In: Statistics and Probability
All Greens is a franchise store that sells house plants and lawn and garden supplies. Although All Greens is a franchise, each store is owned and managed by private individuals. Some friends have asked you to go into business with them to open a new All Greens store in the suburbs of San Diego. The national franchise headquarters sent you the following information at your request. These data are about 27 All Greens stores in California. Each of the 27 stores has been doing very well, and you would like to use the information to help set up your own new store. The variables for which we have data are detailed below.
x1 = annual net sales, in thousands of
dollars
x2 = number of square feet of floor display in
store, in thousands of square feet
x3 = value of store inventory, in thousands of
dollars
x4 = amount spent on local advertising, in
thousands of dollars
x5 = size of sales district, in thousands of
families
x6 = number of competing or similar stores in
sales district
A sales district was defined to be the region within a 5 mile radius of an All Greens store.
x1 | x2 | x3 | x4 | x5 | x6 |
231 | 3 | 294 | 8.2 | 8.2 | 11 |
156 | 2.2 | 232 | 6.9 | 4.1 | 12 |
10 | 0.5 | 149 | 3 | 4.3 | 15 |
519 | 5.5 | 600 | 12 | 16.1 | 1 |
437 | 4.4 | 567 | 10.6 | 14.1 | 5 |
487 | 4.8 | 571 | 11.8 | 12.7 | 4 |
299 | 3.1 | 512 | 8.1 | 10.1 | 10 |
195 | 2.5 | 347 | 7.7 | 8.4 | 12 |
20 | 1.2 | 212 | 3.3 | 2.1 | 15 |
68 | 0.6 | 102 | 4.9 | 4.7 | 8 |
570 | 5.4 | 788 | 17.4 | 12.3 | 1 |
428 | 4.2 | 577 | 10.5 | 14.0 | 7 |
464 | 4.7 | 535 | 11.3 | 15.0 | 3 |
15 | 0.6 | 163 | 2.5 | 2.5 | 14 |
65 | 1.2 | 168 | 4.7 | 3.3 | 11 |
98 | 1.6 | 151 | 4.6 | 2.7 | 10 |
398 | 4.3 | 342 | 5.5 | 16.0 | 4 |
161 | 2.6 | 196 | 7.2 | 6.3 | 13 |
397 | 3.8 | 453 | 10.4 | 13.9 | 7 |
497 | 5.3 | 518 | 11.5 | 16.3 | 1 |
528 | 5.6 | 615 | 12.3 | 16.0 | 0 |
99 | 0.8 | 278 | 2.8 | 6.5 | 14 |
0.5 | 1.1 | 142 | 3.1 | 1.6 | 12 |
347 | 3.6 | 461 | 9.6 | 11.3 | 6 |
341 | 3.5 | 382 | 9.8 | 11.5 | 5 |
507 | 5.1 | 590 | 12.0 | 15.7 | 0 |
400 | 8.6 | 517 | 7.0 | 12.0 | 8 |
(a) Generate summary statistics, including the mean and standard deviation of each variable. Compute the coefficient of variation for each variable. (Use 2 decimal places.)
x | s | CV | |
x1 | % | ||
x2 | % | ||
x3 | % | ||
x4 | % | ||
x5 | % | ||
x6 | % |
(b) For each pair of variables, generate the correlation coefficient r. For all pairs involving x1, compute the corresponding coefficient of determination r2. (Use 3 decimal places.)
r | r2 | |
x1, x2 | ||
x1, x3 | ||
x1, x4 | ||
x1, x5 | ||
x1, x6 | ||
x2, x3 | ||
x2, x4 | ||
x2, x5 | ||
x2, x6 | ||
x3, x4 | ||
x3, x5 | ||
x3, x6 | ||
x4, x5 | ||
x4, x6 | ||
x5, x6 |
(c) Perform a regression analysis with x1 as
the response variable. Use x2,
x3, x4,
x5, and x6 as explanatory
variables. Look at the coefficient of multiple determination. What
percentage of the variation in x1 can be
explained by the corresponding variations in
x2, x3,
x4, x5, and
x6 taken together? (Use 1 decimal place.)
%
(d) Write out the regression equation. (Use 2 decimal places for
x3 and x6. Use 1 decimal
place otherwise.)
x1 = | + x2 | + x3 | + x4 | + x5 | + x6 |
If 12 new competing stores moved into the sales district but the
other explanatory variables did not change, what would you expect
for the corresponding change in annual net sales? (Use 2 decimal
places.)
If you increased the local advertising by 5 thousand dollars but
the other explanatory variables did not change, what would you
expect for the corresponding change in annual net sales? (Use 2
decimal places.)
(e) Test each coefficient to determine if it is or is not zero. Use
a 5% level of significance. (Use 2 decimal places for t
and 3 decimal places for the P-value.)
t | P-value | |
β2 | ||
β3 | ||
β4 | ||
β5 | ||
β6 |
(f) Suppose you and your business associates rent a store, get a bank loan to start up your business, and do a little research on the size of your sales district and the number of competing stores in the district. If x2 = 2.8, x3 = 250, x4 = 3.1, x5 = 7.3, and x6 = 2, use a computer to forecast x1 = annual net sales and find an 80% confidence interval for your forecast (if your software produces prediction intervals). (Use 2 decimal places.)
forecast | |
lower limit | |
upper limit |
(g) Construct a new regression model with x4 as
the response variable and x1,
x2, x3,
x5, and x6 as explanatory
variables. (Use 2 decimal places for the intercept, 4 for
x1, 5 for x3, and 3
otherwise.)
x4 = | + x1 | + x2 | + x3 | + x5 | + x6 |
Suppose an All Greens store in Sonoma, California, wants to
estimate a range of advertising costs appropriate to its store. If
it spends too little on advertising, it will not reach enough
customers. However, it does not want to overspend on advertising
for this type and size of store. At this store,
x1 = 163, x2 = 2.4,
x3 = 188, x5 = 6.6, and
x6 = 10. Use these data to predict
x4 (advertising costs) and find an 80%
confidence interval for your prediction. (Use 2 decimal
places.)
prediction | |
lower limit | |
upper limit |
At the 80% confidence level, what range of advertising costs do you
think is appropriate for this store? (Round to nearest
integer.)
lower limit | $ |
upper limit | $ |
a)
Descriptive Statistics
N Mean Std. Deviation
X1 27 286.5741 192.06172
X2 27 3.3259 2.01105
X3 27 387.4815 191.16774
X4 27 8.1000 3.77451
X5 27 9.6926 5.14003
X6 27 7.7407 4.89578
CV
.67
.60
.49
.47
.53
.63
b)
r R2
.894 .799
.946 .895
.914 .835
.954 .910
-.912 .832
.844 .712
.749 .561
.838 .702
-.766 .587
.906 .821
.864 .746
-.807 .651
.795 .632
-.841 .707
-.870 .757
c)
Model Summary
Model R R Square Adjusted R
Square Std. Error of the Estimate Change
Statistics
R Square Change F
Change df1 df2 Sig. F
Change
1 .997a .993 .992
17.64924 .993 611.590
5 21 .000
a Predictors: (Constant), X6, X2, X4, X5, X3
Coefficientsa
Model Unstandardized
Coefficients Standardized
Coefficients t Sig.
B Std.
Error Beta
1 (Constant) -18.859
30.150 -.626 .538
X2 16.202 3.544
.170 4.571 .000
X3 .175 .058
.174 3.032 .006
X4 11.526 2.532
.227 4.552 .000
X5 13.580 1.770
.363 7.671 .000
X6 -5.311 1.705
-.135 -3.114 .005
a Dependent Variable: X1
99.3% of variation can be explained by the corresponding variations in x2, x3, x4, x5, and x6 taken together.
d)
X1 = -18.9 + 16.2*X2 + 0.17*X3 + 11.5*X4 + 13.6*X5 - 5.31*X6
If 12 new competing stores moved into the sales district but the other explanatory variables did not change:
X1 = -82.62 + 16.2*X2 + 0.17*X3 + 11.5*X4 + 13.6*X5
If you increased the local advertising by 5 thousand dollars but the other explanatory variables did not change:
X1 = 38.6 + 16.2*X2 + 0.17*X3 + 13.6*X5 - 5.31*X6