Question

In: Statistics and Probability

#_Items_Sold TV$ Conference$ JournalAd$ 88,000 87,000 95,000 100,000 80,000 98,000 99,000 99,000 96,000 101,000 103,000 101,000...

#_Items_Sold TV$ Conference$ JournalAd$
88,000 87,000 95,000 100,000
80,000 98,000 99,000 99,000
96,000 101,000 103,000 101,000
76,000 91,000 95,000 93,000
80,000 88,000 102,000 95,000
73,000 84,000 94,000 95,000
58,000 74,000 89,000 80,000
116,000 102,000 112,000 116,000
104,000 105,000 110,000 106,000
99,000 97,000 87,000 105,000
64,000 88,000 90,000 90,000
126,000 108,000 101,000 113,000
94,000 89,000 100,000 96,000
71,000 78,000 85,000 98,000
111,000 109,000 99,000 109,000
109,000 108,000 101,000 102,000
100,000 102,000 93,000 100,000
127,000 110,000 108,000 107,000
99,000 95,000 100,000 108,000
82,000 90,000 69,000 95,000
67,000 85,000 95,000 91,000
100,000 103,000 91,000 114,000
78,000 80,000 80,000 93,000
115,000 104,000 85,000 115,000
83,000 83,000 105,000 97,000
125,000 100,000 98,000 110,000
92,000 85,000 100,000 90,000
72,000 80,000 86,000 95,000
Part A - 10pts) Write down the regression equation for this problem.
Part B - 10pts) How good is this model? Explain your answer.
Part C - 10pts) Which promotional mix strategy below is predicted to produce most Items Sold (answer using all three input variables)?
TV$ Conference JournalAd
Strategy A) 80,000 70,000 80,000
Strategy B) 90,000 60,000 85,000
Strategy C) 55,000 85,000 100,000
Strategy D) 75,000 80,000 70,000
Strategy E) 95,000 70,000 65,000
Strategy F) 65,000 70,000 75,000
Part D - 10pts) Which variable is the most significant to the Target (Items Sold)?

Solutions

Expert Solution

1) Regression equation is Y = B0 + B1*X1 +B2*X2 +B3*X3

So for this problem regression equation is JournalAd$ = -1.091e+05 + (9.081e-01 * TV) +

(8.722e-02* Conference) + (1.090e+00 * JournalAd)

and the formula of regression in R for this problem is lm(formula = `#_Items_Sold` ~ ., data = train1)

2) This is the good model because Multiple R-squared= 0.849, Adjusted R-squared = 0.8223, both values of R square and Adjusted R square are greater thatn 80%.

You can see the summery of this model below

summary(Model1)

Call:
lm(formula = `#_Items_Sold` ~ ., data = train1)

Residuals:
     Min       1Q   Median       3Q      Max 
-16402.8  -4576.6   -601.8   4974.3  17125.4 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -1.091e+05  2.391e+04  -4.565 0.000275 ***
TV           9.081e-01  2.950e-01   3.078 0.006818 ** 
Conference   8.722e-02  2.205e-01   0.395 0.697410    
JournalAd    1.090e+00  3.622e-01   3.010 0.007887 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 8535 on 17 degrees of freedom
Multiple R-squared:  0.849,     Adjusted R-squared:  0.8223 
F-statistic: 31.86 on 3 and 17 DF,  p-value: 3.355e-07

3) according to this model the

Strategy A) value is 56853.4

Strategy B) value is 70512.2

Strategy C) value is 57259.2

Strategy D) value is 42285.1

Strategy E) value is 54124.9

Strategy F) value is 37781.9

4) TV and JournalAd are the most significant variables, you can see the results in summary, only TV and JournalAd have two stars.


Related Solutions

Two equipments A and B have initial costs of $100,000 and $120,000 and expected to generate annual savings during the first year of $88,000 and $98,000 respectively.
Two equipments A and B have initial costs of $100,000 and $120,000 and expected to generate annual savings during the first year of $88,000 and $98,000 respectively. The value of these annual savings is expected to increase by 10% per year (over previous period). Assume service life of 2 years, operating hours per year of 4500, Use the NPW method to determining 5 savings/ hour for each equipment. Select the optimal equipment based on your results.  
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