Slide 1Test 3Practice ProblemsAdditional Reading and ExamplesSlide 5Motivating ExampleDescribing RelationshipsExample 26Correlation CoefficientTI-83/84 CalculatorAnscombe Data – Page 111Slide 12Data Set 1Data Set 2Data Set 3Data Set 4Top Hat 2C. Regression LineRegression LineRegression LineRegression LineEquation of a LineEquation of a LinePrediction EquationResidualMethod of Least SquaresLeast Squares EstimatesExample 27Example 27Example 27Example 27Example 27Example 27Example 27Example 27TI-83/84 CalculatorMotivating ExampleMotivating ExampleMotivating ExampleMotivating ExampleSTAT 210Lecture 15Describing Relationships Between VariablesSeptember 29, 2017Test 3Friday, October 6Covers chapter 5, pages 99 – 138Combination of multiple choice questions and short answer questions and problems.Formulas provided, please bring calculator and writing instrument.Practice ProblemsPages 130 through 137Relevant problems: V.3 (c), V.4 (c), V.7 (b), V.9 (d), and V.10 (c) Recommended problems: V.3 (c), V.7 (b), V.9 (d), and V.10 (c)Additional Reading and ExamplesRead pages 120 and 121Top Hat 2Motivating ExampleWatching television also often means watching or dealing with commercials, and of interest is to describe the relationship between the number of hours of television watched per day and the number of commercials watched.Describing RelationshipsTo describe the relationship between two variables we must describe the direction, form, and strengthof the relationship.A scatterplot and the correlation coefficient are twostatistical tools that can be used to help describe therelationship.Example 26x y x2 y2 xy 6 15 36 225 9020 31 400 961 620 0 10 0 100 014 16 196 256 22425 28 625 784 70016 20 256 400 32028 40 784 1600 112018 25 324 625 45010 12 100 144 120 8 15 64 225 120145 212 2785 5320 3764S x S y S x2 S y2 S xyCorrelation CoefficientSxx = S x2 - ( S x )2 nSyy = S y2 - ( S y )2 nSxy = S xy - ( S x )(S y ) nTI-83/84 Calculator1. First, turn on diagnostics. Hit 2ND and 0, bringing up the Catalog. Scroll down to DiagnosticOn and hit Enter twice. You only need to do this the first time.2. Enter the data into two lists, say L1 and L2.3. Hit STAT, then CALC, and choose option 8:LinReg(a+bx)4. Enter the list containing the X data (say L1), then comma (,), then the list containing the Y data (say L2). Hit Enter and r = is the correlation coefficient.Anscombe Data – Page 111As directed in class, compute the correlation coefficient for the set of data you are assigned. Determine the value to two decimal places.You can and are encouraged to work together.Data Set 1 x y Data Set 2 x yData Set 3 x yData Set 4 x y10 8.04 8 6.95 13 7.58 9 8.8111 8.33 14 9.96 6 7.24 4 4.26 12 10.84 7 4.82 5 5.68 10 9.14 8 8.1413 8.74 9 8.77 11 9.26 14 8.10 6 6.13 4 3.10 12 9.13 7 7.26 5 4.74 10 7.46 8 6.77 13 12.74 9 7.11 11 7.81 14 8.84 6 6.08 4 5.39 12 8.15 7 6.42 5 5.73 8 6.58 8 5.76 8 7.71 8 8.84 8 8.47 8 7.04 8 5.25 19 12.50 8 5.56 8 7.91 8 6.89Data Set 1Anscombe data set10246810120 5 10 15xyData Set 2Data Set 3Data Set 4Top Hat 2C. Regression LineNow our goal is to determine the equation of the line that best models (explains) the relationship between X and Y. This is referred to as the regression line.Regression Line Y . . . . . . . . . . . . . . . . XRegression Line Y . . . . . . . . . . . . . . . . XRegression Line Y . . . . . . . . . . . . . . . . XEquation of a LineY = intercept + slope(X)The intercept is the predicted value of Y when x = 0. Hence when x = 0, the predicted y is the intercept value.The slope is the amount that Y changes (increases or decreases) when X is increased by one unit. Hence if x increases by 1 unit, the predicted y increases or decreases by slope units.Equation of a LineY = intercept + slope(X)where the intercept is the predicted value of Y when X = 0and the slope is the amount that Y changes (increases or decreases)when X is increased by one unit.Example: weight (in pounds) = 6 + 10 * age (in years) intercept = 6 : when a child is 0 years old, the child is predicted to weigh 6 poundsslope = 10: if a child’s age increases by one year, then his or her weight is predicted to increase by 10 pounds (weight increases 10 pounds each year).Prediction EquationOnce we determine the intercept and the slope, we can use the line Y = intercept + slope(X) to predict values of Y given values of X.The prediction equation is Y = intercept +slope(X)Residual y = the observed value of the dependent variable y = the predicted value of the dependent variable y - y is called a residual and our goal is to make the residuals as small as possible.Method of Least SquaresDetermine values of the intercept and slope such that the sum of the squared residuals is minimized:minimize S (Y - Y)2 which is equivalent tominimize S [Y – intercept – slope(X)]2Least Squares Estimatesslope = Sxy = r sy Sxx sxintercept = y - slope x where Sxx and Sxy are defined on page 102 and r is the correlation between X and Y. Again x and y are the means of the X and Y data, respectively, and sx and sy are the standard deviations of the X and Y data, respectively.Example 27From example 26:Sxx = 682.5 Syy = 825.6 Sxy = 690x = S x = 145/10 = 14.5 ny = S y = 212/10 = 21.2 nExample 27slope = Sxy / SxxExample 27slope = Sxy / Sxx = 690 / 682.5 = 1.011Example 27slope = Sxy / Sxx = 690 / 682.5 = 1.011This implies that as the number of ads run increases by one ad, the predicted number of cars sold increases by 1.011 cars.Increases since positive.Example 27intercept = y - slope xExample 27intercept = y - slope x = 21.2 - 1.011 (14.5)= 21.2 - 14.66= 6.54Example 27intercept = y - slope x = 21.2 -
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