Week 7 Test

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Week 7 Test - Results X Attempt 1 of 2 Written Oct 16, 2023 10:37 PM - Oct 17, 2023 12:40 AM Attempt Score 18 /20-90% Overall Grade (Highest Attempt) 18 /20 -90 % Question 1 1/ 1 point The least squares regression line for a data set isy"= -4.6+1.56x and the standard deviation of the residuals is .52 Does a case with the values x = -1.12, y = -8 qualify as an outlier? ) Cannot be determined with the given information O NG v Yes W Hide question 1 feedback Plug in -1.12 for x. y=-46+ 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 2 1/ 1 point A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour. How much is it if you rent the SCUBA equipment for 45 minutes? O %23 - $375 () $35 v() $34.38 ¥ Hide question 2 feedback The equation is Dollars = $12.5(hour) + $25 45 minutes = .75 of an hour Dollars = $12.5(.75) + $25 Question 3 1/ 1 point The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling.

Rollin Weight ns (Ibs) 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 99 Ibs. is: ) 43.982 587213 v 443761

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() 45.6723 w 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 the weight of the child. You want to predict the distance of the bike. Once you get the Regression output, look under the Coefficients to find the values to use for the regression 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 1/ 1 point The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling. Rolli Weight c.> ns Distance (Ibs.) (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48

921 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 Find the 99% prediction interval for rolling distance when a child riding the bike weighs 99 1bs. (round to 4 decimal places) <y< Answer for blank # 1: 19.3556 (50 %) Answer for blank # 2;: 69.3966 (50 %) w 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 distance and the x-variable is the weight. How far the bike will travel will depend on the weight of the child. You want to predict

the distance of the bike. Once you get the Regression output, look under the Coefficients to find the values to use for the regression equation. y = 10.3364819 + 0.343834842 (x) Plug 99 in for x and solve. y = 10.3364819 + 0.343834842 (99) y =44.37613122, this is our y-hat value. This is the equation to use for the prediction interval 1 (33() — 21_3)2 yEt*x (SE) 14+ — + J (SE) n (n—1)SDx? T-Critical Value =T.INV.2T(.01, 22) = 2.818756061 The SE we get from the Regression output and you can use Excel to find the Average and SD of the Weight variable. LL =44.37613122- 2.818756061%8.573835284" 1 (99 — 85.6667)2 1+— + : 24 (24 —1) % 16.00724 UL =44.37613122+ 2.818756061*8.573835284" 1 (99 — 85.6667)? 1+ — + . 24 (24 — 1) % 16.00724

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Question 5 1/ 1 point The marketing manager of a large supermarket chain would like to use shelf space to predict the sales of pet food. For a random sample of 15 similar stores, she gathered the following information regarding the shelf space, in feet, devoted to pet food and the weekly sales in hundreds of dollars. . Store Shelf Space 1 5 2 5 3 5 4 10 5 10 6 10 7 15 8 15 9 15 10 20 11 20 12 20 Weekly Sales 1.3 1.6 1.4 1.7 1.9 2.3 2.2 2.4 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 w 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 Weekly Sales and the x-variable is the Shelf Space. You want to predict the dollar amount of the weekly sales. When you highlight and input these columns in the Regression Analysis make sure you include AND click on Labels and Click OK. Once you get the Regression output, look under the Significance F value for the correct p-value to use to make your decision. Yes, there is a significant relationship p-value = 0.000013 Question 6 1/ 1 point Which of the following describes how the scatter plot appears? Select all that apply.

v | weak ‘/Cj linear v ) nonlinear v negative \/[] Positive Question 7 1/ 1 point Data for a sample of 30 apartments in a particular neighborhood are provided in the worksheet. You want to see if there is a direct relationship between Size of the Apartment and Rent. Using the estimated regression equation found by using Size as the predictor variable, find a point estimate for the average monthly Rent for apartments having 1,000 square feet of space. Place your answer, rounded to the nearest whole dollar, in the blank. When entering your answer do not use any labels, commas or symbols. Simply provide the numerical value. For example, 1234 would be a legitimate entry. Please see Attached Excel for Data. Apartments data Answer: 1091 w Hide question 7 feedback

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You will run a Simple Linear Regression Analysis in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression Highlight Rent for the Y Input: Highlight Size 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 Equation Coefficients Intercept 221.3667335 Size 0.8692387 Rent = 221.367 + 0.869(Size) Plug in 1000 for Size Rent = 221.367 + 0.869(1000) Question 8 1/ 1 point A company want to find out if there is a linear relationship between indirect labor expense (ILE), in dollars, and direct labor hours (DLH). 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 ILE when using DLH as the predictor variable. Please see attached Excel for data. ILE and DLH data () ILE = 47.5504 + 4.8996(DLH)

\/<> ILE = 171.7567 + 9.3172(DLH) () ILE = 0.4291 + 59.4174(DLH) O ILE = 60.0419 + 2.1355(DLH) ¥ Hide question 8 feedback You will run a Simple Linear Regression Analysis in Excel using the Data Analysis ToolPak. 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 Regression Equation Coefficients Intercept 171.7566705 DLH(X) 9.317247081 ILE = 171.7567 + 9.3172(DLH) Question 9 1/ 1 point 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. TemperatureKilowatts (x) (y) /3 680 /8 760 85 910 98 1510 93 1170 83 888 92 923 81 837 76 600 105 1800 ) Kilowatts = 371.223 + 4.269(Temperature) ) Kilowatts = 0.945 + 0.893(Temperature) ¢<> Kilowatts = -2003.896 + 34.858(Temperature) > Kilowatts = 132.031 + 34.858(Temperature) w Hide question 9 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression

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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 Equation Coefficients Intercept -2003.895859 (Tf)mperat“re 34.85759097 Kilowatts = -2003.895859 + 34.85759097(Temperature) Question 10 1/ 1 point A company want to find out if there is a linear relationship between indirect labor expense (ILE), in dollars, and direct labor hours (DLH). Data for direct labor hours and indirect labor expense for 25 months are given. Approximately what percentage of the variation in indirect labor expenses is explained by the regression model you derived? Place your answer, rounded to 2 decimal places, in the blank. Do not use any stray punctuation marks or a percentage sign. For example, 78.91 would be a legitimate entry. __ % Please see attached Excel for data. ILE_and_DLH data

Answer: 45.29 w Hide question 10 feedback The R-squared value is the amount of explained variance in the data points in the model. You convert this decimal to a percent. You will run a Simple Linear Regression Analysis in Excel using the Data Analysis ToolPak. 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 Multiple R 0.672942351 R Square 0.452851408 45.29% of variation in the ILE is accounted for by DLH in this model. Note: Correlation is a value between -1 and 1. This STAYS a decimal. R-square gets converted from a decimal to a percentage. The Correlation IS NOT a percent, leave it as a decimal. Question 11 0/ 1 point The city of Oakdale wishes to see if there is a linear relationship between the temperature and the amount of electricity used (in kilowatts). TemperatureKilowatts (x) (y)

73 680 /8 760 85 9210 98 1510 93 1170 83 888 92 923 81 837 76 600 105 1800 Based on your results, If the temperature increases by 1 degree, Kilowatts, on average, increases by approximately how much? Round to 3 decimal places. Answer: 38.858 ¢ (34.858) w Hide question 11 feedback You are interpreting the slope for this problem. 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 Equation Coefficients Intercept -2003.895859 Temperature 34.85759097 (x)

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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. In our regression out the slope is 34.858 Kilowatts. As Temp. increases by 1 degree, Kilowatts will increase by 34.858. Question 12 0/ 1 point An object 1s thrown from the top of a building. The following data measure the height of the object from the ground for a five-second period. Calculate the correlation coefficient using technology (you can copy and paste the data into Excel). Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. Seconds | Height 0.5 112.5 1] 110.875 1.5 106.8 2| 100.275 2.5 91.3 3] 81.235 3.5 71.553 4 61.83 4.5 45.65 5 0 Answer: Answer: -0.9305 ¢ (-0.9304) w Hide question 12 feedback

Use =CORREL function in Excel Question 13 1/ 1 point A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed. HW3 Midterm 13.3 59.811 21.9 87.539 9.7 53.728 25 96.283 54 39.174 13.2 66.092 20.9 89.729 18.5 78.985 20 86.2 154 73.274 25 93.25 9.7 52257 6.4 43.984 20.2 79.762 21.8 84.258 23.1 92911 23 87.82 114 45.034 14.9 71.869 184 76.704 15.1 60.431 15 65.15 16.8 77.208

Based on the data in the table, is there a significant linear relationship between HW3 and the Midterm grade? v() Yes, because the p-value = 0.0000 () No, because the p-value = 0.0000 () Yes, because the p-value = 0.05 > Cannot determine w Hide question 13 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression Highlight Midterm for the Y Input: Highlight HW3 for the X Input: Make sure you click on Labels and Click OK If done correctly then Significance F 1.99934E- 14 Significance F or p-value = 0.00000000000001999 Because the p-value < .05, Reject Ho. Yes, there is a significant relationship, between MW3 and the Midterm grade. Question 14 1/ 1 point Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the

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bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. Bone Mineral Density (g) 0.8777 0.8925 0.8898 0.8769 0.8999 0.8634 0.8762 0.8888 0.8552 10 0.8546 11 0.8762 Cola Consumed NO OO NN O U A WODN - Based on the data in the table, is there a significant linear relationship between Colas Consumed and Bone Mineral Density? ¢<> No, because p-value = 0.0931 ( > Yes, because the slope = -0.002 ) No, because p-value = 0.0872 > Yes, because p-value = 0.0931 w Hide question 14 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 Significance F 0.093089958 Significance F or p-value = 0.0931 Because the p-value > .05, Do Not Reject Ho. No, there is not a significant relationship, between Colas Consumed and Bone Mineral Density. Question 15 1/ 1 point Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per cubic centimeter. Assume the data is normally distributed. Bone Mineral Density (g) 0.8777 0.8925 0.8898 0.8769 0.8999 0.8634 0.8762 0.8888 0.8552 10 0.8546 11 0.8762 Cola Consumed NV 0O NONU DWW PR Using the estimated regression equation found by using Colas Consumed as the predictor variable, find a point estimate for Bone Mineral Density when

the number of Colas Consumed is 127 ) 0.8767 () 0.8081 ) 0.8089 ¢(> 0.8630 w Hide question 15 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 Regression Equation Coefficients Intercept 0.891716364 Col 01 -0.002389091 Consumed Bone Mineral Density = 0.891716 - 0.002389%(Colas Consumed) Plug in 12 for Colas Consumed Bone Mineral Density = 0.891716 - 0.002389(12) Question 16 1/ 1 point

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A teacher believes that the third homework assignment is a key predictor in how well students will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed. HW3 Midterm 13.3 59.811 21.9 87.539 9.7 53.728 25 96.283 54 39.174 13.2 66.092 20.9 89.729 18.5 78.985 20 86.2 154 73.274 25 93.25 9.7 52.257 6.4 43.984 20.2 79.762 21.8 84.258 23.1 92911 23 87.82 114 45.034 14.9 71.869 18.4 76.704 15.1 60.431 15 65.15 16.8 77.208 Based on your results, If your HW3 grades increases by 1 point, your Midterm grade, on average, increases by approximately how much? Place your answer, rounded to 3 decimal places, in the blank. Do not use any

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stray punctuation marks or a dollar sign. For example, 34.567 would be a legitimate entry. ___ Answer: 2.925 v ¥ Hide question 16 feedback You will be interpreting the slope. Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression Highlight Midterm for the Y Input: Highlight HW3 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 Equation Coefficients Intercept 23.395 HW3 2.925 Midterm = 23.395 + 2.925(HW23) The slope is 2.925 If your HW3 grades increases by 1 point, your Midterm grade, on average, increases by 2.925 points. Question 17 1/ 1 point

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You are thinking about opening up a Starbucks in your area but what to know if it is a good investment. How much money do Starbucks actually make in a year? You collect data to, to help estimate Annual Net Sales, in thousands, of dollars to know how much money you will be making. You collect data on 27 stores to help make your decision. X1 = Rent in Thousand per month X2 = Amount spent on Inventory in Thousand per month X3 = Amount spent on Advertising in Thousand per month x4 = Sales in Thousand per month x5= How many Competitors stores are in the Area Is there a significant linear relationship between these 5 variables and the Annual Net Sales of a Starbucks? If so, what is/are the significant predictor(s) for determining the Annual Net Sales of a Starbucks? See Attached Excel for Data. Starbuck Sales data () No, Rent, p-value = 0.012388536 < .05, Yes, Rent is a significant predictor for Annual Net Sales Advertising, p-value = 0.003599968 < .05, Yes, Advertising is a significant predictor for Annual Net Sales

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Sales per Month, p-value = 0.000876481 < .05, Yes, Sales per Month, is a significant predictor for Annual Net Sales Inventory, p-value = 0.068486021 > .05, No, Inventory is not a significant predictor for Annual Net Sales # Competitor Store, p-value = 0.258240292 > .05, No, # Competitor Stores, is not a significant predictor for Annual Net Sales. > No, Inventory, p-value = 0.068486021 > .05, No, Inventory is not a significant predictor for Annual Net Sales # Competitor Store, p-value = 0.258240292 > .05, No, # Competitor Stores, is not a significant predictor for Annual Net Sales. v( ) Yes Rent, p-value = 0.012388536 < .05, Yes, Rent is a significant predictor for Annual Net Sales Advertising, p-value = 0.003599968 < .05, Yes, Advertising is a significant predictor for Annual Net Sales Sales per Month, p-value = 0.0000876481 < .05, Yes, Sales per Month, is a significant predictor for Annual Net Sales w Hide question 17 feedback You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression Highlight Annual Net Sales for the Y Input: Highlight all 5 columns from Rent to Competitor for the X Input:

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Make sure you click on Labels and Click OK If done correctly then The overall Significance F or p-value = .0000000000000000114933 Because the p-value < .05, Reject Ho. Yes, there is a significant relationship, but which variables are significant? In the ANOVA under the p-value column we see, Rent, p-value = 0.012388536 < .05, Yes, Rent is a significant predictor for Annual Net Sales Inventory, p-value = 0.068486021 > .05, No, Inventory is not a significant predictor for Annual Net Sales Advertising, p-value = 0.003599968 < .05, Yes, Advertising is a significant predictor for Annual Net Sales Sales per Month, p-value = 0.0000876481 < .05, Yes, Sales per Month, is a significant predictor for Annual Net Sales # Competitor Store, p-value = 0.258240292 > .05, No, # Competitor Stores, is not a significant predictor for Annual Net Sales. Question 18 1/ 1 point You decided to join a fantasy Baseball league and you think the best way to pick your players is to look at their Batting Averages. You want to use data from the previous season to help predict Batting Averages to know which players to pick for the upcoming season. You want to use Runs Score, Doubles, Triples, Home Runs and Strike Outs to determine if there is a significant linear relationship for Batting Averages.

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You collect data to, to help estimate Batting Average, to see which players you should choose. You collect data on 45 players to help make your decision. x1 = Runs Score/Times at Bat x2 = Doubles/Times at Bat X3 = Triples/Times at Bat x4 = Home Runs/Times at Bat x5= Strike Outs/Times at Bat Estimate the Batting Average when Run Score = 0.123, Doubles = 0.040, Triples = 0.0045 , Home Runs = 0.009 and Strike Outs = 0.189 See Attached Excel for Data. Baseball data () 0.057 ~0.043 ~0.223 v( 0.229 w Hide question 18 feedback You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression Highlight Baseball Average for the Y Input: Highlight all 5 columns from RS/Times at Bat to SO/Times at Bat for the X Input:

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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 Equation Coefficients Intercept 0.183162857 RS/Ti >/Times at 0.44668017 Bat Doubles/Times 0.990904141 at Bat Triples/times at 0.621603199 bat HR/Ti /Times at 0.27373766 Bat SO/Times at -0.284559939 Bat Batting Average = 0.1832 + 0.4467(RS/Times at Bat) + 0.9909(Doubles/Times at Bat) + 0.6216(Triples/Times at Bat) + 0.2737(HR/Times at Bat) -0.2846(SO/Times at Bat) Plug in, Run Score = 0.123, Doubles = 0.040, Triples = 0.0045 , Home Runs = 0.009 and Strike Outs = 0.189 Batting Average = 0.1832 + 0.4467(0.123) + 0.9909(0.040) + 0.6216(0.0445) + 0.2737(0.009) -0.2846(0.189) Question 19 1/ 1 point You are thinking about opening up a Starbucks in your area but what to know if it is a good investment. How much money do Starbucks actually make in a year? You collect data to, to help estimate Annual Net Sales, in thousands, of dollars to know how much money you will be making. You collect data on 27 stores to help make your decision.

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X1 = Rent in Thousand per month x2 = Amount spent on Inventory in Thousand per month X3 = Amount spent on Advertising in Thousand per month x4 = Sales in Thousand per month x5= How many Competitors stores are in the Area Interpret the slope(s) of the significant predictors for Annual Net Sales (if there are any). See Attached Excel for Data. Starbuck Sales data () There are no significant predictors. > When you hold Inventory, Advertising, Sales per month and #of ~ Competitor constant, as Rent increases by $1000, your Annual Net Sales will increase by $2.74 in $1000. When you hold Rent, Inventory, Sales per Month and #of Competitor constant, as Advertising increases by $1000, your Annual Net Sales will increase by $3.28 in $1000. When you hold Rent , Inventory, Advertising, and # of Competitor constant, as Sales per Month increases by $1000, your Annual Net Sales will increase by $4.84 in $1000. ¢(> When you hold Inventory, Advertising, Sales per month and #of ~ Competitor constant, as Rent increases by $1000, your Annual Net Sales will increase by $15.16 in $1000. When you hold Rent, Inventory, Sales per Month and #of Competitor constant, as Advertising increases by $1000, your Annual Net Sales will increase by $12.18 in $1000.

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When you hold Rent, Inventory, Advertising, and # of Competitor constant, as Sales per Month increases by $1000, your Annual Net Sales will increase by $14.30 in $1000. > When you hold Inventory, Advertising, Sales per month and #of ~ Competitor constant, as Rent increases by $1000, your Annual Net Sales will increase by $5.40 in $1000. When you hold Rent, Inventory, Sales per Month and #of Competitor constant, as Advertising increases by $1000, your Annual Net Sales will increase by $3.72 in $1000. When you hold Rent, Inventory, Advertising, and # of Competitor constant, as Sales per Month increases by $1000, your Annual Net Sales will increase by $2.95 in $1000. w Hide question 19 feedback You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression Highlight Annual Net Sales for the Y Input: Highlight all 5 columns from Rent to Competitor 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 Equation Coefficients Intercept -49.04488287 Rent/$1000 15.15714777 Inventory/$1000 0.1743754 Advertising/$1000 12.17790597

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Sales per month /$1000 # Competior Stores in Area Annual Net Sales = -49.0449 + 15.1571(Rent) + 0.1744(Inventory) + 12.1779(Advertising) + 14.2972(Sales) - 2.9766(Competitor) 14.29717285 -2.976567478 You need to interpret the slope coefficients for all significant predictors. Look at the p-values for the coefficients to find the significant predictors. When you hold Inventory, Advertising, Sales per month and #of Competitor constant, as Rent increases by $1000, your Annual Net Sales will increase by $15.16 in $1000. When you hold Rent, Inventory, Sales per Month and #of Competitor constant, as Advertising increases by $1000, your Annual Net Sales will increase by $12.18 in $1000. When you hold Rent, Inventory, Advertising, and # of Competitor constant, as Sales per Month increases by $1000, your Annual Net Sales will increase by $14.30 in $1000. Question 20 1/ 1 point You decided to join a fantasy Baseball league and you think the best way to pick your players is to look at their Batting Averages. You want to use data from the previous season to help predict Batting Averages to know which players to pick for the upcoming season. You want to use Runs Score, Doubles, Triples, Home Runs and Strike Outs to determine if there is a significant linear relationship for Batting Averages. You collect data to, to help estimate Batting Average, to see which players you should choose. You collect data on 45 players to help make your decision. x1 = Runs Score/Times at Bat

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x2 = Doubles/Times at Bat x3 = Triples/Times at Bat x4 = Home Runs/Times at Bat x5= Strike Outs/Times at Bat Approximately what percentage of the variation in Batting Average is accounted for by these 5 variables in this model? See Attached Excel for Data. Baseball data > 88.67% of variation in Batting Average is accounted for by Runs Score, ~ Doubles, Triples, Home Runs and Strike Outs in this model. O 92.74% of variation in Batting Average is accounted for by Runs Score, ~ Doubles, Triples, Home Runs and Strike Outs in this model. > 1.74% of variation in Batting Average is accounted for by Runs Score, ~ Doubles, Triples, Home Runs and Strike Outs in this model. ¢<> 86.01% of variation in Batting Average is accounted for by Runs Score, - Doubles, Triples, Home Runs and Strike Outs in this model. w Hide question 20 feedback The R-squared value is the amount of explained variance in the data points in the model. You convert this decimal to a percent. You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression Highlight Baseball Average for the Y Input:

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Highlight all 5 columns from RS/Times at Bat to SO/Times at Bat for the X Input: Make sure you click on Labels and Click OK If done correctly then R Square 0.860127109 Adi djusted R 0.842194687 Square R-squared: 86.01% of variation in Batting Average is accounted for by Runs Score, Doubles, Triples, Home Runs and Strike Outs in this model. If you want to give a more conservative estimate, you can use the Adjusted R- squared. This can make sure you don't over promise on what the model can do. But the interpretations are the same. Adjusted R-squared: 84.22% of variation in Batting Average is accounted for by Runs Score, Doubles, Triples, Home Runs and Strike Outs in this model. Note: Correlation is a value between -1 and 1. This STAYS a decimal. R-square gets converted from a decimal to a percentage. The Correlation IS NOT a percent, leave it as a decimal. Done

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