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3.3 Least-Squares Regression.  Calculate the least squares regression line  Predict data using your LSRL  Determine and interpret the coefficient of.

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Presentation on theme: "3.3 Least-Squares Regression.  Calculate the least squares regression line  Predict data using your LSRL  Determine and interpret the coefficient of."— Presentation transcript:

1 3.3 Least-Squares Regression

2  Calculate the least squares regression line  Predict data using your LSRL  Determine and interpret the coefficient of determination  Calculate and graph residual plots

3  If there is a linear relationship, we summarize this overall pattern by drawing a line through the scatterplot.  Least-squares regression is a method for finding a line that summarizes the relationship between two variables, but only in a linear setting.

4  A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. We often use a regression line to predict the value of y for a given value of x.  Regression, unlike correlation, requires that we have an explanatory variable and a response variable.  ** We use this line to show a linear trend. (Ex: We can find the correlation Between your GPA and # of siblings, but we don’t find a regression line to predict data. Do you really want your GPA predicted from the # of siblings you have!!!)

5 Error= Observed – Predicted  Example: We predict the height of a ladder to be 4.9ft, but it is actually 5.1 ft. Find the error. Error= O- P= 5.1- 4.9 = 0.2 ft. A better term for your error is residual. ***A positive residual=prediction was to small***

6  The Least-squares regression line of y on x is the line that makes the sum of the squares of the vertical distances of the data points from the line as small as possible. (This is because we are comparing the observed y value to the predicted y value-not our x values)

7 y= observed response and predicted response (“y hat”) How to find the LSRL on your calculator.

8  Input: x into L1 y into L2 Stat-calc-8(linReg(ax+b)) *we use 8 not 4 like you did in Algebra In Algebra we used y=mx+b where b=y-int. In stat we use y=a+bx where b=slope. # of sneakers owned Amount of exercise per week 12 35 716 1330

9 TI-84: Stat-calc-8 (use 8 not 4) Xlist:L1 Ylist:L2 FreqList: (leave this blank) Store RegEQ: Y1 (vars-Yvars-1-1) Calculate TI-83: Stat-calc-8 (use 8 not 4) then type in L1,L2, Y1, enter -1.11+2.39x where x=#of sneakers y=amt. of exercise # of sneaker s owned Amount of exercise per week 12 35 716 1330

10  Use this data to predict the amount of exercise per week when you own 7 pairs of sneakers. y(7)= -1.11+2.39(7)=15.64 The easy way to do this on your calc since you already told the calc to put the eq. into your y1: Go to a clear home screen. Vars-yvars-function-y1 On your home screen you’ll have a y1. Type y1(7), hit enter  What is the residual? (Remember Residual=Obs. –Pred.) Residual= 16-15.64= 0.36 # of sneakers owned Amount of exercise per week 12 35 716 1330

11  Interpret the slope and y-intercept in context of the problem: Slope: On average, for every change in x, y changes by b. Y-intercept: When x=0, we predict y to be a. *we need to put the words on average and we predict because the line doesn’t touch every point, so the slope and y-int are not exactly what is happening, just estimates.

12 -1.11+2.39x Slope: On average, for every pair of sneakers you own, there is an 2.39 increase in the number of hours you exercise per week Y-intercept: When you don’t own any sneakers, we predict you to exercise -1.11 hours Does the y-int make sense in this problem? NO!! You can’t exercise negative hours

13  We are given: r, *** The point always falls directly on your line of best fit****  The least-squares regression line is the line:  With slope  And intercept

14  Suppose we have an explanatory (standing reach)and response variable (jumping reach) and we know: Even though we don’t know the actual data, we can still construct the equation for the least-squares line and use it to make predictions.

15 b= 0.9952(.9/.6) = 1.4928 a= 7.3-(1.4928)(5.1)= -.3133  therefore -.3133 + 1.4928x  (Hint: Don’t forget to define your variables!) x= standing reach y= jumping reach

16 Slope: On average, for every increase of your standing reach, there is an 1.49 increase in your jumping reach Y-intercept: When your standing reach is 0, we predict your jumping reach to be -0.333.

17 The coefficient of determination,, is the fraction of the variation in the values of y that is explained by least-squares regression of y on x.  In other words: % of the change in y can be explained by the change in x.

18 # of hours studied Test grade 172 280 387 494 590

19 # of hours practiced # of wins 04 13 21 48 69

20  Fact 1: we find slope and intercept by using means, standard deviations, an correlation.  Fact2: we use the regression line to predict y for any given x  Fact 3:recognize outliers and potentially influential points  Fact 4:we use the regression line to calculate residuals. (we look for patterns)

21  Predict data outside our realm of data given. Example: Predict gas prices for 2012 yearAve. gas price 2001$1.09 2004$2.89 2007$3.15 2011$3.70

22  - the difference between an observed value of the response variable and the value predicted by the regression line. That is:  Residual= observed y – predicted y =

23  The sum of the residuals is always 0!!

24 Step 1: Input Data into L₁ and L₂ Step 2: STAT/CALC/8/ENTER Step 3: LinReg(a+bx) L₁,L₂, Y₁ Then go back into your lists and input L3 L3= Obs-Pred=L2-y1(L1) Plot: L₁, L₃ Don’t forget to label your x axis with your x and your y-axis is your residuals. L₁L₂L₃ XYL₂- Y₁(L₁)

25 Follow each step exactly!! # of hours practiced # of wins 04 13 21 48 69

26  1- A curved pattern  2- Increasing or Decreasing pattern  3- Outliers or influential points Any pattern in a residual plot tells us it’s NOT a good linear fit Here’s what to look for when you examine residuals, using either a scatterplot of the data or a residual plot.

27  Outlier-  Influential-


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