Optimal Level of Advertising
The optimal level of advertising is that amount where the last dollar spent on advertising results in only 1 dollar of additional sales (we are assuming here that the marginal cost of producing and selling another burger is zero!). Find the level of level of advertising, adverto, that solves:
Plugging in the least squares estimates from the model and solving for adverto can be done in gretl. A little algebra yields
The script in gretl to compute this follows.
open "@gretldirdatapoeandy. gdt" square advert
ols sales const price advert sq_advert
scalar Ao =(1-$coeff(advert))/(2*$coeff(sq_advert))
which generates the result:
? scalar Ao =(1-$coeff(advert))/(2*$coeff(sq_advert)) Generated scalar Ao (ID 7) = 2.01434
This implies that the optimal level of advertising is estimated to be approximately $2014. To test the hypothesis that $1900 is optimal (remember, advert is measured in $1000)
Ho : Д3 + 2в41.9 = 1
Hi : Дз + 2Д41.9 = 1
you can use a t-test or an F-test. Following the regression, use restrict
b + 3.8*b=1 end restrict
Remember that b refers to the coefficient of the third variable in the regression (A) and b to the fourth. The output from the script is shown in Figure 6.9. The F-statistic is =0.936 and has a p-value of 0.33. We cannot reject the hypothesis that $1900 is optimal at the 5% level.
A one-tailed test would be a better option in this case. Andy decides he wants to test whether the optimal amount is greater than $1900.
Ho : & + 3.8^4 < 1 Hi : вз + 3.8^4 > 1
A one-sided alternative has to be tested using a t-ratio rather than the F-test. The script below computes such a test statistic much in the same way that we did in chapter 5.
1 # One-sided t-test
2 ols sales const price advert sq_advert —vcv
3 scalar r = $coeff(advert)+3.8*$coeff(sq_advert)-1
4 scalar v = $vcv[3,3]+((3.8)"2)*$vcv[4,4]+2*(3.8)*$vcv[3,4]
5 scalar t = r/sqrt(v)
6 pvalue t $df t
Notice that in line 3 we had to compute the variance of a linear combination of parameters. This was easily done in the script. The results are:
t(71): area to the right of 0.967572 = 0.168271 (two-tailed value = 0.336543; complement = 0.663457)
The t-ratio is.9676 and the area to the right is 0.168. Once again, this is larger than 5% and the hypothesis cannot be rejected at that level.
Finally, Big Andy makes another conjecture about sales. He is planning to charge $6 and use $1900 in advertising and expects sales to be $80,000. Combined with the optimality of $1900 in
advertising leads to the following joint test:
H0 ф3 + 3.8^4 = 1 and ві + 6^2 + 1.9вз + 1.92в4 = 80 H1 : not H0
The model is estimated and the hypotheses tested:
1 ols sales const price advert sq_advert
5 end restrict
The result is shown in Figure 6.10 below. Andy is disappointed with this outcome. The null
Figure 6.10: Andy muses about whether $1900 in advertising is optimal and whether this will generate $80000 in sales given price is $6. It is not supported by the data.