Homework 7 Attempt 3

5 2.95 Hide question 3 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 Inches 0.397827379 Winter Severity 0.249286765 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 4 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.

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. Strike Outs/Times at Bat , p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average. Question 5 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

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 Find the estimated regression equation which can be used to estimate Fawn Count when using these 3 variables are predictor variables. See Attached Excel for Data. Deer data.xlsx Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) + 0.2493(Winter Severity) Fawn Count = 0.8643 + 0.0853(Adult Count) + 0.0908(Annual Rain) + 0.0568(Winter Severity) Fawn Count = 0.9661 + 0.9886(Adult Count) + 0.9774(Annual Rain) + 0.1105(Winter Severity) Fawn Count = -6.4320 + 3.5626(Adult Count) + 4.3813(Annual Rain) + 4.3878(Winter Severity)

0 ±∞ -1 Hide question 9 feedback 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. 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 (lbs.) Rolling Distance (m.) 59 26 84 43 97 48 56 20 103 59 87 44

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 11 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 Question 12 1 / 1 point Which of the following describes how the scatter plot appears? Select all that apply.

Standard Error 0.5963 Observations 11 ANOVA             df SS MS F Significance F Regression 1 20.6611 20.6611 58.0968 3.25116E-05 Residual 9 3.2007 0.3556    Total 10 23.8618        Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 3.5850 1.4137 2.5359 0.0319 0.3871 6.7830 Femur Length 0.5899 0.0774 7.6221 0.0000 0.4148 0.7650 A strong linear correlation was found between the two variables. Find the standard error of estimate. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal.

What are the hypotheses for testing to see if a correlation is statistically significant? H 0 :  ρ  = ±1 ; H 1 : ρ   ≠ ±1 H 0 :  r  = 0   ; H 1 :  r   ≠ 0 H 0 :  r  = ±1   ; H 1 :  r   ≠ ± 1 H 0 :   ρ   = 0 ; H 1 : ρ ≠ 0 H 0 :  ρ  = 0   ; H 1 : ρ   =1 Question 16 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 0.5 112.5 1 110.875 1.5 106.8 2 100.275 2.5 91.3 3 79.875 3.5 70.083

4 59.83 4.5 30.65 5 0 Answer:___ Answer: -0.9422 Hide question 16 feedback Use =CORREL function in Excel Question 17 1 / 1 point The correlation coefficient, r, is a number between: 0 and ∞ -10 and 10 -∞ and ∞ 0 and 10 0 and 1 -   1 and 1 Question 18 1 / 1 point Select the correlation coefficient that is represented in the following scatterplot.

0.83

-0.15 -0.50 -0.82 Hide question 18 feedback This has a negative direction with a moderate to strong correlation. Question 19 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. A regression equation for the following data is  ŷ= 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 every additional average weekly soda consumption, a person's bone density decreases by 0.8893 grams per cubic centimeter. 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. For every additional average weekly soda consumption, a person's bone density increases by 0.0031 grams per cubic centimeter. Hide question 19 feedback You want to interpret the slope.  ŷ= 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. Question 20 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 Multiple R 0.7938 R Square 0.6301

Adjusted R Square0.6182 Standard Error 3.8648 Observations 33   CoefficientsStandard Errort Stat P-value Intercept31.6304 5.2538 6.02051.16E-06 x 5.4496 0.7499 7.26733.55E-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 question 20 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) y = 69.776