Question

Stanton (1969) investigated the problem of paroling criminal offenders. He studied the behavior of all male criminals paroled from New York’s correctional institutions to original parole supervision during 1958 and 1959 (exclusive of those released to other warrants or to deportation). The parolees were observed for 3 years following their releases or until they exhibited some delinquent parole behavior. In a study involving a very large number of subjects, Stanton considered criminals convicted of crimes other than first- or second-degree murder. He found that approximately 60% of these parolees did not have any delinquent behavior during the 3 years following their releases.

During the same period, Stanton found that 56 of the 65 paroled murderers (first- or second-degree murderers who were also original parolees) in the study had no delinquent parole behavior. Let a success correspond to a male murderer on original parole who does not exhibit any delinquent parole behavior in the 3-year observation period. Note that we could question Assumptions A2 in this context; parolees convicted of first-degree murder may have a different success probability than parolees convicted of second-degree murder. Even the parolees in the first-degree (or second-degree) group may have different individual success probabilities. For pedagogical purposes, we proceed as if Assumption A2 is valid and denote the common success probability by p.

It is of interest to investigate whether murderers are better risks as original parolees than are criminals convicted of lesser crimes. This suggests testing H₀: p = .6 against the alternative p >.6. Perform this test using the large-sample approximation to procedure

b)suppose we only had 5 of the 10 paroled murderers had no delinquent parole

behavior, what will be your conclusion?

Answer 1

(a) The hypothesis being tested is:

H₀: p = 0.6

H_a: p > 0.6

p̂ = 56/65 = 0.86

The test statistic, z = (p̂ - p)/√p(1-p)/n

z =

The p-value for z = 4.30 is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that murderers are better risks as original parolees than are criminals convicted of lesser crimes.

(b) The hypothesis being tested is:

H₀: p = 0.6

H_a: p > 0.6

p̂ = 5/10 = 0.5

The test statistic, z = (p̂ - p)/√p(1-p)/n

z =

The p-value for z = -0.65 is 0.7407.

Since the p-value (0.7407) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we cannot conclude that murderers are better risks as original parolees than are criminals convicted of lesser crimes.