   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 adverto

The script in gretl to compute this follows.

which generates the result:

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. Figure 6.9: Testing whether \$1900 in advertising is optimal using the restrict statement.

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

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

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:

2 restrict

3 b+3.8*b=1

4 b+6*b+1.9*b+3.61*b=80

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.

hypothesis is rejected since the p-value associated with the test is 0.0049 < .05. Sorry Andy!