Question

In: Statistics and Probability

A national chain of women’s clothing stores with locations in the large shopping malls thinks that...

A national chain of women’s clothing stores with locations in the large shopping malls thinks that it can do a better job of planning more renovations and expansions if it understands what variables impact sales. It plans a small pilot study on stores in 25 different mall locations. The data it collects consist of monthly sales, store size (sq. ft), number of linear feet of window display, number of competitors located in mall, size of the mall (sq. ft),and distance to nearest competitor (ft). USING EXCEL FUNCTIONS

  1. Plot the residuals versus the actual values. Do you think that the model does a good job of predicting monthly sales? Why or why not?
  2. Do you think that this model will be useful in helping the planners? Why or why not?
  3. Test the individual regression coefficients. At the 0.05 level of significance, what are your conclusions?
  4. If you were going to drop just one variable from the model, which one would you choose? Why?
Sales Size Windows Competitors Mall Size Nearest Competitor
4453 3860 39 12 943700 227
4770 4150 41 15 532500 142
4821 3880 39 15 390500 263
4912 4000 39 13 545500 219
4774 4140 40 10 329600 232
4638 4370 48 14 802600 257
4076 3570 37 16 463300 241
3967 3870 39 16 855200 220
4000 4020 44 21 443000 188
4379 3990 38 16 613400 209
5761 4930 50 15 420300 220
3561 3540 34 15 626700 167
4145 3950 36 14 601500 187
4406 3770 36 12 593000 199
4972 3940 38 11 347100 204
4414 3590 35 10 355900 146
4363 4090 38 13 490100 206
4499 4580 45 16 649200 144
3573 3580 35 18 685900 178
5287 4380 42 15 106200 149
5339 4330 40 10 354900 231
4656 4060 37 11 598700 225
3943 3380 34 16 381800 163
5121 4760 44 17 597900 224
4557 3800 36 14 745300 195

Solutions

Expert Solution

a) Plot of residuals versus the actual values

The residuals "bounce randomly" around the 0 line. This suggests that the assumption that the relationship is linear is reasonable.This also suggests that the variances of the error terms are equal.

Do you think that the model does a good job of predicting monthly sales? Why or why not?

Regression Statistics
Multiple R 0.913723178
R Square 0.834890046
Adjusted R Square 0.791440058
Standard Error 244.883447
Observations 25

Here R-square is 0.8349 which near to 1 indicates our model explains 83.49% information about data. So our fitted model is good for predicting sales.

b)

Do you think that this model will be useful in helping the planners? Why or why not?

Yes the fitted regression model is good because in this model we have 0.8349 which near to 1which indicates model is adquate for prediction purpose with higher accuracy. So for predicting monthly sale one can use this regression model can make the arrangment for future planning about how to increase sales.

c)

Test the individual regression coefficients. At the 0.05 level of significance, what are your conclusions?

Coefficients Standard Error t Stat P-value
Intercept 1506.801793 672.1868031 2.241641439 0.037118
Size 0.919372513 0.30062727 3.058180697 0.00647
Windows 9.075980803 28.82342842 0.31488207 0.756283
Competitors -67.68552636 21.95287933 -3.083218622 0.00612
Mall Size -0.000902855 0.000280615 -3.217412572 0.004534
Nearest Competitor 2.095892895 1.594431056 1.314508324 0.204323

by observing above table

For size p-value = 0.00647 <0.05 so we reject the null hypothesis and conclude that size is significant variable

For Windows p-value = 0.756283 > 0.05 so we accept the null hypothesis and conclude that windows is not significant variable

For Competitors p-value = 0.00612 <0.05 so we reject the null hypothesis and conclude thatCompetitors is significant variable

For Mall Size p-value = 0.004534 <0.05 so we reject the null hypothesis and conclude that Mall Size is significant variable

For Nearest Competitor  p-value = 0.204323 > 0.05 so we accept the null hypothesis and conclude that Nearest Competitor is not significant variable.

d)

we will drop windows variable because For Windows p-value = 0.756283 ( which is highest than remainnig variable)  > 0.05 so we accept the null hypothesis and conclude that windows is not significant variable and statndard error is also high.


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