Inference

Paulo Fagandini

Nova SBE

Review

Review

• Moments and distribution for \(\hat{\beta}\)

Exercises

https://moodle.novasbe.pt/mod/folder/view.php?id=4021

Exercise 3.1

One of the most important decisions business leaders face concerns the allocation of the firm’s budget to the different departments. To help in the decision of finding out how much money to allocate to the R&D department you decide to collect data for 32 firms in the chemical industry on the following variables: rdintensity (expenditures on r&d as % of sales), profmarg (profits as % of sales), and sales (in millions of USD). You then estimate the following equation (se in parenthesis):

\[ \widehat{profmarg}=\underset{(6.3043)}{11.76}+\underset{(0.7233)}{0.79}rdintensity-\underset{(0.8814)}{0.63}\log(sales) \]

With \(R^2=0.0464\)

Exercise 3.1

1. Interpret the coefficient on rdintensity. Is this an economically significant effect?
2. Does R&D intensity have a statistical significant effect on profit margin? Do the 2-sided test at a 10% level.
3. Steve Jobs once said: “Innovation has nothing to do with how many R&D dollars you have. When Apple came up with the Mac, IBM was spending at least 100 times more on R&D. It’s not about money. It’s about the people you have, how you’re led, and how much you get it.” Does this exercise provide evidence in favor or against Steve Jobs’ claim?

t-student

1. If rdintensity increases by 1 pp profit margin increases by 0.79pp on average, ceteris paribus.
2. \(H_0: \beta_{rintensity}=0\), \(H_a: \beta_{rintensity}\neq0\). \(\alpha=10\%\), \(df=32-2-1=29\), \(t^c=1.699\). \(t=\frac{0.79-0}{0.7233}=1.09<t^c\). Fail to reject \(H_0\) at 10%.
3. We do not find statistically significant effect of r&d on profit, maybe it is not only about the money.

Exercise 3.3

Consider the estimated equation from a model explaining college GPA with the high school GPA (hsGPA), the achievenment test score (ACT) and average number of lectures missed per week (skipped) as explanatory variables (usual standar errors in parenthesis below the estimates):

\[ \widehat{colGPA}=\underset{(0.33)}{1.39}+\underset{(0.094)}{0.412}hsGPA + \underset{(0.011)}{0.015}ACT-\underset{(0.026)}{0.083}skipped \]

Exercise 3.3

1. Using the standard normal approximation, find the 95% confidence interval for \(\beta_{hsGPA}\).
2. Can you reject the hypothesis \(H_0:\beta_{hsGPA}=0.4\) against the twosided alternative at the 5% significance level?
3. Can you reject the hypothesis \(H_0: \beta_{hsGPA}=1\) against the two-sided alternative at the 5% significance level?

t-student

1. \(0.412\pm 1.96\times 0.094\Rightarrow [0.228, 0.596]\)
2. \(H_0: \beta_1=0.4\) cannot be rejected at 5% significance as 0.4 is in the confidence interval.
3. \(H_0: \beta_1=1\) is rejected at 5% significance as 1 falls out of the CI.

Exercise 3.4

Consider the following model:

\[ \log(scrap)=\beta_0+\beta_1 hrsemp + \beta_2 \log(sales)+\beta_3 \log(employ)+u \]

where hrsemp is annual hours of training per employee, sales is annual firm sales (in dollars), and employ is the number of firm employees.

Exercise 3.4

1. Show that the population model can also be written as

   \[ \log(scrap)=\beta_0+\beta_1 hrsemp + \beta_2 \log(sales/employ)+\theta_3 \log(employ)+u \]

   where \(\theta_3=\beta_2+\beta_3\)

2. Interpret the null hypothesis \(H_0: \theta_3=0\).

1. subtract and add \(\beta_2\log(employ)\) and then you get the result.
2. \(H_0: \theta_3=0\) can be written as \(\beta_2+\beta_3=0\) or \(\beta_2=-\beta_3\), that is under the null hypothesis the effects cancel out, or are symmetric.

Exercise 3.4

3. Using 43 observations, you estimate the original equation:

   \[ \widehat{\log(scrap)}=\underset{(4.57)}{11.74}+\underset{(0.019)}{0.043} hrsemp - \underset{(0.370)}{0.951}\log(sales) + \underset{(0.360)}{0.992}\log(employ) \]

   and when the second equation is estimated, we obtain:

   \[ \widehat{\log(scrap)}=\underset{(4.57)}{11.74}+\underset{(0.019)}{0.043} hrsemp - \underset{(0.370)}{0.951}\log(sales/employ) + \underset{(0.205)}{0041}\log(employ) \]

   With \(R^2=0.310\) for both regressions. Test if the effect of annual firm sales (in USD) on scrap rates is the symmetric of the effect of the number of firm employees on scrap rates.

t-student

\(H_0:\theta_3=0\) and \(H_a: \theta_3\neq 0\). For \(\alpha=5\%\) and \(dof=43-3-1=39\) the critical value \(t^c=2.023\). In this case \(t=\frac{0.041}{0.205}=0.2\). Because \(t<t^c\) we fail to reject \(H_0\) at 5%.

Exercise 3.7

With the goal of evaluating the environmental impact of different management policies of Natural Parks, a study has analyzed data for 28 Natural Parks in country X. The following regression was estimated: \[ \hat{y} = \underset{(-1.085)}{-12.666}+\underset{(17.299)}{3.743}x_{1}-\underset{(-2.164)}{9.614}x_{2} \]

where \(y\) is the number of animals and plants in extintion danger in each park, \(x_1\) is the number of visits to the park (in thousands), and \(x_2\) is the expenditures on nature conservation (in million euro). Consider also that \(Cov\left(\hat{\beta}_1,\hat{\beta}_2\right)=-0.8835\). The numbers in parenthesis are t-statistics.

Exercise 3.7

1. Interpret the parameter estimates.
2. Knowing that \(R^2=0.984\), test the overall significance of the regression.
3. The director of the Natural Parks association is worried about the number of vists to the parks: “… if not for the nature, at least in financial terms we have to reduce the number of visits, because in order to compensate for an increase of one thousand visitors we have to spend 0.5 million euros more in maintenance”. Comment on this sentence.

1. If visitors increase by 1000 the N of endangered species increases by 3.73 on average ceteris paribus. If maintenance increases by 1 million euros, the number of endangered species decreases by 9.614, on average, ceteris paribus.
2. \(F=\frac{R^2_{ur}/k}{(1-R^2_{ur})/(n-k-1)}\sim F_{k,n-k-1}\) here \(F=\frac{0.984/2}{(1-0.984)/(28-2-1)}=768.75\) Since \(F>F^c=F_{5\%,2.25}=3.39\) we reject \(H_0\). The coefficients are jointly statistical significant, or the model is statistical significant.
3. We want to test if 0.5 million euro compensate the impact on the number of endangered species for having 1000 more visits to the park: \(H_0: -\beta_1=0.5\beta_2\) or \(H_0=\beta_1+0.5\beta_2=0\). Test for \(\hat{\rho}=\hat{\beta}_1+0.5\hat{\beta}_2\): \(t=\frac{\hat{\rho}}{se(\hat{\rho})}=\frac{3.743-0.5(9.614)}{se(\hat{\rho})}\) \(se(\hat{\rho})=\sqrt{V(\hat{\beta}_1+0.5\hat{\beta}_2)}\) and that is \(\sqrt{V(\hat{\beta}_1)+0.5^2 V(\hat{\beta}_2)+2\times 0.5Cov(\hat{\beta}_1, \hat{\beta}_2)}=2.024\), leading to \(t=-0.526\). Thus, we fail to reject \(H_0\) as \(|t|<|t^c|\), and we have evidence that spending 0.5 million eur in maintenance compensates the impact on the number of endangered species of having 1000 more visitors.

Appendix

t table

df-\(\alpha/2\)	0.50	0.25	0.20	0.15	0.10	0.05	0.025	0.01	0.005	0.001	0.0005
1	0.000	1.000	1.376	1.963	3.078	6.314	12.71	31.82	63.66	318.31	636.62
2	0.000	0.816	1.061	1.386	1.886	2.920	4.303	6.965	9.925	22.327	31.599
3	0.000	0.765	0.978	1.250	1.638	2.353	3.182	4.541	5.841	10.215	12.924
4	0.000	0.741	0.941	1.190	1.533	2.132	2.776	3.747	4.604	7.173	8.610
5	0.000	0.727	0.920	1.156	1.476	2.015	2.571	3.365	4.032	5.893	6.869
6	0.000	0.718	0.906	1.134	1.440	1.943	2.447	3.143	3.707	5.208	5.959
7	0.000	0.711	0.896	1.119	1.415	1.895	2.365	2.998	3.499	4.785	5.408
8	0.000	0.706	0.889	1.108	1.397	1.860	2.306	2.896	3.355	4.501	5.041
9	0.000	0.703	0.883	1.100	1.383	1.833	2.262	2.821	3.250	4.297	4.781
10	0.000	0.700	0.879	1.093	1.372	1.812	2.228	2.764	3.169	4.144	4.587
11	0.000	0.697	0.876	1.088	1.363	1.796	2.201	2.718	3.106	4.025	4.437
12	0.000	0.695	0.873	1.083	1.356	1.782	2.179	2.681	3.055	3.930	4.318
13	0.000	0.694	0.870	1.079	1.350	1.771	2.160	2.650	3.012	3.852	4.221
14	0.000	0.692	0.868	1.076	1.345	1.761	2.145	2.624	2.977	3.787	4.140
15	0.000	0.691	0.866	1.074	1.341	1.753	2.131	2.602	2.947	3.733	4.073
16	0.000	0.690	0.865	1.071	1.337	1.746	2.120	2.583	2.921	3.686	4.015
17	0.000	0.689	0.863	1.069	1.333	1.740	2.110	2.567	2.898	3.646	3.965
18	0.000	0.688	0.862	1.067	1.330	1.734	2.101	2.552	2.878	3.610	3.922
19	0.000	0.688	0.861	1.066	1.328	1.729	2.093	2.539	2.861	3.579	3.883
20	0.000	0.687	0.860	1.064	1.325	1.725	2.086	2.528	2.845	3.552	3.850
21	0.000	0.686	0.859	1.063	1.323	1.721	2.080	2.518	2.831	3.527	3.819
22	0.000	0.686	0.858	1.061	1.321	1.717	2.074	2.508	2.819	3.505	3.792
23	0.000	0.685	0.858	1.060	1.319	1.714	2.069	2.500	2.807	3.485	3.768
24	0.000	0.685	0.857	1.059	1.318	1.711	2.064	2.492	2.797	3.467	3.745
25	0.000	0.684	0.856	1.058	1.316	1.708	2.060	2.485	2.787	3.450	3.725
26	0.000	0.684	0.856	1.058	1.315	1.706	2.056	2.479	2.779	3.435	3.707
27	0.000	0.684	0.855	1.057	1.314	1.703	2.052	2.473	2.771	3.421	3.690
28	0.000	0.683	0.855	1.056	1.313	1.701	2.048	2.467	2.763	3.408	3.674
29	0.000	0.683	0.854	1.055	1.311	1.699	2.045	2.462	2.756	3.396	3.659
30	0.000	0.683	0.854	1.055	1.310	1.697	2.042	2.457	2.750	3.385	3.646
40	0.000	0.681	0.851	1.050	1.303	1.684	2.021	2.423	2.704	3.307	3.551
60	0.000	0.679	0.848	1.045	1.296	1.671	2.000	2.390	2.660	3.232	3.460
80	0.000	0.678	0.846	1.043	1.292	1.664	1.990	2.374	2.639	3.195	3.416
100	0.000	0.677	0.845	1.042	1.290	1.660	1.984	2.364	2.626	3.174	3.390
1000	0.000	0.675	0.842	1.037	1.282	1.646	1.962	2.330	2.581	3.098	3.300
Z	0.000	0.674	0.842	1.036	1.282	1.645	1.960	2.326	2.576	3.090	3.291

Back to Exercise 3.1 Exercise 3.3 Exercise 3.4