MATH302 Week 7 Knowledge Check

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Apr 3, 2024

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Week 7 Knowledge Check Homework Practice Questio… Attempt 1 of 4 Attempt Score 20 / 20 - 100 % Overall Grade (Highest Attempt) 20 / 20 - 100 % Question 1 1 / 1 point Which of the following equations are linear? Hide ques±on 1 feedback Question 2 1 / 1 point y+7=3x y=6x 2 +8 3y=6x+5y 2 y-x=8x 2 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. Due to erosion, a river shoreline is losing several thousand pounds of soil each year. A linear equation that expresses the total amount of soil lost per year is y = 12,000x.
___ 12000___ Hide ques±on 2 feedback Question 3 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: 1 = Warm 2 = Mild 3 = Cold 4 = Freeze 5 = Severe Find the estimated regression equation which can be used to estimate Fawn Count when using these 3 variables are predictor variables. How many pounds of soil does the shoreline lose in a year? Round to a whole number. Don't use any commas or decimals. Answer: The slope is 12,000. The change in the slope impacts the linear equation. Since the slope is 12,000 and this is the change in the line, the shoreline will lose 12,000 pounds of soil each year.
See Attached Excel for Data. Deer data.xlsx Hide ques±on 3 feedback Fawn Count = -6.4320 + 3.5626(Adult Count) + 4.3813(Annual Rain) + 4.3878(Winter Severity) Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) + 0.2493(Winter Severity) Fawn Count = 0.9661 + 0.9886(Adult Count) + 0.9774(Annual Rain) + 0.1105(Winter Severity) Fawn Count = 0.8643 + 0.0853(Adult Count) + 0.0908(Annual Rain) + 0.0568(Winter Severity) 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 Inches 0.397827379
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Question 4 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 Winter Severity 0.249286765 Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) + 0.2493(Winter Severity) 274.65 275.79
Hide ques±on 4 feedback 280.81 270.77 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 /$1000 14.29717285 # Competior Stores in Area -2.976567478 Annual Net Sales = -49.0449 + 15.1571(Rent) + 0.1744(Inventory) + 12.1779(Advertising) + 14.2972(Sales) - 2.9766(Competitor) 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 5 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 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 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 = 0.001303023 < .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 No,
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Hide ques±on 5 feedback Question 6 1 / 1 point 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 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
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 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 Hide ques±on 6 feedback 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. 11.05% 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.
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, leave it as a decimal. Note: Correlation is a value between -1 and 1. This STAYS a decimal.
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Question 7 1 / 1 point Which residual plot has the best linear regression model? Hide ques±on 7 feedback R-square gets converted from a decimal to a percentage. The Correlation IS NOT a percent, leave it as a decimal. a b c d e f A good residual plot has an even distribution of data points above and below the line x = 0.
Question 8 1 / 1 point In the context of regression analysis, what is the definition of an influential point? Question 9 1 / 1 point When r is close to ____, there is either a weak linear relationship between x and y or no linear relationship between x and y. Hide ques±on 9 feedback Question 10 1 / 1 point The closer the correlation coefficient is to 1, the stronger the indication of a negative linear relationship. Observed data points that are far from the least squares line Observed data points that are far from the other observed data points in the horizontal direction Observed data points that are close to the least squares line Observed data points that are close to the other observed data points in the horizontal direction -1 1 0 ±∞ Correlation is a value from -1 to 1. The closer r is to 0, the worst the correlation is. When r = 0, this means there is no correlation (zero) between the two values. True
Hide ques±on 10 feedback Question 11 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 84 43 97 48 56 20 103 59 87 44 88 48 92 46 53 28 66 32 71 39 False The closer it is to -1.
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100 49 Can it be concluded at a 0.05 level of significance that there is a linear correlation between the two variables? Hide ques±on 11 feedback Question 12 1 / 1 point Which of the following describes how the scatter plot appears? Select all that apply. yes no Cannot be determined 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
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 weak strong nonlinear negative
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 CoefficientsStandard 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 ques±on 13 feedback This is given to you in the output
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Question 14 1 / 1 point What are the hypotheses for testing to see if a correlation is statistically significant? Question 15 1 / 1 point 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. Model Summary b Model R R Square Adjusted R Std. Error of Standard Error1.9173 H 0 : r = ±1 ; H 1 : r ≠ ± 1 H 0 : ρ = ±1 ; H 1 : ρ ≠ ±1 H 0 : r = 0 ; H 1 : r ≠ 0 H 0 : ρ = 0 ; H 1 : ρ =1 H 0 : ρ = 0 ; H 1 : ρ ≠ 0
Square 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 value of the test statistic to see if the correlation is statistically significant? Hide ques±on 15 feedback 0.539 7.502 4.436 5.089 6.212 0.734
Question 16 1 / 1 point Select the correlation coefficient that is represented in the following scatterplot. You need to test the slope. The T-stat that corresponds with the slope (or Wrist Circumference) is 6.212 -0.50 -0.82 -0.15 0.83
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Hide ques±on 16 feedback Question 17 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 ques±on 17 feedback This has a negative direction with a moderate to strong correlation.
Question 18 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 Use =CORREL function in Excel.
14.9 71.869 18.4 76.704 15.1 70.431 15 65.15 16.8 77.208 Answer:___ ___ Answer: 0.9846 Hide ques±on 18 feedback Question 19 1 / 1 point 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. The data is displayed below. Regression Statistics Use =CORREL function n in Excel.
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Multiple R 0.7938 R Square 0.6301 Adjusted R Square 0.6182 Standard Error 3.8648 Observations 33 Coefficients Standard Error t Stat P-value Intercept 31.6304 5.2538 6.0205 1.16E-06 x 5.4496 0.7499 7.2673 3.55E-08 What is the predicted height (in inches) for a person with a wrist circumference of 7 inches? Round answer to 4 decimal places. Answer:___ Answer: 69.7776 Hide ques±on 19 feedback Use the values from the Coefficients to write out the regression equation. Coefficients Intercept 31.6304 x 5.4496 y = 31.6304 + 5.4496(7)
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Question 20 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 13.1 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 y = 69.776
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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 ques±on 20 feedback Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression
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Done 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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