week 7 homework questions V2

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Attempt Score 17 / 20 - 85 % Overall Grade (Highest Attempt) 17 / 20 - 85 % stion 1 1 / 1 p Which of the following equations are linear? 4y=8 y 2 =6x 3 +8 3y=6x+5y 2 y-x=8x 2 Hide question 1 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. n 2 1 Which of the following equations are linear? y+7=3x y=6x 2 +8 3y=6x+5y 2 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. n 3 1 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 Is there a significant linear relationship between these 5 variables and the Batting Average? If so, what is/are the significant predictor(s) for determining the Batting Average? See Attached Excel for Data. Baseball data.xlsx Yes, Runs Score/Times at Bat , p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average. Doubles/Times at Bat , p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average. Strike Outs/Times at Bat , p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average. No, Triples/Times at Bat , p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average. Home Runs/Times at Bat,  p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average. No, Runs Score/Times at Bat , p-value = 0.000219186 < .05, Yes, Runs Score is a significant
predictor for Batting Average. Doubles/Times at Bat , p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average. Strike Outs/Times at Bat , p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average. Yes, Triples/Times at Bat , p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average. Home Runs/Times at Bat,  p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average. 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 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, The overall  Significance F  or p-value = 0.0000000000000012607 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, Runs Score/Times at Bat , p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average. Doubles/Times at Bat , p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average. Triples/Times at Bat , p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average. Home Runs/Times at Bat,  p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.
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Strike Outs/Times at Bat , p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average. n 4 1 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 Pressure. Is there a significant linear relationship between Age and Weight and a person's Systolic Blood Pressure? If so, what is/are the significant predictor(s) for Systolic Blood Pressure? See Attached Excel for Data. BP data No, Age, p-value = 0.001303023 < .05, No, Age is not a significant predictor for Systolic BP Weight, p-value = 0.023799395 < .05, No, Weight is not a significant predictor for Systolic BP Yes, Age, p-value = 0.001303023 < .05, Yes, Age is a significant predictor for Systolic BP Weight, p-value = 0.023799395 < .05, Yes Weight is a significant predictor for Systolic BP Yes, Age, p-value = 0.9388 > .05, Yes, Age is a significant predictor for Systolic BP Weight, p-value = 0 .3092 > .05, Yes Weight is a significant predictor for Systolic BP No, Age, p-value = 0.9388 > .05, No, Age is not a significant predictor for Systolic BP Weight, p-value = 0 .3092 > .05, No, Weight is not a significant predictor for Systolic BP Hide question 4 feedback 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, The overall  Significance F  or p-value = .00000000013112 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, Age, p-value = 0.001303023 < .05, Yes, Age is a significant predictor for Systolic BP Weight, p-value = 0.023799395 < .05, Yes Weight is a significant predictor for Systolic BP n 5 1 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 Approximately what percentage of the variation for Fawn Count is accounted for by these 3 variables in this model?
See Attached Excel for Data. Deer data.xlsx 11.05% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model. 98.86% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model. 98.25% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model. 97.74% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model. Hide question 5 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 R Square 0.977400423 Adjusted R Square 0.966100634 R-squared: 97.74% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity 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: 96.61% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity 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,
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leave it as a decimal. 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. n 6 1 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. 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. There are no 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 .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. 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: 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) 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. n 7 0 Which residual plot has the best linear regression model? a b c d e f
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Hide question 7 feedback A good residual plot has an even distribution of data points above and below the line x = 0. n 8 0 The least squares regression line for a data set is yˆ=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? Yes No Cannot be determined with the given information Hide question 8 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. n 9 1 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 84 43
97 48 56 20 103 59 87 44 88 48 92 46 53 28 66 32 71 39 100 49 Using the regression line for this problem, the approximate rolling distance for a child on a bike that weighs 110 lbs. is: 59.2347 58.7213 60.1846 78.4555 Hide question 9 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 = -8.564718163 + 0.611691023 (x)
Plug 110 in for x and solve. y = -8.564718163 + 0.611691023 (110) y = 58.7213 n 10 1 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 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?
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yes no Cannot be determined Hide question 10 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  Significance F  value for the correct p-value to use to make your decision. Yes, there is a significant relationship p-value = 0.000001 n 11 1 The closer the correlation coefficient is to 1, the stronger the indication of a negative linear relationship. True False Hide question 11 feedback The closer it is to -1. n 12 1 Which of the following describes how the scatter plot appears? Select all that apply. positive negative
neither positive or negative Question 13 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 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             df SS MS F Significance F Regression 1 967.0912 967.0912 263.0757 2.09917E-07 Residual 8 29.4088 3.6761    Total 9 996.5        Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 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 Hide question 13 feedback This is given to you in the output Standard Error 1.9173   n 14 1 Body frame size is 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.
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Model Summary b Model R R Square Adjusted R Square Std. Error of the Estimate 1 .734 a .539 .525 4.01409 a. Predictors: (Constant), Wrist Circumference b. Dependent Variable: Height ANOVA a Model Sum of Squares df Mean Square F Sig. 1 Regression 621.793 1 621.793 38.590 .000 b Residual 531.726 33 16.113     Total 1153.519 34       a. Dependent Variable: Height b. Predictors: (Constant), Wrist Circumference Model Unstandardized Coefficients Standardized Coefficients t Sig. B Std. Error Beta     1 (Constant) 38.177 5.089   7.502 .000 Wrist Circumference 4.436 .714 .734 6.212 .000 What is the correct conclusion for testing that there is a significant correlation? There is a significant correlation. There is not a significant correlation. Hide question 14 feedback The p-value for the slope is 0.000, this is less than .05. Yes, it is significant. This is also the same
p-value for the overall ANOVA Sig. value. n 15 1 To test the significance of the correlation coefficient, we use the t-distribution with how many degrees of freedom? n – 2 n + 1 n 1  + n 2  – 2 n n – 1 Question 16 0 / 1 point Choose the correlation coefficient that is represented in the scatterplot.
0.15 -0.82 0.83 Hide question 16 feedback This has a positive direction with a moderate to strong correlation. n 17 1 The correlation coefficient, r, is a number between: 0 and ∞ -10 and 10
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-∞ and ∞ 0 and 10 0 and 1 - 1 and 1 Question 18 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. 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. 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 Answer:___ Answer: -0.8241 Hide question 18 feedback Use =CORREL function in Excel.
n 19 1 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 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   Find the predicted midterm score when the homework 3 score is 15. Do not round until the end, then round answer to 3 decimal places. ___ Answer: 68.534 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
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y = 25.905 + 2.842x y = 25.905 + 2.842(15) Or Midterm = 25.90467802 + 2.841975886(HW3) Midterm = 25.90467802462 + 2.841975886(15) n 20 1 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 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
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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   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: 25.905 (50 %) Answer for blank # 2: 2.842 (50 %) Hide question 20 feedback Copy and Paste the Data into Excel.  Data -> Data Analysis -> Scroll to Regression
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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
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