MATH302 Week 7 Test

pdf

School

American Military University *

*We aren’t endorsed by this school

Course

302

Subject

Statistics

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by alexwhite101

1/33 Week 7 Test - Results Attempt 1 of 2 Attempt Score 18 / 20 - 90 % Overall Grade (Highest Attempt) 18 / 20 - 90 % Question 1 1 / 1 point ___ 10599___ Hide ques±on 1 feedback Question 2 1 / 1 point During the 2008 Recession homeowners lost thousands of dollars on their homes. A linear equation that expressed this amount was Home Price = -10599(Year) As the Years increased how much did people lose on their homes each year? Round to a whole number. Do not use any commas or decimals. Answer: The slope is -10,599. As the Year increases by 1, the price of your home will decrease by 10,599. The negative in the slope tells us this is a decrease. You don't include the negative but state that it is a decrease.

2/33 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? Hide ques±on 2 feedback Question 3 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. . Yes No Cannot be determined with the given information 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.

3/33 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 Find the 99% prediction interval for the Weekly Sales when the shelf space is 30. (round to 3 decimal places) 2.094 < y < 3.927

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

4/33 Hide ques±on 3 feedback 2.000 < y < 4.000 2.325 < y < 3.232 2.165 < y < 3.915 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 Coefficients value for the regression equation. y = 1.2 + 0.061333333 (x) Plug 30 in for x and solve. y = 1.2 + 0.061333333 (30) y = 3.04 -> this is y-hat This is the equation to use for the prediction interval T-Critical Value =T.INV.2T(.01, 13) = 3.012275839 The SE we get from the Regression output and you can use Excel to find the Average and SD of the Weight variable. LL = 3.04 - 3.012275839*0.248482574*

5/33 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. Weight (lbs.)Rolling Distance (m.) 59 26 83 43 97 49 56 20 103 65 87 44 UL = 3.04 + 3.012275839*0.248482574*

6/33 88 48 91 42 52 39 63 33 71 39 100 49 89 55 103 53 99 42 74 33 Using the regression line for this problem, the approximate rolling distance for a child on a bike that weighs 106 lbs. is: Hide ques±on 4 feedback 54.2378 58.7213 55.8742 54.9610 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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

7/33 Question 5 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 (lbs.) Rolling Distance (m.) 59 26 83 43 97 49 56 20 103 65 87 44 88 48 91 42 52 39 63 33 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 = -0.508294634 + 0.52329484 (x) Plug 106 in for x and solve. y = -0.508294634 + 0.52329484 (106)

8/33 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 4 decimal places) ___< y < ___ Answer for blank # 1: 39.3540 (50 %) Answer for blank # 2: 70.5679 (50 %) Hide ques±on 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 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 = -0.508294634 + 0.52329484 (x) Plug 106 in for x and solve. y = -0.508294634 + 0.52329484 (106)

9/33 Question 6 1 / 1 point Which of the following describes how the scatter plot appears? Select all that apply. y = 54.96095837, this is our y-hat value. This is the equation to use for the prediction interval 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 Weight variable. LL =54.96095837 - 2.144786688*6.679572112* UL =54.96095837 + 2.144786688*6.679572112*

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

10/33 Question 7 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. Based on the data in the table below, is there a significant linear relationship between Direct Labor Hours and the Indirect Labor Expense? Please see attached Excel for data. ILE_and_DLH data strong weak negative positive No, the sample correlation coefficient is equal to 0.878, which provides evidence of a significant linear relationship. Yes, because the p-value = 0.00023 Yes, the sample correlation coefficient is equal to 0.878, which provides evidence of a significant linear relationship.

11/33 Hide ques±on 7 feedback Question 8 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). Based on the data in the table below, is there a significant linear relationship between temperature and the amount of electricity used? Temperature (x) Kilowatts (y) No, because the p-value = 0.00023 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 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.

12/33 73 680 78 760 85 910 98 1510 93 1170 83 888 92 923 81 837 76 600 105 1800 Hide ques±on 8 feedback Yes, because the p-value =0.00004 Yes, the sample correlation coefficient is equal to 0.893, which provides evidence of a significant linear relationship. No, the sample correlation coefficient is equal to 0.983, which does not provide evidence of a significant linear relationship. No, because the p-value =0.00004 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:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

13/33 Question 9 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. Please see Attached Excel for Data. Apartments data.xlsx Using Size as a predictor for Rent, use regression to fit a straight line to all 30 data points. What values for the intercept (a) and slope (b) do you obtain? Place your answers, rounded to 3 decimal places, in the blanks provided. Do not use any stray punctuation marks. For example, 34.567 would be a legitimate entry. Make sure you put the 0 in front of the decimal. intercept (a) =___ slope(b) =___ ___ Answer for blank # 1: 221.367 (50 %) Answer for blank # 2: 0.869 (50 %) Make sure you click on Labels and Click OK If done correctly then Multiple R = 0.944907859 R Square = 0.892850862 Significance F or p-value = 0.00004 Because the p-value < .05, Reject Ho. Yes, there is a significant relationship.

14/33 Hide ques±on 9 feedback 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. Based on your results, If direct labor hours (DLH) increases by one hour, the indirect labor expense (ILE), on average, increases by approximately how much? Place your answer, rounded to 2 decimal places, in the blank. Do not use any stray punctuation marks or a dollar sign. For example, 34.56 would be a legitimate entry. ___ Please see attached Excel for data. ILE_and_DLH data 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)

15/33 Answer: 9.32 Hide ques±on 10 feedback 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). Using that data, find the estimated regression equation which can be used to estimate Kilowatts when using Temperature as You are interpreting the slope for this problem. 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 DHE 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 171.76 DLH(X) 9.32 ILE = 17.76 + 9.32(DLH) The slope is 9.32 If direct labor hours (DLH) increases by one hour, the indirect labor expense (ILE), on average, increases by approximately 9.32

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

16/33 the predictor variable. Temperature (x) Kilowatts (y) 73 680 78 760 85 910 98 1510 93 1170 83 888 92 923 81 837 76 600 105 1800 Hide ques±on 11 feedback Kilowatts = 0.945 + 0.893(Temperature) Kilowatts = 132.031 + 34.858(Temperature) Kilowatts = -2003.896 + 34.858(Temperature) Kilowatts = 371.223 + 4.269(Temperature) Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression

17/33 Question 12 1 / 1 point An object is 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 0.5 111.75 1 110.875 1.5 105.18 2 100.275 2.5 91.3 3 79.875 3.5 71.083 4 57.83 4.5 32.65 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 (x) 34.85759097 Kilowatts = -2003.895859 + 34.85759097(Temperature)

18/33 5 0 Answer:___ ___ Answer: -0.9417 Hide ques±on 12 feedback Question 13 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. Cola Consumed Bone Mineral Density (g) 1 0.8777 2 0.8925 3 0.8898 4 0.8769 5 0.8999 6 0.8634 7 0.8762 8 0.8888 9 0.8552 10 0.8546 11 0.8762 Using that data, find the estimated regression equation which can be used to estimate Bone Mineral Density when using Colas Consumed as the predictor variable. Use =CORREL function in Excel Bone Mineral Density = 0.008627 + 0.001272(Colas Consumed)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

19/33 Hide ques±on 13 feedback Question 14 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 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) 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 Cola Consumed -0.002389091 Bone Mineral Density = 0.891716 - 0.002389(Colas Consumed)

20/33 21.9 87.539 9.7 53.728 25 96.283 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 23 87.82 11.4 45.034 14.9 71.869 18.4 76.704 15.1 60.431 15 65.15 16.8 77.208 Find the y-intercept and slope for the regression equation using technology (you can copy and paste the data into Excel). Round answer to 3 decimal places. ŷ=___+___x ___ Answer for blank # 1: 23.395 (50 %) Answer for blank # 2: 2.925 (50 %) Hide ques±on 14 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression

21/33 Question 15 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 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 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(HW3)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

22/33 9.7 52.257 6.4 43.984 20.2 79.762 21.8 84.258 23.1 92.911 23 87.82 11.4 45.034 14.9 71.869 18.4 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? Hide ques±on 15 feedback No, because the p-value = 0.0000 Yes, because the p-value = 0.0000 Yes, because the p-value = 0.05 Cannot determine 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

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

23/33 Question 16 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 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 23 87.82 11.4 45.034 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.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

24/33 14.9 71.869 18.4 76.704 15.1 60.431 15 65.15 16.8 77.208 Approximately what percentage of the variation in the Midterm grade is accounted for by the HW3 grade in this model? Place your answer, rounded to 1 decimal place, in the blank. Do not use any stray punctuation marks or a percentage sign. For example, 78.9 would be a legitimate entry. ___% ___ Answer: 94.2 Hide ques±on 16 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 you should get, Multiple R 0.970321333 R Square 0.94152349 94.2% of variation in the Midterm grade is accounted for by the HW3 grade 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.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

25/33 Question 17 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 Find the estimated regression equation which can be used to estimate Batting Averages when using these 5 variables are predictor variables. See Attached Excel for Data. Baseball data Batting Average = 10.6855+ 4.07425(RS/Times at Bat) + 3.1695(Doubles/Times at Bat) + 1.0704(Triples/Times at Bat) + 1.6164(HR/Times at Bat) -5.4965(SO/Times at 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) Batting Average = 10.6855+ 0.9274(RS/Times at Bat) + 0.8601(Doubles/Times at Bat) + 0.8422(Triples/Times at Bat) +

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

26/33 Hide ques±on 17 feedback 0.0174(HR/Times at Bat) -5.4965(SO/Times at Bat) Batting Average = 0.0171 + 0.1096(RS/Times at Bat) + 0.3131(Doubles/Times at Bat) + 0.5807(Triples/Times at Bat) + 0.1693(HR/Times at Bat) + 0.5177(SO/Times at Bat) 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: 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/Times at Bat 0.44668017 Doubles/Times at Bat 0.990904141 Triples/times at bat 0.621603199 HR/Times at Bat 0.27373766 SO/Times at Bat -0.284559939 Batting Average = 0.1832 + 0.4467(RS/Times at Bat) + 0.9909(Doubles/Times at Bat) + 0.6216(Triples/Times at Bat) +

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

27/33 Question 18 0 / 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: 1 = Warm 2 = Mild 3 = Cold 4 = Freeze 5 = Severe Is there a significant linear relationship between these 3 variables and Fawn Count? If so, what is/are the significant predictor(s) for determining Fawn Count? See Attached Excel for Data. Deer data 0.2737(HR/Times at Bat) -0.2846(SO/Times at Bat) Yes,

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

28/33 Adult Count, p-value = 0.01188964 > .05, Yes, Adult Count is a significant predictor for Fawn Count. Annual Rain in Inches, p-value = 0.004661804 > .05, Yes, Annual Rain in Inches is a significant predictor for Fawn Count. Winter Severity, p-value = 0.00462881 > .05, Yes, Winter Severity is a significant predictor for Fawn Count. Yes, Adult Count, p-value = 0.01188964 < .05, Yes, Adult Count is a significant predictor for Fawn Count. Annual Rain in Inches, p-value = 0.004661804 < .05, Yes, Annual Rain in Inches is a significant predictor for Fawn Count. Winter Severity, p-value = 0.00462881 < .05, Yes, Winter Severity is a significant predictor for Fawn Count. No, Adult Count, p-value = 0.01188964 < .05, No, Adult Count is not a significant predictor for Fawn Count. Annual Rain in Inches, p-value = 0.004661804 < .05, No, Annual Rain in Inches is not a significant predictor for Fawn Count. Winter Severity, p-value = 0.00462881 < .05, No, Winter Severity is not a significant predictor for Fawn Count. No, Adult Count, p-value = 0.01188964 > .05, No, Adult Count is not a significant predictor for Fawn Count. Annual Rain in Inches, p-value = 0.004661804 > .05, No, Annual Rain in Inches is not a significant predictor for Fawn Count. Winter Severity, p-value = 0.00462881 > .05, No, Winter Severity is not a significant predictor for Fawn Count.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

29/33 Hide ques±on 18 feedback Question 19 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 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 The overall Significance F or p-value = 0.0000250343 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, Adult Count , p-value = 0.01188964 < .05, Yes, Adult Count is a significant predictor for Fawn Count. Annual Rain in Inches , p-value = 0.004661804 < .05, Yes, Annual Rain in Inches is a significant predictor for Fawn Count. Winter Severity , p-value = 0.00462881 < .05, Yes, Winter Severity is a significant predictor for Fawn Count.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

30/33 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 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 Hide ques±on 19 feedback 0.229 0.043 0.057 0.223 You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak. Data -> Data Analysis -> Scroll to Regression

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

31/33 Question 20 1 / 1 point With Obesity on the rise, a Doctor wants to see if there is a linear relationship between the Age and Weight and estimating a person's Systolic Blood 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: 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/Times at Bat 0.44668017 Doubles/Times at Bat 0.990904141 Triples/times at bat 0.621603199 HR/Times at Bat 0.27373766 SO/Times at Bat -0.284559939 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)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

32/33 Pressure. Approximately what percentage of the variation in Systolic BP is accounted for by Age and Weight in this model? See Attached Excel for Data. BP data Hide ques±on 20 feedback 97.25% of variation in the Systolic BP is accounted for by Age and Weight in this model. 2.67% of variation in the Systolic BP is accounted for by Age and Weight in this model. 98.48% of variation in the Systolic BP is accounted for by Age and Weight in this model. 96.98% of variation in the Systolic BP is accounted for by Age and Weight in this model. 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 Systolic BP for the Y Input: Highlight Both Age and Weight columns for the X Input: Make sure you click on Labels and Click OK If done correctly then R Square 0.969825398 Adjusted R Square 0.965183151 R-squared:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

33/33 Done 96.98% of variation in the Systolic BP is accounted for by Age and Weight 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: 96.52% of variation in the Systolic BP is accounted for by Age and Weight 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.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

MATH302 Week 7 Test

Related Documents