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**Section 2.6: Draw Scatter Plots & best-Fitting Lines(Linear Regresion)**

Adv. Algebra Section 2.6: Draw Scatter Plots & best-Fitting Lines(Linear Regresion) A ______________________ is a graph used to determine whether there is a relationship between paired data. The relationship between the data however does not form a perfectly straight line, but the graph will show a linear relationship. When this is the case, a ________________________ can be drawn splitting the data in half. A ______________________ can be found from the line of best-fit to model the data. A prediction equation is used when a rough estimate is a sufficient solution. If the data follows these patterns, then the data shows a linear relationship. If ____ tends to increase as ____ increases, then there is a ________________________. scatter plot line of best-fit prediction equation y x positive correlation Line has positive slope.

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**If ____ tends to decrease as ____ increases, then there is a ________________________.**

y x negative correlation Line has a negative slope. If the points show no linear pattern(very dispersed), then there is relatively _______________________. no correlation Since there is no correlation between the data we can not draw a line of best-fit or write a prediction equation.

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**Steps to Drawing the Line of Best-Fit**

1) Draw a scatter plot of the data. Draw a line of best-fit that intersects two points from the scatter plot, but there should be as many points above this line as below this line. 3) Find the slope of the line by using the two points it intersects. 4) Write an equation of this line from the slope and a point on the line. (Prediction Line) Example 1: The table shows the number of U.S. production of beef from 1990 to 1997, where x is years since 1990 and y is billions of beef in pounds. Production of Beef (In pounds) Years , (x) Beef, (y)

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**A) Graph the data and approximate the best-fitting line.**

Since the line of best-fit pass through (0, 22.7) and (7, 25.5), we will use these points to write your prediction equation. # i of n b b e i e l f l o n s 25 24 23 22 𝑚 = 25.5 − − 0 = = 0.4 # of years B) Use the fitted line to estimate the beef production in the year 2005. 𝑦 − 𝑦 1 = 𝑚 (𝑥 − 𝑥 1 𝑦 − 22.7 = 0.4 𝑥 − 0 𝑦 − 22.7 = 0.4𝑥 𝑦 = 0.4𝑥 + 22.7 Let x = 15 Y = 0.4(15) = = 28.7 Billion lbs. of Beef

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goodness of fit Data that are linear in nature will have varying degrees of ________________________ to the lines of fit. Various formulas are used to find a _________________________ that describes the nature of the data. The more closely the data fit a line, the closer the correlation coefficient r approaches _____ or _____. Statisticians normally use precise procedures relying on computers to determine the correlation coefficients. The graphing calculator uses the _________________________________ which is represented by r. When using these methods, the best-fit line is often called a ________________. In the next example we will use our graphing calculator to find to find the regression line and make a prediction with its equation. correlation coefficient 1 -1 Pearson Product-Moment Correlation regression line

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Example 2: The table contains the carbohydrate and calorie content of 8 foods, ranked according to carbohydrate content. Food Carbohydrates (g) Calories Cabbage (boiled) 1.1 9 Frozen Peas (boiled) 4.3 41 Orange (peeled) 8.5 35 Apple (Raw) 11.9 46 Potatoes 19.7 80 Rice 29.6 123 White Bread 49.7 233 Whole Wheat Flour 65.8 318

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**a. Use a graphing Calculator to find the equation of the regression line and the**

Pearson product-moment correlation. b Use the equation from step A to predict the # of calories in a food with 75.9 g of carbohydrates. We need to make sure your correlation coefficient is on. 1. Go to Catalog 2. Scroll down to Diagnostic ON, hit enter. Hit the STAT button, Hit 1: Edit, Put the carbohydrates in L1 and Calories in L2. Draw a scatter plot of the data. Make sure the scatter plot is on. Then graph. Hit the STAT button, Use the rt. Arrow key to move to CALC. , Then hit 4: LinReg y = 4.70x – 1.43 r = .99 y = 4.70(75.9) – 1.43 = – 1.43 = 355.3 Calories

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