Question

7. An electric utility company must develop a reliable model for projecting the number of resi- dential electricity customers. To this effect, they study the effect of changing population on the number of customers. This information is shown in the table below. Since an area with 0 population would have 0 residential electric customers, one could argue that the regres- sion through the origin is appropriat<e (a) Fit the model y-e to the data (b) Is there evidence that x contributes information for the prediction of y? Test using α-0.01. (c) Now fit the more general model y-Ao + A x + є to the data. Is there evidence (at a 0.01) that x contributes information for the prediction of y? (d) Which model would you recommend? POPULATION IN SERVICE AREA x, hundreds RESIDENTIAL ELECTRICITY CUSTOMERS IN SERVICE AREA YEAR 262 319 361 381 405 439 472 508 547 592 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 14,041 16,953 18,984 19,870 20,953 22,538 23,985 25,641 27,365 29,967

Answer 1

Answer to part a)
enter data of x and y in excel

click on data tab

in it click on data analysis

in it select regression

click ok

The following window appears on screen:

Regression Input $D$2:$D$12 Input Y Range: 262 14041 319 16953 361 18984 381 19870 406 20953 439 22538 472 23985 508 25641 54727365 592 29967 Cancel Input X Range: $C$2:$C$12 Help Labels Confidence Level: Constant is Zero Output options Output Range New Worksheet Ply: O New Workbook Residuals Residuals Standardized Residuals Residual Plots Line Fit Plots Normal Probability Normal Probability Plots

.

In the regression window :

enter the range of Y values

enter the range of X values

check the box for labels

check the box for constant zero (this implies the regression equation will have no constant value)

click ok

the following output is obtained:

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square 0.888516 Standard Error Observations 0.999814 0.999627 457.971 10 ANOVA df F ignificance F MS Regression Residual Total 1 5.06E+09 5.06E+09 24146.71 3.29E-15 9 1887637 209737.5 10 5.07E+09 Coefficientsandard Err P-value #N/A Lower 95%Upper 95%ower 95.0% pper 95.0% t Stat #N/A #N/A #N/A #N/A #N/A 0 #N/A Intercept 51.17322 0.329317155.3921 9.62E-1750.42825 51.91818 50.4282551.91818

The model is: Y = 51.17322 x

.

Part b)

The model test values are: F = 24146.71 , with an extremely low significance value 3.29 x 10^-15

Since the significance value < 0.05 , this implies there is evidence that x contributes information for prediction of y

.

Part c)

In this scenario we repeat all the steps used in part, and do not check the box for constant zero

as follows:

Regression Input OK $D$2:$D$12 Input Y Range: Input X Range: Labels 262 14041 319 16953 361 18984 381 19870 406 20953 439 22538 472 23985 508 25641 54727365 592 2 Cancel $C$2:$C$12 Help Constant is Zero Confidence Level 95 Output options Output Range: New Worksheet Ply: 5 New Workbook 29967 Residuals Residuals Standardized Residuals Residual Plots Line Fit Plots Normal Probability 2 Normal Probability Plots

.

enter the range of data for y

enter the range of data for x

tick mark the check box of label

click ok

the following output appears

SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.999544 0.999089 0.998975 155.8581 10 ANOVA df ignificance F F MS Regression Residual Total 1 2.13E+08 2.13E+08 8771.987 1.89E-13 8 194334 24291.75 9 2.13E+08 Coefficientsandard Err t Stat P-value Lower 95%Upper 95%ower 95.0% pper 95.0% 1845.723 221.0694 8.3490673.21E-05 1335.9362355.51 1335.936 2355.51 47.08182 0.502695 93.65889 1.89E-13 45.92261 48.24104 45.92261 48.24104 Intercept

.

Now again since the P value = 1.89 x 10^-13 < 0.05 , this implies the value of x contributes significant information to predict the value of Y

.

Part d)

The model with the highest R square value must be recommended

Since both the models are significant, the r square value will tell us the percent of variation in Y that can be explained by the model. Hence we compare the R square values for both the models:

R square value for model 1 = 0.999627

R square value for model 2 = 0.999089

Hence model 1 has higher value of R square, thus one must go for model # 1