Question

The data in the accompanying table give the prices​ (in dollars) for gold link chains at the Web site of a discount jeweler. The data include the length of the chain​ (in inches) and its width​ (in millimeters). All of the chains are​ 14-carat gold in a similar link style. Use the price as the response. For one explanatory​ variable, use the width of the chain. For the​ second, calculate the​ "volume" of the chain as π times its length times the square of half the​ width, Volume = π Metric Length x (Width/2)². To make the units of volume mm³ first convert the length to millimeters (25.4 mm=1 inch).

Complete parts​ (a) through​ (e).

Price ($)	Length (inch)	Width (mm)
66.57	16	1.1
98.36	16	1.5
135.69	16	2.3
58.12	18	1.1
157.55	18	1.5
216.28	18	2.3
372.55	18	3.1
614.14	18	4
72.21	20	1.1
128.14	20	1.5
242.81	20	2.3
400.43	20	3.1
688.97	20	4
273.27	22	2.3
449.43	22	3.1
81.68	24	1.1
199.83	24	1.5
291.48	24	2.3
450.53	24	3.1
879.77	24	4
240.38	30	1.9
367.39	30	2.3

​(a) The explanatory variable Volume includes Width. Are these explanatory variables perfectly​ correlated? Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A. Yes

B.​ No, because the correlation is _____ . ​(Round to three decimal places as​ needed.)

Can we use them both in the same multiple​ regression?

A. ​Yes, but with collinearity.

B. ​No, because there is collinearity between the two​ variables, which violates an assumption of the MRM.

C. Yes, because there is no collinearity.

______________________________________________________________________________________________________________________________________

​(b) Fit the multiple regression of Price on Width and Volume. Do both explanatory variables improve the fit of the​ model? Use a=0.05.

A. ​No; only the variable Width improves the fit of the model.

B. ​No; only the variable Volume improves the fit of the model.

C. ​Yes, both explanatory variables improve the fit of the​ model, though the variable Volume just barely improves the model.

D. ​Yes, both explanatory variables improve the fit of the model.

______________________________________________________________________________________________________________________________________

​(c) Find the variance inflation factor and interpret the value that you obtain.

VIF(Width)=​VIF(Volume)=_______ . ​(Round to two decimal places as​ needed.)

Interpret the variance inflation factor. Choose the correct answer below.

A. The collinearity has little effect on the standard errors.

B. The two variables Width and Volume are perfectly collinear. One of the variables should be dropped from the model​ (the one with the lower​ p-value).

C. The collinearity has a significant effect on the standard errors.

D. The collinearity has no effect on the standard errors.

______________________________________________________________________________________________________________________________________

​(d) What is the interpretation of the coefficient of​ Volume?

A. For chains of a given​ length, the retailer loses about $55.04 per additional mm³.

B. For chains of a given​ width, the retailer loses about $55.04 per additional mm³.

C. For chains of a given​ width, the retailer charges about ​$0.13 per additional mm³.

D. For chains of a given​ length, the retailer charges about 0.13 per additional mm³.

______________________________________________________________________________________________________________________________________

​(e) The marginal correlation between Width and Price is 0.94​, but its slope in the multiple regression is negative. How can this​ be?

A. Since the variance inflation factor is close to​ 1, the slope is negative.

B. The correlation coefficient is not indicative of the sign of the slope.

C. The slope is negative because the correlation indicates that the estimate of the slope is not significant.

D. Collinearity between the two explanatory variables causes the negative slope.

Answer 1

a.)

Option B

They are not perfectly correlated. Correlation is 0.965

Option B

You should not use both the variables but with Collinearity

b.)

Model using only width

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.939416956 0.882504217 0.876629428 76.84186245 ANOVA df Significance F MS Regression Residual Total 1 886993.1589 886993.1589 150.218875 9.3491E-11 20118093.4365 5904.671825 21 1005086.595 Lower 95% Coefficients Standard ErrortStat 192.1578137 42.97601264-4.47128065 0.000233913-281.8042052102.511422.2 212.139641617.30850581 12.25638099 9.3491E-11 176.0347312 248.2445521 P-value Upper 95% Intercept Width (mm)

Model Using Volume

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.990877003 0.981837235 0.980929097 30.21187824 ANOVA df Significance F MS Regression Residual Total 1 986831.4436 986831.4436 1081.153921 6.94119E-19 20 18255.15173 912.7575866 21 1005086.595 Coefficients Standard ErrortStat P-value Lower 95% Upper 95% Intercept Volume 24.2031443 10.45059433 2.315958646 0.03129364 2.40358652 46.00270208 0.1041543840.00316762532.88090511 6.94119E-19 0.097546833 0.110761934

Model using both Width and Volume

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.992937043 0.985923972 0.984442285 27.28757882 ANOVA df Significance F MS Regression Residual Total 2 990938.9682 495469.4841 665.4062949 2.57372E-18 19 14147.6272 744.611958 21 1005086.595 Coefficients Standard Error 86.3129715828.0785836 3.073978831 0.00624684627.5438207 145.0821225 -55.03832113 23.43365859-2.348686652 0.029812405104.08553225.991110028 P-value t Stat Lower 95% Upper 95% Intercept Width (mm) Volume 0.1289 0.010907732 11.81512959 3.36617E-10 0.106046123 0.151706415

From above O/Ps its clear that Volume improves the model

Option B is correct

c.)

When we find regression equation between width and volume we get,

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.964988297 0.931202414 0.927762534 0.26038137

So, Vif = 1 / (1 - R²)

Vif = 1/ (1 - 0.9312)

Vif = 14.535

Option B is correct

d.)

Option C is correct

e.)
Option D is correct