Test 7

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Jan 9, 2024

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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
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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