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**2-7 Curve Fitting with Linear Models LESSON PLAN Warm Up (Slide #2)**

Objective and PA State Standards (Slide #3) Vocab (Slides #4 – 8) Lesson Presentation (Slides #9 – 24) Text Questions - NONE Worksheet–2.7A (Slide #25) Lesson Quiz (Slides #26 – 28) Holt Algebra 2

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Warm Up Write the equation of the line passing through each pair of passing points in slope-intercept form. 1. (5, –1), (0, –3) 2. (8, 5), (–8, 7) Use the equation y = –0.2x + 4. Find x for each given value of y. 3. y = y = 3.5

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Objectives Fit scatter plot data using linear models with and without technology. Use linear models to make predictions. Pa. Dept of Educ. Math Standards and Anchors: 2.6.A2.C/E 2.7.A2.C/E M11.E.1.1.2 4.1.1 4.2.1 4.2.2

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**Vocabulary regression correlation line of best fit**

correlation coefficient

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Researchers, such as anthropologists, are often interested in how two or more measurements are related. (ex. Climate, Size, Species) The study of the relationship between two or more variables is called regression analysis.

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Scatter plots like those below are helpful in determining the strength and type of a relationship between variables. Correlation is the strength and direction of the relationship between the two variables.

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If there is a strong linear relationship between two variables, a line of best fit, or a line that best fits the data, can be used to make predictions. Try to have about the same number of points above and below the line of best fit. Helpful Hint

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**The correlation coefficient r is a measure of how well the data set is fit by a model.**

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**Example 1: Meteorology Application**

Albany and Sydney are about the same distance from the equator. Make a scatter plot with Albany’s temperature as the independent variable (the “x” coordinate). Name the type of correlation. Then sketch a line of best fit and find its equation.

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**• • • • • • • • • • • • Example 1 Continued**

Step 1 Plot the data points. Notice that the data set is negatively correlated...as temperature rises in Albany, it falls in Sydney. Step 2 Identify the correlation. Step 3 Sketch a Line of Best Fit o • • • • • • • • • • • •

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**Step 4 Identify two points on the line.**

Example 1 Continued Step 4 Identify two points on the line. For this data, you might select (60,52) and (40,61). Step 5 Find the slope of the line that models the data. 61 – 52 9 -20 -.45 Step 6 Use either point-slope or slope intercept form to develop the equation y= mx + b y – y1= m(x – x1) 52 = -.45(60) + b y – 52 = –0.45(x – 60) 52 = b y – 52 = –0.45x + 27 y = –0.45x = b 79 = b y = –0.45x + 79 y = –0.45x + 79 An equation that models the data is y = –0.45x + 79

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Check It Out! Example 1 Make a scatter plot for this set of data. Identify the correlation, sketch a line of best fit, and find its equation.

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**Check It Out! Example 1 Continued**

Step 1 Plot the data points. Notice that the data set is positively correlated…as time increases, more points are scored Step 2 Identify the correlation. Step 3 Sketch a Line of Best Fit • • • • • • • • • •

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**Check It Out! Example 1 Continued**

Step 4 Identify two points on the line. For this data, you might select (20, 10) and (40, 25). Step 5 Find the slope of the line that models the data. Step 6 Develop the equation: Using point-slope form. Using slope-intercept form. y – y1= m(x – x1) y= m(x) + b 10= .75(20) + b y – 10 = 0.75(x – 20) 10= 15 + b -5= b y – 10 = 0.75x – 15 y = 0.75x – 5 y = 0.75x - 5

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**You can use a graphing calculator to **

graph scatter plots and lines of best fit. Find video tutorial links for both of these calculator functions SCATTER PLOTS and LINES OF BEST FIT in Moodle Algebra 2 Chapter 2

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**Example 2: Anthropology Application**

Anthropologists can use the femur, or thighbone, to estimate the height of a human being. The table shows the results of a randomly selected sample.

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**• • • • • • • • Example 2 Continued**

a. Make a scatter plot of the data with femur length as the independent variable (which means “y”). • • • • • • • •

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Example 2 Continued b. Find the correlation coefficient r and the line of best fit. Interpret the slope of the line of best fit in the context of the problem.

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Example 2 Continued c. A man’s femur is 41 cm long. Predict the man’s height. The equation of the line of best fit is h ≈ 2.91l Use it to predict the man’s height for a 41cm femur. h ≈ 2.91(41) Substitute 41 for l. h ≈ The height of a man with a 41-cm-long femur would be about 173 cm.

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Check It Out! Example 2 The gas mileage for randomly selected cars based upon engine horsepower is given in the table.

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**Check It Out! Example 2 Continued**

a. Make a scatter plot of the data with horsepower as the independent variable. • • • • • • • • The scatter plot is shown on the right. • •

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**Check It Out! Example 2 Continued**

b. Find the correlation coefficient r and the line of best fit. Interpret the slope of the line of best fit in the context of the problem. Enter the data into lists L1 and L2 on a graphing calculator. Use the linear regression feature by pressing STAT, choosing CALC, and selecting 4:LinReg. The equation of the line of best fit is y ≈ –0.15x

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**Check It Out! Example 2 Continued**

c. Predict the gas mileage for a 210-horsepower engine. The equation of the line of best fit is y ≈ –0.15x Use the equation to predict the gas mileage. For a 210-horsepower engine, y ≈ –0.15(210) Substitute 210 for x. y ≈ 16 The mileage for a 210-horsepower engine would be about 16.0 mi/gal.

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**A line of best fit may also be referred to as a trend line.**

Reading Math

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Worksheet 2.7A

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Lesson Quiz: Part I Use the table for Problems 1–3. 1. Make a scatter plot with mass as the independent variable.

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Lesson Quiz: Part II 2. Find the correlation coefficient and the equation of the line of best fit on your scatter plot. Draw the line of best fit on your scatter plot.

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Lesson Quiz: Part I Use the table for Problems 1–3. 1. Make a scatter plot with mass as the independent variable.

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Lesson Quiz: Part II 2. Find the correlation coefficient and the equation of the line of best fit on your scatter plot. Draw the line of best fit on your scatter plot. r ≈ 0.67 ; y = 0.07x – 5.24

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