units across to get to another point on the line. So the slope is approximately 1/3.5 = 2/7 » .29.
When x = 0, y » 0.25. This gives you the value of b.
The equation for Species B is y = (2/7)x + (1/4) or approximately y = 0.29x + 0.25.
m = 2/7 means that the crown width changes, on average, by 2/7 meters each time tree height increases by 1 m. Theoretically, the value of b indicates that trees that are 0 meters tall will have crown widths that are approximately 0.2 m. This is nonsense. The value of b does not have meaning relevant to this context. It may suggest that, in the early stages of growth when the height is near zero, the crown grows rapidly.
Up to this point, most of the linear equations you have worked with have been written in slopeintercept form, meaning as members of the y = mx + b family. Suppose you choose b = 3. Then, all members of the y = mx + 3 family will pass through the point (0,3). (Why?) Or suppose that you wanted your lines to pass through the point (4,3). What equations would you use to describe these lines? The key to the answer is contained in the next question.
The two lines that are the same have equations y – 3 = 2(x – 4) and y = 2(x – 4) + 3.
Figure 5. Data with X at center
When x = 10, then y = m(10 – 10) + 20 = 20.
Sample answer: Select two points on the line as far apart as possible. For example, when x = 0, y » 15 and when x = 20, y » 24. This gives a slope of (24 – 15)/20 = .45
Sample answer:
y = 0.45(x – 10) + 20.
Sample answer:
y = 0.45x + 15.5.
m is the slope.
When x = h, y = k; (h,k) is a point on the line.
It’s a good name because, when you know the slope and a point on the line, you know the line’s equation.
Activity 3—I Predict That 

The data in Table 3 show the forearm lengths and heights of students in a tenthgrade class.
Female 
Male 

Name 
Forearm 
Height 
Name 
Forearm 
Height 
Alice 
24 
157 
Allan 
26.5 
173 
Bia 
24.5 
166 
Brian 
27 
177 
Christi 
27 
164 
Daniel 
27 
174 
Chantalle 
24 
164 
Davis 
31 
192 
CoralAnne 
23 
161 
Eric 
28 
172 
Jennifer 
27.5 
164 
Kevin 
29 
180 
JiHyun 
27 
167 
Lenny 
27 
174 
Kim 
26 
162 
Larry 
28 
175 
Kristen 
26 
175 
Mike 
32 
185 
Nancy 
28.5 
166 
Neil 
30 
185 
Tanner 
26.5 
172 
Rob 
30 
178 
Teresa 
25.5 
176 
Table 3. Heightforearm data from Class A
The heights of the students vary from 157 cm to 192 cm for a total difference of 35 cm.
Recall the general principle that data can be examined graphically. A graphic representation of the height data might help you to assess the amount of variation in student height. Follow the instructions in Item 2 to construct your own dot plot.
= 9, 10, 11, 12
Figure 7. Partial dot plot of height data from Class A
Two sample answers follow. The first answer is more in the spirit of prediction.
Sample answer #1: There were only three people in Class A with height’s below 164 cm and three above 180 cm. The heights of seventeen of the 23 students fall within the 164 to 180cm range. So, if heights of students in Class A are representative of tenthgrade students in general, chances are pretty good that this prediction will be correct. Even if the prediction is wrong, it will be off by at most 12 cm (provided tenthgrade students, in general, have heights similar to those of the students in Class A).
Sample answer #2: This prediction could be wrong. What if the tenth grader is only 157 cm?
The average height is approximately, 172.1 cm.
If the new student is 157 cm, the prediction will be off by 15.1 cm.
If the new student is 192 cm, the prediction is off by 19.9 cm.
Sample answer #1: The prediction in part (a) is near the middle of the height data. If the unknown student’s height lies somewhere in the middle of the heights of students in Class A, this prediction will be very close to the actual height of the student. If the student’s height is as small as the smallest student in Class A or as tall as the tallest student, this prediction will be off by at most 19.9 cm. So this prediction seems reasonable.
Sample answer #2: Perhaps the median height, 173 cm, might be better.
Sample answer #3: Maybe she should use 174.5 cm, midway between the shortest and tallest student’s heights. Then her prediction would be off by no more than 17.5 cm.
Figure 8. Number lines for comparative dot plots
The girl’s heights are between 157 cm and 176 cm; the boys, between 172 cm and 192 cm. Even though a few of the girls in Class A are taller than several of the boys, the boys in Class A tend to be taller than the girls.
Using the average of the girls’ heights, the predicted height is 166.2 cm. If Malisa is 157 cm, this prediction is off by 9.2 cm; if she is 176 cm, this prediction is off by 9.8 cm.
The prediction from Item 4, 172.1 cm, is probably too high if you know that the student is a girl. (Only three girls are at least 172 cm tall.) If Malisa’s height is in the middle of the girls’ heights for students in Class A, using the prediction of 166.2 cm will be off by no more than 9.8 cm. On the other hand, the prediction of 172.1 cm may be off by as much as 15.1 cm.
Sample answer:
Possible predictions: Average of boys’ heights is approximately 178.6 cm; median of boys’ heights is 177 cm. Some students may just eyeball what seems to be a typical boy’s height.
This answer depends on the answer to part a. Sample answer: If the prediction is 178.6 cm, it could be off by at most 13.4 cm.
Using the height data from Class A, you have computed at least two and possibly three different averages: an average of all the data, an average for the girls, and an average for the boys.
The term mean is another name for average. For the remaining items, when you are asked to calculate the mean, just find the sum of the data and then divide by the number of data points. It’s no different from calculating an average.
If your data have been entered into one of your calculator’s lists, you can use a builtin calculator command to compute the mean. However, you will need to know some mathematical shorthand to understand what your calculator is telling you. (See Table 4.)
Shorthand Notation 
Meaning 
å x 
The sum of the data 
n 
The number of data 
The mean, the sum of the data divided by the number of data 
Table 4. Table of shorthand
Figure 9. Onevariable statistics screen
972
6 people
162 cm
Answer depends on your class data.
Answer depends on your class data.
Most likely, the boys will tend to be taller than the girls.
Answer depends on your class data. If your students are older than tenthgrade students, the boys in your class may be taller than the boys in Class A. Perhaps the girls’ heights will be similar to the girls in Class A.
If you find it helpful, you may use your calculator’s builtin statistical capabilities to calculate the means in the remaining problems.
0 
23 
45 
6 
8 
9 
33 
15 
0 
2 
4 
10 
12 
13 
34 
67 
40 
38 
27 
25 
0 
13 
34 
23 
56 
34 
7 
789 
44 
6 
4 
0 
31 
22 
5 
16 
17 
11 
2 
1 
(Students may choose simply to list the largest data point rather than to put a break in the scale.)
Ordered data: 0, 0, 0, 0, 1, 2, 2, 4, 4, 5, 6, 6, 7, 8, 9, 10, 11, 12 , 13, 13, 15, 16, 17, 22, 23, 23, 25, 27, 31, 33, 34, 34, 34, 38, 40, 44, 45, 56, 67, 789
Mean of smallest 10 data points: (0 + 0 + 0 + 0 + 1 + 2 + 2 + 4 + 4 + 5)/10 = 1.8
Mean of largest 10 data points: (34 + 34 + 34 + 38 + 40 + 44 + 45 + 56 + 67 + 789)/10 = 118.1
The first mean, 1.8, appears to be a better prediction than the second mean, 118.1. At least 1.8 is close to several data points, while 118.1 is not near any of the data.
Mean of all data: 1526/40 » 38.2
Mean data with outlier removed: 737/39 = 18.9
When 789 was removed, the mean went down considerably. So, the outlier inflated the prediction.
Prediction: 18.9 (see figure below)
When predicting the number of gray hairs, if the actual number of gray hairs is very low, using 1.8 would be a more accurate estimate. However, if the actual number is 780, using 118 would be the more accurate estimate. But you won’t know the actual number of gray hairs before you make your prediction; therefore, use 18.9 as your prediction because it is closer to more of the data than the other predictions.
Figure 10. Number of gray hairs on the heads of 20yearolds
The mean is 2.6. Prediction: approximately 3 gray hairs.
Because there is less variability in the data for the 20yearolds, you are more likely to predict a number that is very close to the actual number.
If the data have a lot of variability (in other words, the data are very spread out), it is difficult to make precise predictions. If, instead, the variability in the data is small, so that the data are very concentrated, it is much easier to make fairly precise predictions.
Homework 3—Exercising Judgment 

Each of the items in this assignment provides an opportunity to compare data from two groups. When you make comparisons to analyze data, use what you have learned from Activity 3 as well as common sense.
Never smoked 
6.3  7.3  8.2  7.1  7.8  9.7  6.1  9.6  7.4  7.8  9.4  7.6 
Smoked ten or more cigarettes per day 
6.3  6.4  4.2  9.4  7.1  5.9  6.8  8.2  7.8  5.9  5.4  6.3 
Table 5. Babies’ birthweights
Never Smoked: Average weight—approximately 7.9 pounds
Smoked: Average weight—approximately 6.6 pounds
Mothers who smoked had, on average, babies that weighed less than mothers who did not smoke.
The mothers of the four babies who weigh below 6 all smoked. None of the babies whose mothers did not smoke weighed less than 6 pounds.
Does not exercise 
10 
5 
20 
4 
20 
20 
15 
0 
8 
40 
8 
15 
Does exercise 
15 
15 
15 
5 
10 
5 
5 
6 
30 
25 
30 
60 
Table 6. Cost of entertainment (dollars)
The comparative dot plots appear below.
The data for the "exercise" group appear shifted slightly right of the data for the "does not exercise" group. This would indicate that the "exercise" group tends to spend more on entertainment than the "does not exercise" group. The shift is more noticeable if you ignore the largest value in each data set. Then the pattern for the "does not exercise" group spans from $0 to $20 while the pattern for the "exercise" group spans from $5 to $30.
The average of these data is approximately $18.42. However, one person in this group spent $60. If the $60 outlier is removed, the average is only about $14.64. A prediction of $14.64 looks more central to the remaining data.
Sample answer: "I predict that a person from the "does not exercise" group will spend $11.36 on his or her next entertainment. However, given what this group has spent on entertainments in the past, this person might spend as little as $0 or as much as $40. So my prediction might be as far off as $28.64. However, it is more likely that they will be off by no more than $11.36."
If students did not remove the highest value, the average would be $13.75 instead of $11.36.
Sample answer:
We predict that a student who exercises will spend $14.64 on entertainment and a student who does not exercise will spend $11.36 on entertainment. The dot plot indicated that students who exercised tended to spend more than students who did not exercise. So the dot plot confirms the higher prediction for the group that exercises.
Solution (chart below):
The scatter plot doesn’t make any sense in this situation. The spending levels for the "exercise" group and the "does not exercise" group are not paired. In other words, the x and y coordinates were matched up only by the order in the table. They don’t belong together in any meaningful way.
Activity 4—Forearmed Is Forewarned 

= 13, 14
In Activity 1, you were asked to predict the heights of Bones 1 and Bones 2 (and perhaps Bones 3 or more) whose bones were described in Table 1. At that time, your best guess was most likely based on what you know, in general, about people’s heights. In Activity 3, you were asked to predict students’ heights. Your predictions were based on data from a single variable, student heights. Later, you used your knowledge of whether the student was male or female to improve the precision of your predictions. But you have not yet made any connections between the length of someone’s forearm and his or her height.
For this assignment, use the data from Class A shown in Table 3.
Student graphs should be similar to the one below. Circles represent boys, and squares girls.
Sample answer:
There are six female students with forearm lengths between 25 and 27 cm. The heights of these six students are 162, 164, 167, 172, 175, and 176 cm. Use the average of these heights, approximately 169 cm, for the prediction. (Some students may decide to choose the median or some other number between 162 and 175 for their predictions.) If the student’s height turns out to be only 162 cm, this prediction is off by 7 cm. On the other hand, if the student turns out to be 175 cm tall, this prediction is off by 6 cm. So the largest error would be around 7 cm (provided the students in Class A are similar to all tenthgrade students).
Student answers will vary. Here are some important points to consider.
Only one student in Class A has a 28.5cm forearm. The student is 166 cm tall and happens to be a girl. There are, however, two boys with forearms of 28 cm. Their heights are 172 cm and 177 cm. So the average height would be 173.5 cm. The height of a boy with a 29cm forearm is 180 cm. A reasonable prediction might be to select the height that is midway between 173.5 and 180 cm for a predicted height of approximately 177 cm.
Student answers will vary. This forearm length is larger than any in the data collected from Class A. So, to make the prediction you need to follow the pattern of the data beyond the observed data. A prediction of around 194 cm seems reasonable.
The 16% model: y = x/0.16 = 6.25x, where y is height (cm) and x is forearm length (cm).
For the remainder of this activity, this equation will be referred to as the "16% model."
Student graphs should be similar to the one below.
No, this relationship is not true for all students in Class A. If it were true, all the data would have to fall along the line representing the 16% model.
The boys’ data appear to lie closer to the line than the girls’ data. The residual errors appear in the last column of the table that follows.
Girls’ Data
Forearm length (cm) 
Height (cm) 
Predicted height using 16% model (cm) 
Residual error (cm) 
24.3  157  152  5.1 
24.5  166  153  12.9 
27.8  164  174  –9.8 
24.1  164  151  13.4 
23.7  161  148  12.9 
27.5  164  172  –7.9 
27.4  167  71  –4.3 
26.6  162  166  –4.3 
26.1  175  163  11.9 
28.5  166  178  –12.1 
26.6  172  166  5.8 
25.4  176  159  17.3 
Table 7
Boys’ Data
Forearm length (cm) 
Height (cm) 
Predicted height using 16% model (cm) 
Residual error (cm) 
26.5  173  166  7.4 
27.3  177  171  6.4 
27.8  174  174  0.3 
31.4  192  196  –4.3 
28.3  172  177  –4.9 
29.2  180  183  –2.5 
27.3  174  171  3.4 
27.9  175  174  0.6 
31.8  185  199  –13.8 
30.2  185  189  –3.8 
29.8  178  186  –8.3 
Table 8
Average absolute value of the errors for girls’ data: » 9.5 cm; for boys’ data: 5.4 cm.
The heightforearm data from Class A are fairly spread out. Picking a line that describes the data, or makes good predictions for heights based on forearm lengths, is somewhat difficult because of the large amount of variability in the data.
= 15, 16, 17
Method #1:
Pick a point that appears to lie in the middle of the points displayed in your scatter plot. What are the coordinates of this point? Now anchor your line to this point and adjust the slope until you find a line that you think best describes the pattern of the data. What is the equation of the line you have selected. How did you decide what line was best?
Method #2:
Draw two lines in such a way that the points on your handdrawn scatter plot are bounded as tightly as possible between these lines. (The lines don’t have to be parallel.) Now draw one line halfway between the two lines that you have drawn. What is the equation of this line? How did you decide which line was closest to the middle of the two outer lines?
Student answers will vary. Here are sample answers for the two methods.
Method #1: Selected point (27.2,172.6), the mean of the forearm lengths and heights. Using a graphing calculator, we experimented with equations of the form y = m(x – 27.2) + 172.6 until we found one that split the data about in half with about the same number of points above the line as below the line. The equation of our model is y = 4(x – 27.2) + 172.6. A graph of the data, the anchor point marked with a square, and the line are shown below.
Method #2 (see figure above). We drew two lines that bounded the data between them. Then we selected a line between these two lines that had approximately as many of the data points above the line as below the line. The results shown in the calculator screen below are similar to the lines that we drew by hand.
Figure 12
The equation for our model is y = 4(x – 27) + 170.
Student answers will vary depending on their answers to Item 5.
Sample answer:
Method #1: y = 4.0x + 63.8
Method #2: y = 4.0x + 62.0
In this case, the two models differed in only the yintercept. The graph of Model #1 is 1.8 units higher than the graph of Model #2.
Sample answer:
In this case it is difficult to tell which model fits the data better just by looking at the models and the data. For Model #1, 12 of the data points lie above the line, one lies on the line, and 10 lie below the line (note that (27,174) appears twice in the data). For Model #2, 14 data points lie above the line and 9 below the line. So, if we use the criterion of equal numbers of points above and below the line, Model #1 is better.
Students could also analyze the prediction errors.
Sample answer:
The line that was closer to having an equal number of points above and below was the better line.
Some students may decide that the line with the sum of the errors closer to zero was the better line. Still others might calculate the average of the absolute values of the errors and choose the line for which this average was smaller.
The 16% model had 11 points above the line, 2 on (or approximately on ) the line, and 10 below the line. Using the criterion of choosing a line that splits the data into equal numbers of points above and below the line, the 16%model does as well as the model from Method #1. However, the line from Method #1 seems to capture the pattern of the data better than the 16% model.
Sample answer:
Using the model from Method #1, y = 4x + 63.8, the predicted height is 171.8 cm. Students with forearms of length 27 cm varied in height from 164 cm to 177 cm. If the student is actually 164 cm tall, the prediction will be off by 7.8 cm.
Sample answer:
Using the model from Method #1, y = 4x + 63.8, the predicted height is about 196 cm. There are no data from students with forearms as long as 33 cm. Hence, it is difficult to assess how far off this prediction might be.
Sample answer:
Using the model from Method #1, y = 4x + 63.8, the predicted difference in height for the students whose forearm lengths differ by one centimeter is 4 cm. If the students forearm lengths differed by 2 cm, the predicted difference in height would be 8 cm.
For each 1cm increase in forearm length, you should expect about a 4cm increase in height.
The heightforearm data from Class A were fairly scattered about the line that you chose for your model. The amount of scatter (variability in the data) made determining a model that "best" describes the data somewhat difficult.
Homework 4—The Nature of Our Relationship 

In this assignment you will make scatter plots of data sets and, in some cases, fit lines to the scatter plots and make predictions. Pay particular attention to the characteristics of the relationship between the variables.
Do the points in the scatter plot appear to be scattered on either side of a straight line? If so, the scatter plot has linear form and it makes sense to describe it with a linear equation (a member of the y = mx + b family).
Does the pattern made by the points move upward as you look from left to right? If so, the two variables are positively related (as one variable increases the other tends to increase). If the pattern drifts downward, the two variables are negatively related (as one variable increases the other tends to decrease).
Month  Sep  Oct  Nov  Dec  Jan  Feb  Mar  Apr  May 
Outdoor temperature ºF  48  46  38  29  26  28  49  57  65 
Gas used per day _ 100 cu ft  5.1  4.9  6.0  8.9  8.8  8.5  4.4  2.5  1.1 
Table 7. Gas usage and temperature data
The independent variable is temperature and the dependent variable is the gas used per day. Generally, people want to predict the amount of fuel they use. Generally, temperature is a key factor in determining fuel usage (see chart below).
There is very little scatter around a straight line. The direction of the line is downward, so the variables have a negative relationship. An increase in the mean monthly temperature produces a decrease in the monthly gas consumption.
Sample answer:
Using the points (26,8.8) and (65,1.1): y = –0.20x + 14.
Sample answer based on equation in c above: 3.7
Player 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
Practice 
89 
90 
87 
95 
86 
81 
105 
83 
88 
91 
79 
Competition 
94 
85 
89 
89 
81 
76 
89 
87 
91 
88 
80 
Table 8. Golf scores
Answer (chart):
The points are loosely scattered around a straight line with a positive association between the variables. This indicates that the better golfer should have lower scores for both practice and competition.
The outlier is circled in the scatter plot in (a). It is not possible to tell if this player is a good player who had a bad practice round or a weaker player who had an unusually good competition. You would need more data on this player to decide.
The graph below shows the data with the outlier omitted and the line y = x superimposed. It appears to do a reasonable job in describing these data.
Prediction: 95 in competition round.
When the relationship between two variables is strong and has linear form, the points in a scatter plot will fall very close to a line. For weaker relationships, the data are more scattered.
Figure 11. Two scatter plots showing height versus ulna length
Display 2: The data in Display 2 appear to lie in a narrower band about the line than the data in Display 1.
y = 60.2 + 0.42x
(where x is ulna length in mm and y is height in cm)
to predict the height of the deceased whose ulna length of 290 mm was recorded in Figure 1. What did scientists predict as his or her height?
182 cm
Sample Answer:
Display 1: The actual height might be as low as 165 cm and as high as 195 cm.
Display 2: The actual height might be as low as 175 cm and as high as 185 cm.
The prediction is more precise if the real data are from Display 2.
When the data in a scatter plot form a narrow band about the linear model, the predictions are more precise than when the data form a wide band about the linear model. You can make more precise predictions when you are dealing with a strong relationship than with a weaker one.
Activity 5—Dangerous Waters 

= 15, 16
On the coast of Florida lives the manatee, a large, appealingly ugly, and friendly marine mammal. However, the gentle Florida manatee does not live a carefree life. It is one of the most endangered marine mammals in the United States. One major threat to the manatee’s survival is the large numbers of manatees killed each year by powerboats.
Should the Florida Department of Environmental Protection limit the number of registered boats in order to protect the manatee population? Before deciding to limit the number of registrations, the department will have to present a convincing argument to the public.
Table 9 contains data on the number of powerboats registered in Florida (in thousands) and the number of manatees killed.
Year 
Powerboat Registrations (in thousands) 
Manatees Killed 
1977  447  13 
1978  460  21 
1979  481  24 
1980  498  16 
1981  513  24 
1982  512  20 
1983  526  15 
1984  559  34 
1985  585  33 
1986  614  33 
1987  645  39 
1988  675  43 
1989  711  50 
1990  719  47 
Table 9. Powerboat registration and manatee deaths in Florida 19771990.
In this activity, you are asked to make a convincing case for whether or not to restrict powerboats. The main steps in building a convincing argument to present to the authorities are:
Your graphing calculator (as well as most scientific calculators and spreadsheets) has builtin commands that calculate the equation of the leastsquares line, a line that statisticians frequently select as the bestfitting line.
Leastsquares criterion: A line’s score is the sum of the squared (residual) errors. Statisticians frequently refer to this number as the SSE. The best line (the "leastsquares line") has the smallest SSE.
However, you probably won’t see the term "leastsquares line" anywhere on your calculator. The general name for determining an equation that fits the data is regression. Because, in this case, you are looking for a linear relationship, the technique used in fitting the "best" line to data is called linear regression.
Leastsquares equation: y = 0.125x – 41.4
(The values for slope and yintercept have been rounded. More decimal places have been retained for the slope. You can round predictions to an appropriate number of decimal places later.)
The slope tells you that, each time you increase the number of powerboats by one unit (1000 boats), you can expect 0.125 additional manatee deaths. The value of the yintercept is meaningless in this context.
In the plot below, there appear to be about as many points above the line as below the line. Most of the dots are fairly close to the line. The line appears to fit the data fairly well.
 The sum of the errors, also called residuals or residual errors, should be close to 0.
 The data in the scatter plot should be randomly scattered above and below the line.
 Set your calculator to compute a list of the residuals (errors) for the leastsquares line. Then calculate the sum of the residuals. Is the sum of the residuals close to 0? (How can you use the statistical features on your calculator to compute this sum most efficiently?)
Sample answer:
We used the onevariable statistical feature on the calculator to get the sum of the data in the list containing the residuals. The value of the sum of the residuals is –3 × 10–12.
 Does the leastsquares line appear to satisfy criterion (2) above? What does this imply about the residuals?
Yes, the dots appear to be randomly scattered above and below the line. This means that the residuals are randomly scattered between those with positive sign and those with negative sign.
 Using your calculator, make a scatter plot of the residuals versus the number of powerboat registrations. Do the dots in your plot look randomly scattered above and below the horizontal line y = 0?
The dots in the scatter plot below appear randomly scattered.
A plot of the residuals versus the independent variable, such as the plot you have made for Item 4(c), is called a residual plot. The residual plot gives you information about how well your model describes the data. For example, the residual plot can help you decide if your model satisfies criterion (2) above.
 Explain how the residual plot can help you decide if your model satisfied criterion (2).
Each dot in the residual plot that lies above the line y = 0 is a positive error and hence comes from a point on the scatter plot that lies above the regression line. Each dot in the residual plot that is negative and hence lies below the line y = 0, came from a point on the scatter plot that lies below the regression line. If the points in the residual plot are randomly scattered above and below the line y = 0, that means the points in the original scatter plot are randomly scattered above and below the regression line.
If the points in the residual plot do not appear randomly scattered, but instead form a strong pattern, your linear model cannot adequately describe the data. In this case, you may need to search for another model, perhaps a new slope or new intercept, or maybe a model that is not linear.
If your model does not adequately describe the data, you probably don’t want to use it to make predictions.
Yes, the residual plot appears to be randomly scattered.
Sample answer: x = 571.2 units or about 571,000 registrations.
Solve 30 = 0.125x – 41.4 for x.
Sample answer:
y = 0.125(700) – 41.4 = 46.1 or approximately 46 manatees would be killed by powerboats. To assess the precision of this prediction, we look at the data. Since x = 700 is between observations for x = 675 and x = 711, we can base our answer on observations (675,43) and (711,50). The actual numbers of manatees killed might be as small as 43 or as large as 50, even though we have predicted 46 deaths.
Every time the limit on boat registrations is raised by 50,000 registrations, one can predict that, on average, an additional ________ manatees would be killed by powerboats each year.
Justify your response.
Because the units for the powerboats are in terms of 1000 boats, a 50,000 increase in the number of powerboats corresponds to an increase of 50 units for the independent variable. Using our model, we would predict an increase of (0.125)(50) = 6.25 deaths. Based on this calculation, we will respond to objectors as follows:
Every time the limit is raised by 50,000 registrations, we predict that, on average, an additional 6 manatees will be killed by powerboats each year.
Homework 5—Anscombe’s Data 

Tables 10–13 present four sets of data prepared by statistician Frank Anscombe.
x 
10 
8 
13 
9 
11 
14 
6 
4 
12 
7 
5 
y 
8.04 
6.95 
7.58 
8.81 
8.33 
9.96 
7.24 
4.26 
10.84 
4.82 
5.68 
Table 10. Data Set A
x 
10 
8 
13 
9 
11 
14 
6 
4 
12 
7 
5 
y 
9.14 
8.14 
8.74 
8.77 
9.26 
8.10 
6.13 
3.10 
9.13 
7.26 
4.74 
Table 11. Data Set B
x 
8 
8 
8 
8 
8 
8 
8 
8 
8 
8 
19 
y 
6.58 
5.76 
7.71 
8.84 
8.47 
7.04 
5.25 
5.56 
7.91 
6.89 
12.50 
Table 12. Data Set C
x 
10 
8 
13 
9 
11 
14 
6 
4 
12 
7 
5 
y 
7.46 
6.72 
12.74 
7.11 
7.81 
8.84 
6.08 
5.39 
8.15 
6.42 
5.73 
Table 13. Data Set D
[Anscombe 1973]
All the equations are approximately y = 0.5x + 3
Answer: See charts below.
The only data set of the first three that has a linear form is Data Set A. You shouldn’t even try to fit a line unless the data set appears to have a linear form. So, for Data Set A, you could predict that the value of y when x = 14 is 10. Data set D has an outlier that appears to tilt the line more steeply than the general flow of the data.
He wanted students to see the dangers of not first checking to see if the data have linear form or an outlier before determining the leastsquares equation.
y = 0.35x + 4.0
If you ignore the outlier, this line appears to fit the data almost perfectly.
Activity 6—The Plot Thickens 

There are times when it is possible to make predictions using two or more models, each based on a different independent variable. You will then need to choose which model to use. In this activity, you will learn one method for selecting among these models the one that results in the most precise predictions.
First, how would you determine which relationship will result in the more precise predictions: height versus stride length or height versus forearm length? (The asterisks(*) indicate missing data.)
Suppose you collected the data in Table 14 from a ninthgrade class.
Height (cm) 
Stride Length (cm) 
Forearm Length (cm) 
166.0  58.2  28.5 
164.5  55.9  27.2 
175.0  59.1  28.6 
184.0  68.9  30.5 
161.0  72.5  26.5 
164.0  *  28.2 
171.0  *  28.4 
Table 14. Data on height, stride length, and forearm length
Answer: See charts below.
Sample answer:
Height versus forearm length is the stronger of the two relationships. The dots in the heightforearm scatter plot fall closer to fitting on a line than the dots in the heightstride scatter plot.
Height = 160 + 0.167(stride length)
Height = 10.9 + 5.59 (forearm length)
Average for heightstride: 342.252/5 » 68.5
Average for heightforearm: 73.29/7 » 10.5
The average squared error represents the typical squared distance from a point to the leastsquares line. The smaller this distance the stronger the relationship.
Sample answer:
Heightforearm model: Height = 10.9 + 5.59(27) » 161.8 cm
Heightstride model: Height = 160 + 0.167(73) » 172.2
The forearm estimate is more reliable. As was found in part b), the relationship between height and forearm length is stronger than the relationship between height and stride length.
Select the model associated with the smaller average squared residuals. This should be the model for which the data appear most concentrated along the leastsquares line.
Now that you have a method for selecting between two models based on different independent variables, read the following scenario and, together with the members of your group, solve the problem.
A school’s mascot is stolen (see Figure 12). The thief has left clues: a plain black sweater and a set of footprints under a window. The footprints appear to have been made by a man’s sneaker.
The distance between the footprints, from the back of the heel on the first footprint to the back of the heel on the second, reveals that the thief’s steps are approximately 58 cm long. The thief’s forearm length can be estimated from the sweater by measuring from the center of a worn spot on the elbow to the turn at the cuff. The thief’s forearm is between 26 and 27 cm.
Figure 12. The missing manatee
School officials suspect that the thief is a student from a rival high school. You have two tasks: Gather data on stride lengths of students in your class in order to estimate the thief’s height from his or her stride length. Then use the data collected from Activity 2 to predict, as accurately as possible, the height of the thief from the length of his or her forearm.
See sample answer to Homework 2.
Sample answers for this activity are based on class data contained in the notes to the teacher.
The mean stride length for all students in this class is 62 cm. One stride length of 79.5 cm tends to inflate the mean somewhat. The average stride length for girls is 60.34 cm or a bit more than 60 cm; for boys it’s approximately 64 cm. However, if you compare dot plots for the boys’ stride lengths and the girls’ stride lengths, a cluster of girls had stride lengths below 55.5 cm. Very few of the boys fell in this category.
Sample answer:
There is a cluster of students whose stride lengths were around 58 cm. Some were boys and some girls. The stride length does appear to provide helpful information about the gender of the thief.
Sample answer:
Four people from this class had forearms with measurements between 26 and 27 cm. All of these students were girls. So, the forearm data tend to indicate that the thief was a girl!
Now predict the height of the thief. You have two possible variables to consider for your prediction, stride length and forearm length. Which is better? Divide the work in Items 6 and 7 among the members of your group.
Sample answer:
Class data: y = 75.6 + 3.324x
Boys’ data: y = 87.2 + 2.954x
Girls’ data: y = 84.4 + 2.980x
The slopes for the boys’ data and girls’ data are fairly close. However, the boy’s line is shifted vertically from the girls’ line. It appears, according to these two models, that boys are, on average, 2.8 cm taller than the girls with similar forearm lengths.
Sample answer:
To assess the strength of the linear relationship, we used the average of the squared errors.
Class data: 720.678/27 » 27
Boys’ data: 361.158/12 » 30
Girls’ data: 342.513/15 » 23
The girls’ model did better than the boys’ or the class models.
Sample answer:
You may do better using the class data even though you are told the thief is male. If you are fairly certain that the thief is female, use the girls’ model.
Answer: See charts below.
Sample answer:
Class model: y = 145 + 0.399x
Boys’ model: y = 137 + 0.590x
Girls’ model: y = 165 + 0.019x
Because the slope of the girls’ model is so small, it offers very little help in predicting height from stride length. Because we have decided that it is likely that the thief is a girl, we stop here and use the girls’ model from Item 6.
I predict that the thief is ______ cm tall. But the thief might be as short as _____ or as tall as _____.
Sample answer:
Girls’ model based on forearm length: y = 84.4 + 2.980x
I predict that the thief is between 162 and 165 cm tall. But the thief might be as short as 155 cm or as tall as 168 cm. We came to this conclusion from looking at heights corresponding to girls with forearm lengths near 26 or 27 cm.
Homework 6—You Are What You Eat 

Selecting a line according to the leastsquares criterion often produces a line with good properties. That’s why selecting a line using the leastsquares criterion is so popular. However, sometimes this line does a terrible job in describing the pattern of the data. In such cases, you may have to adjust your model so that it better describes the pattern in the data.
Food 
Guessed calories 
Correct calories 
8 oz whole milk 
196  159 
5 oz spaghetti with tomato sauce 
394  163 
5 oz macaroni with cheese 
350  269 
One slice of wheat bread 
117  61 
One slice white bread 
136  76 
2oz candy bar 
364  260 
Saltine cracker 
74  12 
Mediumsize apple 
107  80 
Mediumsize potato 
160  88 
Creamfilled snack cake 
419  160 
Table 15. Guessed calories and actual calories
[USA TODAY, October 12, 1983]
Answer: See plot below.
There is a positive relationship between the variables. However, there are two points that appear to be outliers. Those have been circled. The pattern of the rest of the points appears to have a linear form.
Regression line fit to entire data set: There are two outliers that appear to be pulling the regression line up. Most of the residuals are negative. The regression line does not appear to adequately describe these data.
Yes, the entries for creamfilled snack cake and spaghetti. People guessed that they had many more calories than they actually did.
Regression line after removal of the outliers. The residual plot looks more randomly scattered.
The size of the largest error was approximately 30 calories. The line appears to do a good job in predicting the guessed calories from the actual calories, provided the spaghetti and creamfilled snack cakes are removed from the data.
The slope was 1.14. This means that the guessed calories will increase by 1.14 each time the actual calories increase by one.
Race Number 
25Yard Butterfly Time 
1  60.81 
2  66.11 
3  47.32 
4  42.69 
5  43.40 
6  44.82 
7  42.67 
8  45.17 
9  41.20 
10  43.68 
11  42.47 
12  41.74 
13  40.40 
14  42.90 
Table 16. Butterfly times
As race number increases, time tends to decrease. The data look fairly linear with the exception of two points in the upper left corner.
The charts below represent (from left to right) regression equation, leastsquares line, and residual plot.
The dots in the residual plot appear to have a strong pattern (sort of like a check mark). This indicates that this model is not adequate to describe the relationship between time and race number.
Since the leastsquares line tries to make the sum of the squared distances between the data and the line as small as possible, these outliers have the effect of tilting the line in their direction.
If these points were removed, the slope of the line would increase. (The absolute value of the slope is smaller, but because this slope is negative, the slope increases.)
The dots in the residual plot appear to be randomly scattered.
Therefore this linear equation adequately describes the relationship between time and race number.
Time for 15th race, 41.08; for 16th race, 40.75.
Her times go down by approximately 0.33 second (about 1/3 of a second).
Eventually, her times must start to level off at some value. Otherwise she would eventually end up swimming the length of a pool in a negative amount of time.
Assessment—Cattle Stocks 

Some archaeologists gather information about the dietary habits of ancient societies. Animal bones give the archaeologists clues about the kinds of meat that were eaten by such groups. Sometimes scientists find enough bones to determine the size of the whole animal. Most of the time, however, only a few bones are found. If they find the leg bone (metacarpus) of a cow, archaeologists can predict its height (Figure 1).
Figure 1. "Reconstructing" a cow from its leg bone
Figure 2. Linear relationship between h and m
The drawn line is based on data. For other values of m the relation is not known. Smaller and larger values of m may be impossible: the height of a cow is limited. The most extreme value (m = 0) should predict a value of 60 for h !!
h = 3*m + 60
111 cm
3m = 25, so m = 25/3. An answer giving the nearest integer (8) is correct also.
In a certain excavation, archaeologists found bones of cattle that didn’t fit their expectations. The relationship between the metacarpus and the total height of the cow was different from that for the other cow bones. These animals were apparently of another stock. To distinguish between the two stocks, call the original stock "A" and the new stock "B."
The equation that describes the relationship between metacarpus and height for stock B is:
h = 5m + 15.
This equation is reliable only for 5 £ m£ 30.
Answer: Chart below
The difference between A and B is 35 for m = 5. For every extra cm, the difference shrinks by 2 cm, so it takes 17.5 steps to arrive at the same value of h. Therefore, the values are: m = 22.5 and h = 127.5. An accurate estimation from the graph (together with a check on the answer) is good too. We don’t expect students to solve a "system of equations" in the traditional way.
In reality, the relationship between metacarpus and height is not as well defined as is suggested by the graphs of the linear equations. Therefore, it is possible that the length of the metacarpus and the height of a stockA or B cow don’t exactly fit the linear equation of stock A or B. Remember that the equations of the two straight lines were found as the leastsquares lines from scatter plots.
Stock A. The expected value should be 105 for stock A and 90 for stock B. So the real value is the nearest to the one of stock A.
The value of h is now between the predicted values for stock A and B: 90 and 65. It is nearer to the value of stock B.
Because of the scatter, it is sometimes difficult to tell to which stock a cow belongs. For these data, the scatter around the leastsquares line is relatively small. Ignoring a few outliers, residual errors rarely exceed 2 cm (positive or negative). So you can say that there is a region around the line where all dots of the scatter can be found.
Figure 3. Magnification of a section of the graph from Figure 2
Figure 3 shows an enlarged portion of Figure 2. The line for cattle stock A is drawn, together with the region of variability around that line.
The top line for the shaded area is parallel to the given line for stock A. the vertical distance to this line is exactly 2 cm. This can be checked for m = 21: the value of h is 125, and that is 2 more than the value on the given line. The lower straight line is exactly 2 cm below the given line. So the shaded area is showing the region.
The part of the picture that is shaded twice is the "problematic" region. From Figure 2 it is clear that problems arise for values of m between 20.5 and 24.5 We don’t expect exact borders, although they can be found by intersecting two lines (two times) with equations that can be derived from the given lines.
Archaeologists are interested not only in the kinds of food that ancient people consumed, but also in the quantities. Using the height of a cow, archaeologists can roughly estimate the total amount of meat. The relationship between height and total amount of meat is almost linear. Figure 4 shows the data for cattle of the two stocks, A and B.
Height (cm) 
Amount of meat for stock A (kg) 
Amount of meat for stock B (kg) 
110  400  380 
120  470  435 
Figure 4. Amount of meat given the height and stock
Linear extrapolation using the table. For stock A: m = 21 means h = 123 cm. For every cm increment of the height, the amount of meat grows by 7 kg. So, for h = 123, there is 470 + 3*7 = 491 kg. In the same way you find for stock B: h = 120 and the amount of meat is given directly in the table: 435 kg.
Unit Project—Who Am I? 
Reconsider the data from Bones 1 and Bones 2 (in the preparation reading). The table in Figure 1 contains the information about those bones.
Bones 1 
Bones 2 
Uncertain 
Femur: 413, 414 
Femur: 508 
Skull: 230 
Ulna: 228 
Ulna: 290 
Humerus: 357 
Radius: 215 

Tibia: 416 
Figure 1. Classification of bones discussed in preparation reading
Table 1 contains actual data from the Forensic Anthropology Data Bank (FDB) at the University of Tennessee.
The FDB contains metric, nonmetric, demographic, and other kinds of data on skeletons from all over the United States. These individuals most likely came through the medicolegal channels as unidentified bodies, then went to forensic anthropologists for analysis and identification.
1 168 307 240 258 448 384 368 1 178 336 247 261 463 404 390 1 161 294 213 227 413 335 322 1 155 324 262 279 465 395 375 1 165 314 243 258 432 364 364 1 168 303 223 244 441 355 342 1 165 311 231 254 436 362 360 1 173 312 248 266 483 405 401 1 165 322 229 246 448 368 352 1 163 298 221 245 443 355 361 1 153 280 218 234 410 345 344 1 165 294 220 235 448 354 353 1 170 311 235 253 440 360 347 1 160 316 214 226 437 356 348 1 159 292 223 233 419 346 336 1 163 315 228 251 438 356 347 1 165 303 237 249 451 356 348 1 165 308 234 248 439 348 344 1 165 315 227 240 448 363 353 1 175 316 244 260 473 390 374 1 180 333 256 278 475 391 381 1 168 321 230 248 450 365 362 1 163 299 219 236 435 357 339 1 165 304 246 264 467 392 383 1 160 309 236 248 432 364 358 1 158 319 246 268 442 371 364 1 165 325 242 250 448 378 365 1 170 335 248 263 474 400 382 1 182 334 254 273 514 420 407 
1 165 307 230 248 452 363 355 1 163 297 240 260 435 356 356 1 143 282 216 233 398 334 318 1 154 297 228 248 423 344 334 1 171 342 272 290 485 418 407 1 162 303 237 262 433 367 364 1 150 308 220 247 383 352 341 1 157 288 201 215 429 363 350 1 158 314 239 263 432 371 358 1 162 306 250 268 444 355 352 1 159 310 238 255 449 362 352 2 169 337 254 273 460 396 385 2 153 296 223 243 407 337 338 2 175 339 256 271 470 390 381 2 179 343 242 263 464 378 371 2 179 352 253 269 484 407 397 2 198 354 263 292 508 417 412 2 173 327 256 276 463 383 387 2 180 357 268 278 494 401 390 2 178 344 254 269 464 371 366 2 175 339 245 272 456 374 366 2 177 343 250 266 483 361 365 2 180 353 260 281 490 420 415 2 170 303 235 249 435 366 361 2 191 364 263 278 511 430 417 2 188 349 269 288 498 427 423 2 179 323 256 276 486 398 400 2 180 350 263 280 480 419 418 2 181 350 263 282 488 391 381 
2 178 337 272 272 475 393 390 2 172 344 255 281 470 400 393 2 188 360 269 283 510 422 416 2 189 347 272 283 547 432 445 2 177 330 246 262 462 386 370 2 166 322 242 258 442 373 374 2 186 332 267 283 478 391 388 2 177 322 245 265 457 397 395 2 176 332 259 274 458 382 378 2 180 323 251 275 448 390 387 2 173 335 253 273 497 404 389 2 175 330 253 274 470 384 382 2 169 313 252 265 472 391 385 2 175 336 256 274 464 388 377 2 181 390 284 303 521 440 435 2 193 356 297 318 522 451 433 2 182 362 275 293 499 424 405 2 169 322 249 266 426 366 356 2 180 337 265 281 482 412 399 2 185 363 286 302 520 429 420 2 180 355 274 292 490 422 424 2 170 378 272 291 512 404 390 2 180 370 278 292 523 429 420 2 175 333 260 273 484 398 386 2 168 342 262 280 484 404 385 2 170 347 269 291 476 396 393 2 166 315 240 260 456 377 362 2 185 363 295 309 524 446 427 2 191 382 299 316 537 479 466 
Table 1. Data from Forensic Anthropology Data Bank (FDB)
Key to Data (in order from left to right): sex (1 = female, 2 = male), height (cm), humerus (mm), radius (mm), ulna (mm), femur (mm), tibia (mm), fibula (mm)
Use the data in Table 1 to answer the following items. Present your findings in a formal report. All of your conclusions must be supported by statistical analysis using the data in Table 1.
Mathematical Summary—Scatter plots 
In many situations, one is confronted with questions such as "Are values of quantity 1 related to values of quantity 2?" For example, a forensic scientist might ask, "Is height related to femur length?" In general, such questions suggest the use of graphs called scatter plots.
Since the question implies that one quantity might help predict values of the other quantity, it is common to refer to the quantities as the independent and dependent variables, respectively. A scatter plot is a graph in which the dependent variable’s values are represented on the vertical axis and the independent variable’s values are represented on the horizontal axis. This is also referred to as a graph of the dependent variable versus the independent variable.
A scatter plot is an ideal tool in looking for patterns in a relationship between two quantities.
Linear relationships between two variables can be described by graphs, equations, tables, and arrow diagrams.
the slopeintercept form, y = mx + b,
and,
the pointslope form, y – k = m(x  h).
Given any two points on a line, you can determine the value of m, the slope of the line, by computing the ratio D y/D x between the two points. Note that, in each of these forms, the slope appears as the number multiplying the independent variable, x.
Two linear equations are equivalent if they have the same slope and both pass through the same point.
For example, the graph of the equation
y – 5 = 3(x – 1)
is a line that passes through the point (1,5) and has slope 3. The equation
y = 3x + 2
is an equivalent equation because the slope is 3 and the point (1,5) is a solution.
y = 3(1) + 2
y = 5
also passes through the point (1,5). You could verify that fact by substituting the pair (1,5) for x and y into each equation.
The main question of this unit is, "How can you identify and describe a relationship between two variables so that you can predict values of one variable from values of the other?"
First, collect data on the two variables. As noted above, a scatter plot is a useful display for gaining insight into possible relationships. From the scatter plot, check the direction (positive, negative, or neither) and the form (linear or nonlinear) of the relationship.
If a scatter plot has a linear form, you can "fit’ a line to the data and use the equation of your line to make predictions. The principal tool in evaluating the fit of your line is the set of residual errors–the differences between the actual and predicted values of the dependent variable. Different criteria based on the residual errors can be used to determine the "bestfitting" line. Unfortunately, the "bestfitting" line according to one criterion is not always the best according to another. However, a "good" fit should always have residuals that are randomly scattered around the horizontal axis.
One of the most commonly used criteria for determining the "bestfitting" line is called the leastsquares criterion. The leastsquares line has the smallest sum of the square errors (residuals). Also referred to as the regression line, it is popular because it generally does a good job of describing data that have a linear form. However, when outliers are present or when the scatter plot does not have a linear form, the leastsquares line, or any other line, can do a very poor job of describing the pattern of a scatter plot.
A plot of the residuals versus the independent variable can be very helpful in spotting outliers or nonlinear data. Such plots can display outliers more prominently than a scatter plot of the original data. Also, if the data have a nonlinear form, a residual plot will show a strong pattern.
When outliers are present, removing the outliers and refitting a linear model to the remaining data may produce a better prediction model. However, when data have a nonlinear form, no line will adequately describe the pattern of the data. In this situation, look for a different kind of model.
The precision of a prediction is linked to the variability inherent in the data. For example, suppose you had the following data on student heights (in cm): 150, 152, 154, 156, 158. If you were asked to predict the height of a student in this group, you might decide to chose the mean height of 154 cm for your prediction. In this case, the actual height could be as short as 150 cm or as tall as 158 cm; so you could be as far off as 4 cm. You can use a similar approach when dealing with relationships between two variables by examining the variability in the residuals.
In some situations, you may have two independent variables that are linearly related to the same dependent variable. In this case, it is generally best to base your predictions on the independent variable that has the stronger linear relationship with the dependent variable. Strong relationships have low variability, so one way of determining the strength of the linear relationship is to use the sum of the squared errors. For example, you could select the leastsquares line associated with the independent variable that has the smaller sum of square residuals. If the data on the two independent variables contain different numbers of observations, select the leastsquares line associated with the independent variable that has the smaller average squared error.
Key Concepts 
Dot plot: Display in which dots are placed above a number line to represent the values of data for a single variable
Independent variable: The variable on which a prediction is based; the variable that "explains" the dependent variable. Mathematicians frequently use the letter "x" to represent this in noncontextual situations.
Leastsquares criterion: Choose the line with the smallest sum of squared errors (SSE).
Leastsquares line: The line that satisfies the leastsquares criterion
Linear equation: An equation relating two variables, x and y, that can be put in the form y = mx + b
Linear form: The form of a scatter plot for which it is possible to draw a line that describes the general flow of the data
Linear regression: Fitting a line to data using the leastsquares criterion
Negative relationship: A relationship between two variables in which one variable tends to decrease while the other increases
Nonlinear form: The form of a scatter plot on which the general flow of the data is not well described by a straight line
Outlier: In a collection of data, an individual data point that falls outside the general pattern of the other data
Pointslope form: y – k = m(x – h); a form for a linear equation where (h, k) is a point on the line and m is the slope of the line
Positive relationship: A relationship between two variables in which both variables tend to increase together
Regression: Fitting lines or curves to data
Residual errors: Actual value of the dependent variable minus the predicted value
Residual plot: A scatter plot of the residuals versus the independent variable
Dependent variable: The variable that is to be predicted; the variable that "responds" to changes in the independent variable. Mathematicians frequently use the letter "y" to represent this in noncontextual situations.
Scatter plot: A plot of ordered pairs of data
Slopeintercept form: y = mx + b; a form for a linear equation where m is the slope and b is the yintercept
SSE: The sum of the squared errors
Strong relationship: A scatter plot of the data lies in a narrow band.
Weak relationship: A scatter plot of the data does not lie in a narrow band; they are more scattered.
Versus: When used in the phrase y versus x, it describes a scatter plot of y and x in which y is the dependent variable and x is the independent variable.
Solution to Short Modeling Practice 
Solution to Christmas Tree Farming 
Table 3 

Elevation 
Mean Max Temp 
Mean Min Temp 
1,000' 
67.3°F 
40.0°F 
2,000'  63.8°F  36.5°F 
3,000' 
60.3°F 
33.0°F 
4,000' 
56.8°F 
29.5°F 
5,000' 
53.3°F 
26.0°F 
6,000' 
49.8°F 
22.5°F 
7,000' 
46.3°F 
19.0°F 
2.
Yes, both the maximum and minimum temperatureelevation relationships are linear.
Yes
Tract 1 
2,000ft elevation 
mean max temp = 63.8°F mean min temp = 36.5°F 
3,100ft elevation 
mean max temp = 60.0°F mean min temp = 32.7°F 

Tract 2 
3,200ft elevation 
mean max temp = 59.6°F mean min temp = 32.3°F 
4,200ft elevation 
mean max temp = 56.1°F mean min temp = 28.8°F 

Tract 3 
5,500ft elevation 
mean max temp = 51.6°F mean min temp = 24.3°F 
Tract 1 = Douglas fir
Tract 2 = Douglas fir and noble fir
Tract 3 = Noble fir
Yes, Tract 2 is a favorable site for both of these species of trees.
Solutions to Practice and Review Problems 
=
=
80 mg * = = 14.9 milligrams
0.2F_{1} = 500D_{2}
F_{1} = = 2500D_{2}
4000 – 44C = –1340 + 134C
134C + 44C = 4000 + 1340
178C = 5340
C » 30
=
35x = 250
x = 7.14
between $5 and $10.
Water Pressure Versus Depth 
Air Pressure Versus Altitude 

Depth (feet) 
Pressure (Atm) 
Altitude (1000 ft) 
Pressure (Atm) 

0  1  0  1.00  
33  2  10  0.69  
66  3  20  0.46  
100  4  30  0.30  
133  5  40  0.19  
166  61  50  l0.11  
200  7  60  0.07  
300  10  
400  13  
500  16 
The mean is 9.805 ¸ 12 » 0.817 or about 81.7%.
W = 2.7F
F = W
7
10
where P is the cost to purchase the mice in dollars, and
n is the number of mice purchased.
4.90n = 830 + 0.60 n
Then solve for n. Subtract 0.60 n from both sides.
4.90n – 0.60 n = 830 + 0.60 n – 0.60 n
430n = 830
Then, divide both sides by 4.30.
n = 193 mice (rounded)
B = 830 + 0.60 (200)
B = 950.00, or $950.00
while the cost for purchasing would be
P = 4.90 (200)
P = 980.00, or $980.00
Thus, it would be cheaper to breed the mice if more than 193 mice per week were needed.
where L is the cost for labor and equipment in dollars, and
T is the time required, in hours.
Number of bags = (1 bag × 60 min) Û 10 min
Number of bags = 6 bags
Thus, since 6 bags are used each hour, the sand will cost $24 (6 bags × $4 per bag) each hour. So, …
S = 24 T
where S is the cost for sand in dollars, and
T is the number of hours the sandblaster is used.
C = 24T + 40T + 110
C = 64T + 110
which is in standard form. The variables are C and T, the coefficient is 64, and the constant is 110.
C = 64 (2) + 110
C = 238
and the cost is $238.
For 4 hours, T = 4,
C = 64 (4) + 110
C = 366
and the cost is $366.
For 6 hours, T = 6,
C = 64 (6) + 110
C = 494