Week 7 Knowledge

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Feb 20, 2024

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Week 7 Knowledge Check Homework Practice Questio... X Attempt 1 of 4 Written Oct 15, 2023 10:21 PM - Oct 16, 2023 12:49 AM Attempt Score 19/20-95% Overall Grade (Highest Attempt) 19/20-95% Question 1 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. What is the dependent variable? ) resort location () diver certification ¢<f‘> fee amount w Hide question 1 feedback The equation is Dollars = $12.5(hour) + $25 Dollars depends on how many hours you use the equipment for. Dollars is the dependent variable and time (in hours) is the independent variable.

Question 2 1/ 1 point Which of the following equations are linear? \/<> y+/7=3X [ y=6x°+8 ) By=6x+5y? () y-X=8x%2 ¥ Hide question 2 feedback A linear equation is a linear line. If a problem has a squared or a cubed term, it isn't linear. It is a quadratic equation. Question 3 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. 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

Estimate the Annual Net Sales of a Starbucks when Rent = 2.5, Inventory = 430, Advertising =7.75, Sales = 9.89 and Number of Competitors = 8. See Attached Excel for Data. Starbuck Sales data.xlsx () 270.77 () 280.81 \/C) 275.79 ) 274.65 w Hide question 3 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 Plug in, Rent = 2.5, Inventory = 430, Advertising = 7.75, Sales = 9.89 and Number of Competitors = 8 Annual Net Sales = -49.0449 + 15.1571(2.5) + 0.1744(430) + 12.1779(7.75) +14.2972(9.89) - 2.9766(8) Question 4 1/ 1 point You move out into the country and you notice every Spring there are more and more Deer Fawns that appear. You decide to try and predict how many Fawns there will be for the up coming Spring. You collect data to, to help estimate Fawn Count for the upcoming Spring season. You collect data on over the past 10 years. x1 = Adult Deer Count x2 = Annual Rain in Inches x3 = Winter Severity « Where Winter Severity Index: o 1 =Warm o 2 = Mild o 3 =Cold o 4 = Freeze o 5 = Severe Estimate Fawn Count when Adult Deer Count = 10, Annual Rain = 13.5 and Winter Severity = 4

See Attached Excel for Data. Deer data.xlsx w Hide question 4 feedback You can run a Multiple Linear Regression Analysis using the Data Analysis ToolPak in Excel. Data -> Data Analysis -> Scroll to Regression Highlight Fawn Count for the Y Input: Highlight columns Adult Count to Winter Severity 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 -5.559106707 Adult Count 0.303715877 Annual Rain in 0.397827379 Inches Wi inter 0.249286765 Severity Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) + 0.2493(Winter Severity) Plug in, Adult Deer Count = 10, Annual Rain = 13.5 and Winter Severity = 4

Fawn Count = -5.5591 + 0.3071(10) + 0.3978(13.5) + 0.2493(4) Question 5 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. 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 Find the estimated regression equation which can be used to estimate Annual Net Sales when using these 5 variables are predictor variables. See Attached Excel for Data. Starbuck Sales data.xlsx O Annual Net Sales = -49.0449 + 0.9911(Rent) + 0.9823(Inventory) + ~ 0.9782(Advertising) + 27.6181(Sales) - 2.9766(Competitor) v() Annual Net Sales = -49.0449 + 15.1571(Rent) + 0.1744(Inventory) + © 12.1779(Advertising) + 14.2972(Sales) - 2.9766(Competitor) O Annual Net Sales = -1.1039+ 2.736(Rent) + 1.9204(Inventory) + ~ 3.2767(Advertising) + 4.8384(Sales) - 1.1620(Competitor)

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) Annual Net Sales = 44.4256 + 5.5406(Rent) + 0.0979(Inventory) + ~ 3.7165(Advertising) + 2.9543(Sales) - 2.5614(Competitor) w Hide question 5 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 Sales per month 1429717285 /$1000 # Competior -2.976567478 Stores in Area Annual Net Sales = -49.0449 + 15.1571(Rent) + 0.1744(Inventory) + 12.1779(Advertising) + 14.2972(Sales) - 2.9766(Competitor) Question 6 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 x2 = Doubles/Times at Bat X3 = Triples/Times at Bat x4 = Home Runs/Times at Bat x5= Strike Outs/Times at Bat Interpret the slope(s) of the significant predictors for Batting Average (if there are any). See Attached Excel for Data. Baseball data.xlsx ) When you hold Double, Triples, Home Runs and Strike Outs constant, ~as Runs Score increases by 1, your Batting Average will increase by .225/Times at Bat. When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .358/Times at Bat. When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .389/Times at Bat.

() There are no significant predictors. v() When you hold Double, Triples, Home Runs and Strike Outs constant, - as Runs Score increases by 1, your Batting Average will increase by 447/Times at Bat. When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .991/Times at Bat. When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .285/Times at Bat. - When you hold Double, Triples, Home Runs and Strike Outs constant, as Runs Score increases by 1, your Batting Average will increase by .109/Times at Bat. When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .313/Times at Bat. When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .052/Times at Bat. w Hide question 6 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 Tri : riples/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) 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 Double, Triples, Home Runs and Strike Outs constant, as Runs Score increases by 1, your Batting Average will increase by .447/Times at Bat. When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .991/Times at Bat. When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .285/Times at Bat.

Question 7 1/ 1 point The least squares regression line for a data set is y"'=2.3-0.1x and the standard deviation of the residuals is 0.13. Does a case with the values x = 4.1, y = 2.34 qualify as an outlier? v()Yes ONe () Cannot be determined with the given information ¥ Hide question 7 feedback Plug in 4.1 for x. y=2.3-.1(4.1) y=1.89 Residual is y-given - y-predicted. 2.34 - 1.89 = .45 -> this is the residual value. To see if it is an outlier take 2 and multiply it by .13 2*.13 = .26 45 is greater than .26, Yes, it is an outlier. Question 8 1/ 1 point The least squares regression line for a data set is y"'=5+0.3x and the standard deviation of the residuals is 0.4. Does a case with the values x = 2, y = 6.2 qualify as an outlier? () Cannot be determined with the given information

v( ) No O Yes ¥ Hide question 8 feedback Plug in 2 for x. y=5+.3*%2 y=5.6 Residual is y-given - y-predicted. 6.2 - 5.6 = .6 -> this is the residual value. To see if it is an outlier take 2 and multiply it by .4 2*.4 = 8. .6 is less than .8, No it is not an outlier. Question 9 1/ 1 point A negative linear relationship implies that larger values of one variable will result in smaller values in the second variable. v () True ) False w Hide question 9 feedback For a negative relationship, as one value increases, the other value decreases.

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Question 10 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. Weight (1bs.) Rolling Distance (m.) 59 26 84 43 97 48 56 20 103 59 87 44 88 48 92 46 53 28 66 32 71 39 100 49 Can it be concluded at a 0.05 level of significance that there is a linear correlation between the two variables? ‘/C> YES

() Cannot be determined w Hide question 10 feedback Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression. The y-variable 1s 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 Significance F value for the correct p-value to use to make your decision. Yes, there is a significant relationship p-value = 0.000001 Question 11 1/ 1 point Whenris close to ____, there is either a weak linear relationship between x and y or no linear relationship between x and y. w Hide question 11 feedback Correlation is a value from -1 to 1. The closer r is to O, the worst the correlation is. When r = O, this means there is no correlation (zero) between the two values. Question 12 1/ 1 point

Which of the following describes how the scatter plot appears? Select all that apply. ¢<‘> strong () nonlinear ) negative Question 13 7 8 9 1/ 1 point A new fad diet called Trim-to-the-MAX is running some tests that they can use in advertisements. They sample 25 of their users and record the number of days each has been on the diet along with how much weight they have lost in pounds. The data are below. Days on Diet Weight Lost 7 5 12 7 16 12 19 15 25 20 34 25 39 24 43 29

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44, 33 49 35 Regression Statistics Multiple R 0.9851 R Square 0.9705 Adjusted R Square 0.9668 Standard Error 1.9173 |[Observations 10 ANOVA i ag MS . Significance IF Regression 1 967.0912] 967.0912] 263.0757]2.09917E-07 Residual 8 29.4088 3.6761 Total 9 996.5 Cocfficients|Standard Error|t Stat P-value |Lower 95% [Upper 95% I[ntercept 0.4912 1.3746 0.3574 0.7301 -2.6785 3.6610

Days on Diet 0.6947 0.0428] 16.2196 0.0000 0.5960 0.7935 A strong linear correlation was found between the two variables. Find the standard error of estimate. Round answer to 4 decimal places. Answer: 1.9173 w Hide question 13 feedback This is given to you in the output Standard Error{1.9173 Question 14 1/ 1 point What are the hypotheses for testing to see if a correlation is statistically significant? > Hyp: p=+1; Hy:p#+1 ~ ) Hoip= 0;H;:pp=1 () Hoir=0;Hp:r#0 () i =03 i 0 () Hpir=*1Hyir# 1 Question 15 1/ 1 point Body frame size 1s determined by a person's wrist circumference in relation to height. A researcher measures the wrist circumference and height of a random sample of individuals.

80.00- 75.007] 65.00] o 60.00 55.00 T T T T T T 550 6.00 6.50 7.00 7.50 8.00 Wrist Circumference Model Summaryb Std. Adjusted Error of R R the Model R Square Square Estimate 1 7348 .539 525 4.01409 a. Predictors: (Constant), Wrist Circumference b. Dependent Variable: Height ANOVA? Sum of Mean Model Squares df Square F Sig. 1 Regression 621.793 1 621.793 38.590 .000° Residual 531.726 33 16.113 Total 11563.519 34 a. Dependent Variable: Height b. Predictors: (Constant), Wrist Circumference Unstandardized Standardized Coefficients Coefficients t Sig. Std. Model B Error Beta 1 (Constant) 38.177 5.089 7.502 .000

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Wrist 4.436 714 734 6.212 .000 Circumference What 1s the value of the test statistic to see if the correlation is statistically significant? > 0.539 > 7.502 ) 4.436 > 5.089 ¢<> 6.212 (O 0734 w Hide question 15 feedback You need to test the slope. The T-stat that corresponds with the slope (or Wrist Circumference) is 6.212 Question 16 1/ 1 point Choose the correlation coefficient that is represented in the scatterplot.

| o © | o ¢ i O O o Oo lDO O | (@] ® — — — — 4 1 . O | —— — o O o O @ O O ” | 0 0o T @] o | O | o » o) + o ‘ o] v 083 () -0.82 w Hide question 16 feedback This has a positive direction with a moderate to strong correlation. Question 17 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. 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. HW3 | Midterm 13.1 59.811 21.9 87.539 8.8 53.728 24.3 95.283 5.4 39.174 13.2 66.092 20.9 89.729 18.5 78.985 20 86.2 15.4 73.274 25 93.25 9.7 52.257 6.4 43.984 20.2 79.762 21.8 84.258 23.1 92.911 22 87.82 11.4 54.034 14.9 71.869 18.4 76.704 15.1 70.431 15 65.15 16.8 77.208

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Answer: Answer: 0.9846 w Hide question 17 feedback Use =CORREL function n in Excel. Question 18 1/ 1 point The correlation coefficient, r, 1S a number between: > 0 and o : -10 and 10 j, -00 and oo \) 0 and 10 () 0and1 ‘/C -land 1 Question 19 0/ 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 13.1 59.811 21.9 87.539 8.8 53.728

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24.3 95.283 5.4 39.174 13.2 66.092 20.9 89.729 18.5 78.985 20 86.2 15.4 73.274 25 93.25 9.7 52.257 6.4 43.984 20.2 79.762 21.8 84.258 23.1 92.911 22 87.82 11.4 54.034 14.9 71.869 18.4 76.704 15.1 70.431

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15 65.15 77.208 16.8 Find the predicted midterm score when the homework 3 score 1s 15. Do not round until the end, then round answer to 3 decimal places. Answer: 2.925 x (68.532, 68.535, 68.533, 68.534, 68.536) W Hide question 19 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression Highlight Midterm Score 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 25.90467802 HW3 2.841975886 y = 25.905 + 2.842x y = 25.905 + 2.842(15) Or Midterm = 25.90467802 + 2.841975886(HW3) Midterm = 25.90467802462 + 2.841975886(15)

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Question 20 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 i1s normally distributed. A regression equation for the following data 1s $=0.8893-0.0031x. Which is the best interpretation of the slope coefficient? X y 1| 0.883 2| 0.8734 3| 0.8898 4 0.8852 5/ 0.8816 6| 0.863 7| 0.8634 8 0.8648 9 0.8552 10| 0.8546 11 0.862 > For an increase of 0.8893 in the average weekly soda consumption, a person's bone density decreases by 0.0031 grams per cubic centimeter. \/<> For every additional average weekly soda consumption, a person's bone density decreases by 0.0031 grams per cubic centimeter.

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> For every additional average weekly soda consumption, a person's bone density increases by 0.0031 grams per cubic centimeter. > For every additional average weekly soda consumption, a person's bone density decreases by 0.8893 grams per cubic centimeter. w Hide question 20 feedback You want to interpret the slope. 5=0.8893-0.0031x The slope value is -0.0031 X is the number of sodas you drink and Y is your bone density. As the number of sodas you drink per week increases by 1, then your bone density will decrease by 0.0031. Your bone density will decrease because it is negative. The "decrease" accounts for the negative. Saying "increase by a -0.0031" isn't correct. If a slope is negative then it decreases. Done

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