Question

In: Statistics and Probability

A firm that operates a​ large, direct-to-consumer sales force would like to put in place a...

A firm that operates a​ large, direct-to-consumer sales force would like to put in place a system to monitor the progress of new agents. A key task for agents is to open new​ accounts; an account is a new customer to the business. The goal is to identify​"superstar agents" as rapidly as​ possible, offer them​ incentives, and keep them with the company. To build such a​ system, the firm has been monitoring sales of new agents over the past two years. The response of interest is the profit to the firm​ (in dollars) of contracts sold by agents over their first year. Among the possible predictors of this performance is the number of new accounts developed by the agent during the first three months of work. Formulate the SRM with Y given by the natural log of Profit from Sales and X given by the natural log of Number of Accounts.

Data Table:

Profit_from_Sales   Number_of_Accounts
23248.39 46
35347.43 10
27384.29 83
19946.44 7
8201.63 12
35757.84 169
19386.39 10
39931.42 22
18165.13 16
17813.03 15
16141.71 14
33833.27 23
35746.25 33
17043.17 19
25808.56 28
25910.96 72
33035.56   21
10157.41 13
59928.04 489
30807.71 136
25407.95 18
4508.74 2
40696.58 37
18013.02 6
14866.23 10
35936.01 31
12825.21 7
48987.22 25
33015.75 26
32611.79 42

If a sales representative opens 150 ​accounts, what level of sales would you predict for this​ individual? Express your answer as an interval.​ (Use 95%.)

__ to __ dollars (Round to two decimals places as needed)

Solutions

Expert Solution

Column1 Column2 log sales log account
23248.39 46 10.05399 3.828641
35347.43 10 10.47298 2.302585
27384.29 83 10.21772 4.418841
19946.44 7 9.900806 1.94591
8201.63 12 9.012088 2.484907
35757.84 169 10.48452 5.129899
19386.39 10 9.872327 2.302585
39931.42 22 10.59492 3.091042
18165.13 16 9.807259 2.772589
17813.03 15 9.787685 2.70805
16141.71 14 9.689162 2.639057
33833.27 23 10.4292 3.135494
35746.25 33 10.4842 3.496508
17043.17 19 9.743505 2.944439
25808.56 28 10.15846 3.332205
25910.96 72 10.16242 4.276666
33035.56 21 10.40534 3.044522
10157.41 13 9.225959 2.564949
59928.04 489 11.0009 6.192362
30807.71 136 10.33552 4.912655
25407.95 18 10.14282 2.890372
4508.74 2 8.413773 0.693147
40696.58 37 10.6139 3.610918
18013.02 6 9.79885 1.791759
14866.23 10 9.606847 2.302585
35936.01 31 10.4895 3.433987
12825.21 7 9.459168 1.94591
48987.22 25 10.79931 3.218876
33015.75 26 10.40474 3.258097
32611.79 42 10.39243 3.73767
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.702472246
R Square 0.493467257
Adjusted R Square 0.475376801
Standard Error 0.401853136
Observations 30
ANOVA
df SS MS F Significance F
Regression 1 4.404976259 4.404976259 27.27776903 1.50845E-05
Residual 28 4.521606407 0.161485943
Total 29 8.926582666
Coefficients Standard Error t Stat P-value Lower 95%
Intercept 8.946916302 0.226362488 39.52473032 4.1491E-26 8.483233764
log account 0.355405202 0.068048625 5.222812368 1.50845E-05 0.216013912

ln sales ^= 8.9469 + 0.3554 ln account

= 8.9469 + 0.3554*ln(150)

= 10.7277

ata = read.csv("sales.csv")
y = data$ï..log.sales
x=data$log.account
model = lm (y~x)
predict(model,data.frame(x=log(150)) ,data= data,conf.level=0.95,interval= "confidence")

 fit     lwr      upr
1 10.72772 10.4276 11.02785


exp(10.4276)
[1] 33779.18
> exp(11.02785)
[1] 61565.07




hence confidence interval


( 33779.18 , 61565.07 )

                    

                                        
                    


        

            


         

             
orchestra answered 2 years ago

         

    

        

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