CHAPTER 17

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CHAPTER 17

SOLUTIONS TO PROBLEMS

1. There are two problems in using ordinary least squares regression when you have a binary or dichotomized dependent variable (a dependent variable coded 0 or 1). One problem occurs because the predicted probabilities from an OLS model may exceed the lower (0) and upper (1) limits of a probability. That is, in using an OLS model to predict the probability of the dependent even occurring,it is possible to get predicted probabilities less than 0 and greater than 1. A second problem concerns the error term. OLS regression assumes that the error term is normally distributed and independent of any given X value. When OLS is used to predict probabilities, however, the error terms are correlated with X, i.e., there is heteroscedasticity.

2. In a logistic regression analysis, the dependent variable is the natural log of the odds that the event will occur (ln P/1-P, where P is the probability of the event occurring). The dependent variable in a probit regression model is an unobserved continuous index Z that is presumed to be normally distributed with a mean of 0 and standard deviation of 1. In neither the logit or probit regression model, therefore, is the dependent variable the probability of an event. We can, however, obtain estimates of those probabilities with both types of models.

3. A logistic or probit regression model would be appropriate whenever the dependent variable is a dichotomy. For example, whether or not someone is arrested or rearrested, whether or not someone is convicted, commits or refrains from committing a criminal act, makes bail or fails to make bail, is sentenced to probation or a prison term, appears at trial or absconds. Anytime you are interested in a dependent variable that has only two outcomes, the logistic and probit regression models would be appropriate. Remember that you can use continuous (interval/ratio) independent variables in logistic and probit regression, and you can have dichotomized independent variables in OLS regression.

4. These problems ask you about probabilities and odds. Once a probability is determined, the odds of an event can be calculated as P/1-P. The odds of an event, then, is the ratio of the probability of an event occurring to the probability of the event not occurring.

a. probability = 250/700 = .36 odds = .36/.64 = .56

b. probability = 500/700 = .71 odds = .71/.29 = 2.45

c. probability = 630/700 = .90 odds = .90/.10 = 9.00

d. probability = 180/700 = .26 odds = .26/.74 = .35

a. The logistic regression coefficient is .3278. This tells us that the log of the odds of a guilty verdict is .3278 higher for males than females. Remember, a logistic regression coefficient is interpreted in terms of a change in the log of the odds of the dependent variable for a one unit change in the independent variable. Since the independent variable is coded 0 for females and 1 for males, a one unit change means going from a female to a male defendant. Changing genders increases the log of the odds of a guilty plea by .3278, which is the size of the logistic regression coefficient for GENDER.

The log of the odds is not something that is easy to interpret. If we take the antilog of the logistic regression coefficient, however, we will have the odds multiplier. The odds multiplier tells us how much greater the odds of the dependent variable are for a one unit change in the independent variable. For example, the antilog of .3278 is e^.3278 = 1.39. This tells us that the odds of a guilty verdict for males is 1.39 times higher than for females.

b. The null hypothesis we wish to test here is that the logistic regression coefficient in the population is equal to zero: H₀: b_GENDER = 0, against the alternative that it is different from zero: H₀: b_GENDER ¹ 0. There are three ways we can determine the significance of the logistic regression coefficient. One way is to simply look at the significance level provided in the table. This significance level is .0166. This tells us that if the null hypothesis is true that b = 0, the probability that we would observe a sample b of .3278 is .0166. Since this is less than our selected alpha of .05, we would reject the null hypothesis. Another way would be to conduct a t test. The t statistic is simply the ratio of the regression coefficient to its standard error. The number of degrees of freedom is n - k, where k is the number of parameters estimated. In our problem, t = .3278/.1369 = 2.39. With a two-tailed alpha of .05 and 52 - 2 = 50 degrees of freedom (remember, the constant counts as one parameter and the independent variable as a second), our critical value of t from the t table is approximately equal to ±2.00. Since our t ratio is 2.39, we would reject the null hypothesis that b = 0. Yet another way to test the null hypothesis is with the Wald statistic, which is shown in the table. The Wald statistic is the square of the t ratio, and has a chi-square distribution. With 1 degree of freedom and an alpha of .05, the critical value of chi-square is 3.841. Since our Wald statistic is 5.736, we would reject the null hypothesis.

c. The predicted log odds that a male would plead guilty would be:

We can now use this predicted log odds for a male to determine the probability of a guilty plea. The predicted probability of a guilty plea for a male would be:

The predicted probability that a male would enter a guilty plea, then, is .595.

The predicted log odds that a female would enter a guilty plea is:

Notice that the difference between the log odds of a guilty plea for a male and for a female .3839 - .0561 = .3278, which is the value of the logistic regression coefficient. An increase of one unit in the independent variable, then, increases the log odds of the dependent variable by .3278.

We can now use this predicted log odds for a female to determine the probability of a guilty plea. The predicted probability of a guilty verdict for a female would be:

The predicted probability that a female would enter a guilty plea, then, is .514.

Now that we have the probabilities, we can calculate the odds of a guilty plea for males and females. For males the odds are .595/.405 = 1.47. The odds of a guilty plea for a female are .514/.486 = 1.06. Notice that if we take the ratio of the male odds to the female odds 1.47/1.06 it will equal 1.39, which is the odds multiplier or the antilog of the regression coefficient (e^.3278 = 1.39). Now you can see that the odds of a guilty plea are 1.39 times higher for males than females (See part a above).

d. Using the independent variable, we have 22 + 21 or 43 correct predictions out of 52 cases. This is equal to 82.7%.

a. The equation for this model is:

b. The coefficient for DESIRE indicates that a one unit increase in the desire of the victim to have an arrest made increases the log of the odds of an arrest by .81477, when holding the age of the victim constant. Substantively, this means that controlling for the age of the victim, the log of the odds of an arrest increases when the victim wants an arrest made. The coefficient for AGE indicates that the log of the odds of an arrest increases by .15930 when age increases by one year, holding constant the expressed desire of the victim for an arrest. Substantively, this means that the log of the odds of an arrest increases when age increases, controlling for victim's desire for an arrest.

c. The estimated log of the odds of an arrest in a case where a victim wants an arrest made and the victim is 50 years old is:

From this log odds, we can now determine the predicted probability that an arrest will be made:

The predicted probability that an arrest will be made is .82.

The odds of an arrest in this case are .82/.18 = 4.55.

d. H₀: b_DESIRE = b_AGE = 0

H₁: b_DESIRE , b_AGE > 0

With a one-tailed alpha of .05 and 50 - 3 = 47 degrees of freedom, the critical value of t is 1.684. We will reject the null hypothesis if t_obt is greater than or equal to 1.684. The t ratio for DESIRE is .868. We cannot reject the null hypothesis that the population regression coefficient for DESIRE is 0. The t ratio for AGE is 3.889. We can reject the null hypothesis that the population regression coefficient for AGE is equal to zero.

e. The coefficient for DESIRE is not significant once AGE is controlled. The coefficient for AGE is significant even with DESIRE controlled. On the basis of this, we would conclude that the age of the victim has more of an effect on the log odds of an arrest than the desire of the victim that an arrest be made.

a. The probit equation is:

b. The coefficient for JOB is -1.0941. This tells us that a change of one unit on JOB decreases the unmeasured variable Z by 1.0941 standard deviation units. This is not easy to comprehend, but we can translate this into probabilities.

c. We can determine the predicted probability of rearrest for someone with no job after release from the probit equation. First we obtain the predicted Z score:

The Z score for a person with no job is .63. To find the predicted probability, go to the z table and find the area of the curve up to a z score of .63. We know that .50 of the curve lies to the left of the mean, and from the table we see that .2357 of the curve lies to the right of the mean to a z score of .63. The probability of a z score of .63, then, is the sum of these two values: .50 + .2357 = .7357. The predicted probability of someone being rearrested who has no job on release is .73.

We can follow the same procedure to find the predicted probability of rearrest for someone with a job on release. First we find the z score:

To find the predicted probability of this z score, go to the table. We know that .50 of the area of the curve lies to the left of the mean, and from the table we see that .1772 of the curve lies from the mean to a z score of -.46. The probability of a z score of -.46, then, is .50 - .1772 = .3228. The predicted probability of rearrest for someone with a job on release is .32.

d. H₀: b_Job = 0 H₁: b_Job < 0

With a one-tailed alpha of .01 and 50 - 2 = 48 degrees of freedom, our critical value of t from the t table is -2.423. Our decision will be to reject the null hypothesis if the t ratio for JOB is less than or equal to -2.423. Since the t_obt ratio is -2.820, we decide to reject the null hypothesis.

a. The probit equation is:

b. The coefficient for JOB is -1.0186. This tells us that a increase of one unit on JOB decreases the unmeasured variable Z by 1.0186 standard deviation units, controlling for the number of prior arrests. The coefficient for PRIORS is .18247. This tells us that a change of one arrest increases the unmeasured variable Z by .18247 standard deviation units, controlling for the existence of a job after release.

c. We can determine the predicted probability of rearrest for someone with no job after release and five prior arrests from the probit equation. First we obtain the predicted Z score:

The Z score for a person with no job is .65. To find the predicted probability, go to the z table and find the area of the curve up to a z score of .65. We know that .50 of the curve lies to the left of the mean, and from the table we see that .2422 of the curve lies to the right of the mean to a z score of .65. The probability of a z score of .65, then, is the sum of these two values: .50 + .2422 = .7422. The predicted probability of someone being rearrested who has no job on release and five prior arrests is .74

We can follow the same procedure to find the predicted probability of rearrest for someone with a job on release and 1 prior arrest. First we find the z score:

To find the predicted probability of this z score, go to the z table. We know that .50 of the area of the curve lies to the left of the mean, and from the table we see that .3621 of the curve lies from the mean to a z score of -1.09. The probability of a z score of -1.09, then, is .50 - .3621 = .1379. The predicted probability of rearrest for someone with a job on release and one prior arrest is approximately .14.

d. H₀: b_Job = b_PRIORS = 0

H₁: b_Job ¹ 0; b_PRIORS ¹ 0

With a two-tailed alpha of .05 and 50 - 3 = 47 degrees of freedom, our critical value of t from the t table is ±2.021. Our decision will be to reject the null hypothesis if the t ratio for JOB and PRIORS is either greater than or equal to 2.021 or less than or equal to -2.021. Since the t ratio for JOB is -2.371 and that for PRIORS is 3.031, we decide to reject the null hypothesis for both coefficients. We conclude that both having a job and prior arrests are related to the likelihood of arrest after release from a prison boot camp.

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