Understanding Outliers in Least Squares Regression Analysis

n 1 1 The least squares regression line for a data set is yˆ= -2.3−0.33x and the standard deviation of the residuals is 0.26. Does a case with the values x = -3.33, y = -1.27 qualify as an outlier? Yes No Cannot be determined with the given information Hide question 1 feedback Plug in -3.33 for x. y = -2.3 -.33(-3.33) y = -1.2011 Residual is y-given - y-predicted. -1.27 - (-1.2011) -1.27 + 1.2011 = -.2011 -> this is the residual value. To see if it is an outlier take -2 and multiply it by .26 -2*.26 = -.52 -.2011 is greater than -.52, No, it is not an outlier because if it inside the range of the -2 to 2. Question 2 The least squares regression line for a data set is yˆ= -4.6+1.56x and the standard deviation of the residuals is .5 Does a case with the values x = -1.12, y = -8 qualify as an outlier? Cannot be determined with the given information No Yes Hide question 2 feedback Plug in -1.12 for x. y = -4.6 + 1.56(-1.12) y = -6.3472 Residual is y-given - y-predicted.

-8 - (-6.3472) -8 + 6.3472 = -1.6528 -> this is the residual value. To see if it is an outlier take -2 and multiply it by .52 -2*.52 = -1.04 -1.6528 is less than -1.04, Yes, it is an outlier because if it outside of the -2 to 2 range. Question 3 The following data represent the weight of a child riding a bike and the rolling distance achieved after going d Weight (lbs.) Rolling Distance (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48 91 42 52 39 63 33 71 39 100 49 89 55 103 53 99 42 74 33 75 30 89 30 102 40 103 33

99 33 102 35 86 37 85 37 Using the regression line for this problem, the approximate rolling distance for a child on a bike that weighs 9 43.982 58.7213 44.3761 45.6723 Hide question 3 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on th to predict the distance of the bike. Once you get the Regression output, look under the Coefficients equation. y = 10.3364819 + 0.343834842 (x) Plug 99 in for x and solve. y = 10.3364819 + 0.343834842 (99) y = 44.37613122 Question 4 The marketing manager of a large supermarket chain would like to use shelf space to predict the sales of pet similar stores, she gathered the following information regarding the shelf space, in feet, devoted to pet food of dollars. . Store Shelf Space Weekly Sales 1 5 1.3 2 5 1.6 3 5 1.4 4 10 1.7

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5 10 1.9 6 10 2.3 7 15 2.2 8 15 2 9 15 1.8 10 20 2.2 11 20 2.4 12 20 2.9 13 25 2.9 14 25 2.7 15 25 2.5 Can it be concluded at a 0.01 level of significance that there is a linear correlation between the two variables no, because the p-value = .000013 yes, because the p-value = .00053 yes, because the p-value = .00053 yes, because the p-value = .000013 Hide question 4 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable is the Weekly Sales and the x-variable is the Shelf Space. You want to predict the dollar am you highlight and input these columns in the Regression Analysis make sure you include AND click on La the Regression output, look under the Significance F value for the correct p-value to use to make your deci Yes, there is a significant relationship p-value = 0.000013 Question 5 The following data represent the weight of a child riding a bike and the rolling distance achieved after goin

Weight (lbs.) Rolling Distance (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48 91 42 52 39 63 33 71 39 100 49 89 55 103 53 99 42 74 33 Find the 95% prediction interval for rolling distance when a child riding the bike weighs 106 lbs. (round to ___< y < ___

Answer for blank # 1: Answer for blank # 2: Hide question 5 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on to predict the distance of the bike. Once you get the Regression output, look under the equation. y = -0.508294634 + 0.52329484 (x) Plug 106 in for x and solve. y = -0.508294634 + 0.52329484 (106) y = 54.96095837, this is our y-hat value. This is the equation to use for the prediction interval y^±t ∗ (SE)1+1n+(x0−x¯)2(n−1)SDx2 T-Critical Value =T.INV.2T(.05, 14) = 2.144786688 The SE we get from the Regression output and you can use Excel to find the Average and SD of the Weigh LL =54.96095837 - 2.144786688*6.679572112*1+116+(106−82.1875)2(16−1) ∗ 17.440262 UL =54.96095837 + 2.144786688*6.679572112*1+116+(106−82.1875)2(16−1) ∗ 17.440262 Question 6 Which of the following describes how the scatter plot appears? Select all that apply.

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strong weak negative positive The city of Oakdale wishes to see if there is a linear relationship between the temperature and the amount of electricity used (in kilowatts). Using that data, find the estimated regression equation which can be used to estimate Kilowatts when using Temperature as the predictor variable. Temperat ure (x) K il o w at ts (y ) 73 6 8 0 78 7 6

0 85 9 1 0 98 1 5 1 0 93 1 1 7 0 83 8 8 8 92 9 2 3 81 8 3 7 76 6 0 0 105 1 8 0 0 Kilowatts = -2003.896 + 34.858(Temperature) Kilowatts = 0.945 + 0.893(Temperature) Kilowatts = 371.223 + 4.269(Temperature)

Kilowatts = 132.031 + 34.858(Temperature) Hide question 7 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression Highlight Kilowatt for the Y Input: Highlight Temperature for the X Input: Make sure you click on Labels and Click OK If done correctly then you look under the Coefficients for the values to write out the Regression Equati Coefficients Intercept Temperature (x) Kilowatts = -2003.895859 + 34.85759097(Temperature) Question 8 A company want to find out if there is a linear relationship between indirect labor expense (ILE), in dollars direct labor hours (DLH). Data for direct labor hours and indirect labor expense for 25 months are given. Based on the data in the table below, is there a significant linear relationship between Direct Labor Hours a the Indirect Labor Expense? Please see attached Excel for data. ILE_and_DLH data No, the sample correlation coefficient is equal to 0.878, which provides evidence of a significant linear relationship. No, because the p-value = 0.00023 Yes, the sample correlation coefficient is equal to 0.878, which provides evidence of a significant linear relationship. Yes, because the p-value = 0.00023 Hide question 8 feedback You will run a Simple Linear Regression Analysis in Excel using the Data Analysis ToolPak.

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Data -> Data Analysis -> Scroll to Regression Highlight ILE for the Y Input: Highlight DLH for the X Input: Make sure you click on Labels and Click OK If done correctly then Significance F 0.000227794 Significance F or p-value = 0.00023 Because the p-value < .05, Reject Ho. Yes, there is a significant relationship, between DHL and ILE. Question 9 The city of Oakdale wishes to see if there is a linear relationship between the temperature and the am electricity used (in kilowatts). Temperature (x) Kilowatts (y) 73 78 85 98 93 83 92 81 76 105 Based on your results, If the temperature increases by 1 degree, Kilowatts, on average, increases by a decimal places.

A n s w e r: 3 4 . 8 5 8 Hide question 9 feedback You are interpreting the slope for this problem. Copy and Paste the Data into Excel. Highlight Kilowatt for the Y Input: Highlight Temperature for the X Input: Make sure you click on Labels and Click OK If done correctly then you look under the Coefficients for the values to write out the Regressio Intercept Temperature (x) Kilowatts = -2003.895859 + 34.85759097(Temperature) The slope is 34.858 At the Temperature increases by 1 degree, then Kilowatts will change by whatever the slope is Kilowatts. As Temp. increases by 1 degree, Kilowatts will increase by 34.858. Question 10 The city of Oakdale wishes to see if there is a linear relationship between the temperature and electricity used (in kilowatts). Temperature (x) 73 78 85 98 93 83 92

81 76 105 Approximately what percentage of the variation in Kilowatts is accounted for by Temperature Place your answer, rounded to 1 decimal place, in the blank. Do not use any stray punctuation would be a legitimate entry. ___% A n s w e r: 8 9 . 3 Hide question 10 feedback The R-squared value is the amount of explained variance in the data points in the model Copy and Paste the Data into Excel. Highlight Kilowatt for the Y Input: Highlight Temperature for the X Input: Make sure you click on Labels and Click OK If done correctly then Multiple R = 0.944907859 R Square = 0.892850862 89.3% of variation in the Kilowatts is accounted for by Temperature in this model. Note: Correlation is a value between -1 and 1. R-square gets converted from a decimal to a percentage. Question 11 A company want to find out if there is a linear relationship between indirect labor expen Data for direct labor hours and indirect labor expense for 25 months are given. Using that data, find the estimated regression equation which can be used to estimate IL Please see attached Excel for data. ILE_and_DLH data ILE = 0.4291 + 59.4174(DLH)

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ILE = 171.7567 + 9.3172(DLH) ILE = 60.0419 + 2.1355(DLH) ILE = 47.5504 + 4.8996(DLH) Hide question 11 feedback You will run a Simple Linear Regression Analysis in Excel using the Data Analysis T Data -> Data Analysis -> Scroll to Regression Highlight ILE for the Y Input: Highlight DLH for the X Input: Make sure you click on Labels and Click OK If done correctly then you look under the Coefficients for the values to write out the R Intercept DLH(X) ILE = 171.7567 + 9.3172(DLH) Question 12 A teacher believes that the third homework assignment is a key predictor in how well third homework score and y the midterm exam score. A random sample of last terms s below. Assume scores are normally distributed. Calculate the correlation coefficient u Excel). Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. HW3 12.9 21.9 8.8 24.3

6.6 13.2 21.9 19.2 20 15.4 25 12.7 6.4 20.2 22.8 23.1 22 11.9 14.9 18.2 15.1 15.2 17.1 Answer:___ ___ Answer: Hide question 12 feedback Use =CORREL function n in Excel. Question 13 Bone mineral density and cola consumption has been recorded for a sample of patients week and y the bone mineral density in grams per cubic centimeter. Assume the data i

Cola Consumed 1 2 3 4 5 6 7 8 9 10 11 Using that data, find the estimated regression equation which can be used to estimate B the predictor variable. Bone Mineral Density = 0.891716 - 0.002389(Colas Consumed) Bone Mineral Density = 0.201737 +0.01334(Colas Consumed) Bone Mineral Density = 103.3549 - 1.87809(Colas Consumed) Bone Mineral Density = 0.008627 + 0.001272(Colas Consumed) Hide question 13 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression Highlight Bone Mineral Density for the Y Input: Highlight Colas Consumed for the X Input: Make sure you click on Labels and Click OK If done correctly then you look under the Coefficients for the values to write out the

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