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 |
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)? |
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.