Question

Can you do problem 20 and 21, problem 18 and and 19 is there for reference, Thank you!

Question 18 3 pts Consider the following OLS multiple regression results from Table 2 of “The Impact of Light Skin on Prison Time for Black Female Offenders" (The Social Science Journal 48 (2011), p. 256]. The dependent variable is the natural logarithm of time served, where time served is measured as the number of days served in prison. At the time of admission to prison, correctional officers noted whether female African- American inmates had light skin tones or not, and denoted whether a prisoner was thin (in terms of body weight) or not. Another explanatory variable is the total sentencing components ("the number of counts for a crime an individual is charged with and sentenced for. For example, an individual who breaks into someone's house can be simultaneously charged with breaking and entering and criminal trespass, as well as, potentially, violation of probation. Each of these individual charges can carry a separate sentence component that contributes to the total sentence length"). The other explanatory variables should be self-evident. For some reason, the authors do not report the intercept coefficient, but the coefficient estimates (with the standard errors in parentheses), are reported below. Dependent Variable: Natural Logarithm of Time Served, In(time served) Explanatory Variables Light skin Thin Total sentencing components Parole arrest Conviction date Homicide Robbery Habitual felon Total infractions in prison n R2 Coefficients and Standard Errors in Parentheses -0.109 (0.040) -0.099 (0.021) 0.057 (0.004) 0.674 (0.047) -0.001 (0.000) 1.496 (0.070) 0.764 (0.046) 1.973 (0.111) 0.053 (0.002) 11,093 0.280 How many of the explanatory variables are dummy variables? O 2 O 6 08 0 1 Question 19 3 pts Consider the multiple regression output from the question above regarding prison time. The 99-percent confidence interval for the coefficient on homicide is approximately: 01.196 to 1.796 1.359 to 1.633 O 1.426 to 1.546 O 1.316 to 1.676

Answer 1

20.

H0: All coefficients on explanatory variables(except intercept) are jointly equal to 0

H1: At least one of the coefficients of explanatory variables(except intercept) is not equal to 0

In order to test this hypothesis, we use the F-test.

The value of F-statistic is given as:

$F = \frac{\frac{R^2}{k-1}}{\frac{(1-R^2)}{(n-k)}}$

where k = (No. of explanatory variables + 1) = (9 + 1) = 10

n = sample size

R² = R-squared value

Substituting the values of R², n and k in the above formula, we get:

$F = \frac{\frac{0.28}{9}}{\frac{(0.72)}{(11083)}}$

Ans) 479

21.

We know that in order to check the significance of the coefficients of explanatory variables, we use the t-test.

The value of t statistic is calculated as:

$t = \frac{\widehat{\beta} - \beta_0}{se(\widehat{\beta})}$

where $\widehat{\beta}$ = estimated value of coefficient

$\beta_0$ = value of the coefficient under null hypothesis

$se(\widehat{\beta})$ = standard error of the estimated coefficient

For all our coefficients, the hypothesis for t-test will be:

$H_0: \beta = 0$

$H_1: \beta \neq 0$

So, in all cases, $\beta_0$ will be equal to 0. So, the t-statistic will simply be the ratio of estimated coefficients and their standard errors. So, based on the values in the table, we'll calculate t-values for all variables.

Exp Var	Coefficients	Std Errors	t
Light skin	-0.109	0.04	-2.725
Thin	-0.099	0.021	-4.71429
Total sentencing comp	0.057	0.004	14.25
Parole arrest	0.674	0.047	14.34043
Conviction date	-0.001	0	-
Homicide	1.496	0.07	21.37143
Robbery	0.764	0.046	16.6087
Habitual felon	1.973	0.111	17.77477
Total infractions	0.053	0.002	26.5

Note: Since the standard error of the estimated coefficient for conviction date is 0, the t-value will be a very large value and hence, is not displayed in the table.

Since this is a 10 variable model(9 explanatory variables and 1 dependent variable), the degrees of freedom for the t-test = n - 10.

So, df = 11093 - 10 = 11083

Since the sample size is very high, at 5% level of significance and the above df, we find that the critical t-value for a two tailed test is 1.96.

If we take the absolute values of the calculated t- values, we find that for all explanatory variables, the calculated t-values is greater than the critical t-value(1.96) at 5% level of significance.Hence, we can reject the null hypothesis and infer that all the estimated coefficients are statistically different from 0 at 5% level of significance.

Ans) All of the estimated coefficients are (individualy) different from zero at 5% level of significance.