a.)What is the equation of the regression line? Y= b.)What is the estimate of th

Question

a.)What is the equation of the regression line?

Y=

b.)What is the estimate of the 1 coefficient?

c.)A 0.1 increase in diamond weight would have what impact on the average price of the diamond?

d.)What is the expected average price of a 0.83 weight diamond?

e.)What is the expected average price of a 1.5 weight diamond?

f.)Is diamond weight a significant predictor of diamond price?

g.)Why is the model DF = 1?

h.) If you were creating a report and you see a p-value in your output that is 0.0000 [like in the regression output from problem 2], what should you put in your report for the p-value?

aragrap We looked at the ability of diamond weight to predict price in a linear regression model 2 points eachl Here is the model output Linear Regression Results Model: Linear Regression Model Dependent Variable: PRICE 308 Number of Observations Read 308 Number of Observations Used Analysis of Variance Mean Sum of F Value Square Source 1 3173248722 3173248722 2540.73 0.0000 306 382178624.5 T 1248949.753 imii Error 3555427347 Corrected Total 1117.56421 R-Square 0.8925 Dependent Mean 5019.48377 AdR-Sg Var 22.26452 Parameter Estimates Standard Parameter Estimate Variable -2298.357602 -14 50 0.0000 11598.88401 230.1106037 50.41 0.0000 Intercept WEIGHT

Explanation / Answer

a) Y (Price)=-2298.357602+11598.88401(Weight)

b) 11598.88401

c) 11598.88401(0.1)=1159.88

d) Y (Price)=-2298.357602+11598.88401(0.83)=7328.7161

e) Y (Price)=-2298.357602+11598.88401(1.5)=15099.96

f) p-value is less than 0.05. Hence, we can say that diamond weight is a significant predictor of diamond price.

g) Because it is testing variance in the dependent variable based on one independent variable.

h) p-value is less than 0.05. Hence, we can say that the variable does have an impact on the dependent variable and that it is a significant variable.

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