EXAM ANSWERS · Web viewEXAM ANSWERS 1. For each of the following statements, write which one is the independent and which is the dependent variable. a. You think there is a relationship

Instructor ResourcePaternoster, Essentials of Statistics for Criminology and Criminal Justice 4e

SAGE Publishing, 2018

EXAM ANSWERS

1. For each of the following statements, write which one is the independent and which is the dependent variable.

a. You think there is a relationship between the age at which someone was first arrested and the number of arrests they have after age 18.

Independent variable ______ age of first arrest______

Dependent variable ______ number of adult arrests______

b. You think there is a relationship between someone’s race/ethnicity and the length of their prison sentence.

Independent variable ______ race/ethnicity ______

Dependent variable ______ length of prison sentence ______

c. You think there is a relationship between the arrest record that the parent has and the number of delinquent offenses committed by the eldest child.

Independent variable ______ parent’s arrest record ______

Dependent variable ______ number of eldest child’s delinquent acts ______

d. You think there is a relationship between how quickly the police in your town respond to a 911 call and the social class of the neighborhood where the call came from.

Independent variable ______ social class of neighborhood ______

Dependent variable ______ speed of response to 911 call ______

2. What is the level of measurement for each of the following variables?

a. The number of drunk driving arrests for a person measured as______ times (the exact number of times)

Interval/ratio ______ level



b. The number of drunk driving arrests for a person measured as0–2 times ______3–5 times ______6–8 times ______9 or more times ______

Ordinal ______ level

c. The type of counsel at trial someone hadRetained counsel (paid for it themselves) ______Public defender ______No counsel at trial ______

nominal ______ level

3. We are examining sentence lengths in months for people convicted of aggravated assault in Maryland. We take a random sample of 50 inmates currently convicted and incarcerated for aggravated assault, their sentence lengths are reported below.

Stated Limits Real Limits f cf % c% m fm m2 fm2

1–6 .5 – 6.5 7 7 14 14 3.5 24.5 12.25 85.75

7– 12 6.5–12.5 12 19 24 38 9.5 114.0

90.25 1083.00

13– 18 12.5–18.5 7 26 14 52 15.5 108.5

240.25 1681.75

19– 24 18.5–24.5 9 35 18 70 21.5 193.5

462.25 4160.25

25– 30 24.5–30.5 5 40 10 80 27.5 137.5

756.25 3781.25

31– 36 30.5–36.5 7 47 14 94 33.5 234.5

1122.25

7855.75

37– 42 36.5–42.5 3 50 6 100

39.5 118.5

1560.25

4680.75

Σ 50 100% 931. 4243.7 23,328.5



0 5 0

a. In the chart above, fill out the columns for real limits, cumulative frequency, percentage, cumulative percentage, and midpoint.

b. What is the modal sentence length?

7–12 months . . . if they say bimodal, 7–12 months and 19.24 months, that is OK

c. Calculate and interpret the median.

d. Calculate and interpret the mean.

e. Calculate the standard deviation and variance.



f. What percentage of the offenders received a sentence length of 2 years or less?

[(9 + 7 + 12 + 7) / 50]*100

[35 / 50]*100 = 70%

g. How many, and what proportion of the offenders received a sentence of 19 months or more?

(9 + 5 + 7 + 3) = 24 24 / 50 = .48

4. The number of thefts and the median sentence lengths (in months) for theft in the state of Maryland for 5 years is reported below.

Year Number of Thefts Population Theft Rate per 10,000

1990

5,782 4,803,550 12.04

1995

7,898 5,670,627 13.93

2000

8,024 5,690,703 14.10



2005

4,153 5,500,875 7.55

2010

6,907 5,876,519 11.75

a. Why would it be inappropriate to directly compare the raw number of thefts for each year? The population sizes are different over time so there is variation in opportunities to commit theft. Taking a rate standardizes for population.

d. Calculate the rates of theft per 10,000 for each year. e.

c. What was the percentage change in the number of thefts from 2000 to 2005?

d. What was the percent change in the number of thefts from 1990 to 2010?

5. Below is the actual number of thefts caught on security cameras in a random sample of 27 department stores in D.C. over the past year. The empty columns are to aid you in your calculations for the remainder of the problem.

x f p cf cp fx x2 fx2 C%

1 3 .11 3 .11 3 1 3 11%

2 7 .26 10 .37 14 4 28 37%

3 2 .07 12 .44 6 9 18 44%

5 10 .37 22 .81 50 25 250 81%

6 3 .11 25 .92 18 36 108 92%

12 2 .07 27 .99 24 14 288 99%



(Not 1.0due torounding)

4 (Not 100%due torounding)

Σ 27 115 219

695

a. Fill out the labeled columns.

b. What is our level of measurement for thefts?

interval/ratio

c. What was the modal number of thefts?

5 thefts

d. Calculate and interpret the mean number of thefts caught on security cameras.

e. Calculate and interpret the median number of thefts. 5 thefts

f. What is the range of the number of thefts? 12 thefts – 1 theft = 11 thefts

g. Calculate the variance.



h. Calculate and interpret the standard deviation of the number of thefts caught on security cameras.

6. The following data represents the responses of 1,000 citizens of South Park to the question, “How satisfied are you with police services provided by the South Park Police Department?”

fVery Unsatisfied 128Unsatisfied 216Satisfied 485Very Satisfied 171

a. What is the level of measurement? ______ Ordinal______

b. Determine, calculate, and interpret the appropriate measure of central tendency. Mode is Satisfied

The highest probability outcome is to be satisfied with police services.

c. Determine, calculate, and interpret the appropriate measure of dispersion.

d. What percentage of the residents were either Satisfied or Very Satisfied with the police services of South Park Police Department?

e. Graph the percent data.



12.8

21.6

48.5

17.1

Attitude Toward South Park Police

Very Unsatisfied Unsatisfied Satisfied Very Satisfied

Or students could do a bar chart on the next page.

Very Unsatisfied

Unsatisfied

Satisfied

Very Satisfied

0 10 20 30 40 50 60

12.8

21.6

48.5

17.1


Sales

Or also this option:



Very Unsatisfied Unsatisfied Satisfied Very Satisfied0

10

20

30

40

50

60

12.8

21.6

48.5

17.1


Satisfaction

7. Social control theory argues that individuals with strong ties to conventional institutions will tend to commit fewer criminal acts. The following contingency table presents information regarding the strength of conventional ties and number of criminal acts for 89 individuals. Use this information to answer the following probability questions:

a. What is the probability that someone has strong conventional ties?44 / 89 = .494

b. What is the probability that someone has committed more than 5 criminal acts?22 / 89 = .247

c. What is the probability that someone had 5 or less criminal acts?67 / 89 = .753

d. What is the probability that someone had weak ties or 3 to 5 criminal acts? Are these mutually exclusive events? Explain.



P(weak ties or 3–5 acts) = (45 + 26 - 13) / 89 = .652

OR

P(weak ties) = 45 / 89 = .506 P(3–5 acts) = 26 / 89 = .292 P(weak ties and 3–5 acts) = 13 / 89 = .146P(weak ties or 3–5 acts) = (.506 + .292) - .146 = .652

These are not mutually exclusive events because an individual can be in both categories.

That is, someone can have weak ties and commit 3 to 5 criminal acts (this is why we need to

subtract the joint probability of weak ties and 3–5 acts = .146).

e. What is the conditional probability that someone had 6 or more criminal acts given that they had strong ties?

6 / 44 = .136

f. What is the probability that someone had weak ties and 0–2 criminal acts?16 / 89 = .180

e. Are conventional ties and criminal acts statistically independent or statistically dependent? Explain.

Test if P(A) = P(A | B) P(strong ties) = 44 / 89 = .494P(ST | 0–2) = 25 / 41= .610 P(ST | 3–5) = 13 / 26 = .500 P(ST | 6 or more) = 6 / 22 = .273Because P(A) = P(A | B), the two variables are statistically dependent

[You only need to calculate one of the conditionals to conclude if it’s independent or dependent.]

This means that the level of conventional ties (i.e., having strong ties) tells us something about the probability of criminal acts. As you can see, the probability of strong ties decreases as the number of criminal acts increases. Individuals with strong ties are less likely to commit higher numbers of criminal acts.

f. What is the probability that someone did not have strong ties and did not commit 6 or more

criminal acts?

P(Weak AND not 6 or more) = 16 / 89 + 13 / 89 = 29 / 89 = .326

If you want to do it the hard way, you can use the multiplication rule . . .Find P(A)*P(B|A) Not having strong ties means having weak ties. P(weak ties) = 45 / 89 = .506



Not committing 5 or more acts means committing either 0 to 2 or 3 to 5 acts. P(0–2 or 3–5 | weak ties) = 29 / 45 = .644The joint probability of weak ties and 0 to 5 acts = .506*.644 = .326

8. You are the colonel of the Maryland State Police, and you are concerned about drinking among state troopers. You take a random sample of 150 police offers and find that the median number of drinks per week is 2 per week and the mean number of drinks per week is 4.7 with a standard deviation of 8.

a. Is your sample skewed, how can you tell, and if so, in what direction? Yes. Positively skewed. The mean is greater than the median.

b. Construct a 90% confidence interval around the appropriate point estimate. Interpret your results.

90 %C . I .=X̄±z α(s√n )=4 .7±1 . 65(8/12 .25 )=4 .7±1 .08

¿3 .62≤μ≤5 .78

You can be 90% confident that the true population mean number of drinks among Maryland state troopers is between 3.62 and 5.78.

c. Construct a 93% confidence interval around your point estimate. Interpret your results. Why did the interval change?

With a 93% c.i., 7% of the distribution is split between the tails, leaving .035 in each tail. .5 - .035 = .465— look this up in the z table and you get a zα of 1.82.

93%C . I .=X̄±z α(s√n )=4 . 7±1 .82(8/12.25 )=4 .7±1 .19

¿3 .51≤μ≤5 . 89

You can be 93% confident that the true population mean number of drinks among Maryland state troopers is between 3.51 and 5.89. Increasing our confidence (from 90% to 93%) results in a wider range. If you want to be surer of including the true population mean in your confidence interval, you must include more values in your interval.

d. What would happen if we increased our confidence level to 99% (You do not need to recalculate the c.i.)? Increasing the confidence level 99% would result in an even wider interval.



e. What would happen to the size of your 90% confidence interval if you increased your sample size from 150 to 500? (You do not need to recalculate the c.i.). Increasing the sample size shrinks the standard error (by increasing the denominator of the fraction) resulting in a narrower interval. Also, as you include more people in your sample, it becomes more and more likely that the sample mean reflects the true population mean.

f. Recalculate a 90% confidence interval but assume that now you only have a sample size of 16 officers (your sample standard deviation is still 8). Interpret this new interval and explain why it is different from the 90% confidence interval you calculated in part b above? Be specific.

.

90 %C . I .=X̄±tα (s√n )=4 .7±1 .753 (8/4 )=4 . 7±3. 51

¿1 .19≤μ≤8 .21

You can be 90% confident that the true mean number of drinks among Maryland state troopers is between 1.19 and 8.21. This interval is considerably wider than the 90% confidence interval calculated in part b above because with a smaller sample size it is necessary to use the t distribution. The shape of the t distribution is a function of the sample size n (incorporated in the formula through the degrees of freedom (df)), and with small sample sizes (such as 16 police officers) the t distribution is flatter and more spread out than the z distribution resulting in a wider confidence interval at the same alpha level.

9. You want to estimate the proportion of adults in your state who think that existing gun laws are too weak and should be made stricter to include full background checks, a 30-day waiting period and other things. You take a sample of 250 people in your state and find that 183 of them are in favor of making gun laws stricter.

a. What is your point estimate of the proportion of adults in your state who favor stricter gun laws?

b. Build a 95% confidence interval around your point estimate and interpret your results.



The proportion of Maryland residents in favor of stricter gun laws is between 67.7% and 78.7% (or between .667 and .787) at a 95% confidence level.

c. Build a 99% confidence interval around your point estimate. What happens to the size of the interval and why? Can you explain this?

The proportion of the Maryland public that favors stricter gun laws is between 66% and 80.4% (or between .660 and .804) at a 95% level of confidence.The interval got larger with a 99% confidence interval because the z value increased—with greater confidence of a 99%c.i., the price to pay is less precision.

10. According to the Gallup Polling Organization, 46% of the U.S. population thinks that the criminal justice (CJ) system is too lenient. You take a sample of 200 University of Maryland students and find that 80 of them think that the criminal justice system is too lenient.

a. Construct a 95% confidence interval around your point estimate.

We can be 95% confident that the true population proportion of Maryland students who think the CJ system is too lenient is between .332 and .468.

b. In a hypothesis test, where the null hypothesis is that the percentage of Maryland students who think that the CJ system is too lenient, is no different than the national average; the alternative hypothesis is that Maryland students are different than the national average. Based on the results from your confidence interval above (i.e., with an alpha of .05), would you reject or fail to reject the null hypothesis? You DO NOT need to do the steps of the test . . . Fail to reject the null—.46 falls within the confidence interval:.332 < .46 < .468, so there is no evidence of a significant difference.



c. What would happen to the size of your interval if you increased your confidence from 95% to 99%? Increasing your level of confidence results in widening your interval.

d. What would happen to the size of your 95% confidence interval if you increased your sample size from 200 to 500? Increasing your sample size reduces your standard error and results in a narrower, more precise, confidence interval.

e. What would happen to the size of your 95% confidence interval if you decreased your sample size from 200 to 50? Conversely, shrinking your sample size increases your standard error, resulting in a wider and less precise confidence interval.

11. Your job as the research director in the Maryland Department of Youth Services is to advise the Director which policies to follow. In the past, you have heard that some institutions have had as many as 80 disturbances in a year. The average number of disturbances (riots, stabbings, fights, etc.) per year per institution, though, is 35, with a standard deviation of 7.0. The distribution is normal. Answer the following questions:

a. What is the z score for an institution that had 28 disturbances in 1 year? What does this z score indicate?

z= x−X̄s

=28−357

=−1

28 disturbances is exactly 1 standard deviation below the mean of 35

b. What is the z score for an institution that had 45 disturbances in 1 year?

z= x−X̄s

=45−357

=1 . 43

c. You want to give raises to the wardens in the prisons that have disturbances in the bottom 10% of the distribution, and fire the wardens in prisons in the top 5%. What number of disturbances will be the cutoff for each of these decisions?For the bottom 10%, we have .10 in the left tail of the distribution, and (.50 - .10) or .40 between the mean and our cutoff. We find the z score associated with a probability of .40, which is -1.28. Now we will solve for -1.28.



−1 .28= x−X̄s

= x−357

=−1 .28(7 )+35=26 .04

For the top 5%, we have .05 in the right tail of the distribution, and (.50 - .05) between the mean and our cutoff. We find the z score associated with a probability of .45, which is 1.65. Now we will solve for 1.65.

1 .65= x−X̄s

= x−357

=1. 65(7 )+35=46 .55

Wardens who have 26.04 disturbances or fewer should get a raise, and wardens with 46.55 disturbances or more should be fired.

d. What is the probability that an institution would have 40 or more disturbances?

z= x−X̄s

=40−357

=. 714

Find z score of .71 in z table = .2611. (.5 - .2611) = .239 or about a 23.9% probability of observing 40 or more disturbances

e. What is the probability that an institution would have 32 or fewer disturbances?

z= x−X̄s

=32−357

=−. 429

Find z score of .43 = .1664; (.5 - .1664) = .334 Or 33.4 % probability of 32 or fewer disturbances

f. Suppose the director wanted to identify the top 20% of institutions for reprimand. How many disturbances would qualify as the cutoff?Identify z score that corresponds to .20 in the right tail of the distribution.

.5 - .2 = .3 Look up .3 in body of z table.z = .84 Then solve the z equation for x.



~ 41 disturbances

g. Suppose the director wanted to identify the bottom 5% of institutions for praise. How many disturbances would qualify as the cutoff? See above explanation (be sure to use -1.65 not 1.65).

−1 .65= x−X̄s

= x−357

=−1 .65 (7 )+35=23. 45 disturbances

12. Suppose you take a sample of 81 offenders from Maryland state prisons and you find that the average number of tattoos is 4.4 with a standard deviation of 1.9. According to Tattoo Me magazine, the average number of tattoos for all incarcerated offenders is 2.9.

a. Test the null hypothesis that Maryland offenders have the same mean number of tattoos as other offenders in the general incarcerated population against the alternative that they have a significantly different mean number of tattoos. Use an alpha level of .05. Interpret your results.

Step 1: H0: µMD tattoos = 2.9H1: μMD tattoos ≠ 2.9

Step 2: z distribution two-tailed

Step 3: Alpha = .05

zcrit = ±1.96 Reject if zobt > 1.96 or zobt < -1.96[or reject of |zobt| > 1.96]

Step 4:

zobt=X̄−μ

( s√n )

=4 . 4−2.9

( 1 .9√81 )

= 1. 5. 211

=7 .11

Step 5: 7.11 > 1.96, so we reject the null hypothesis of no difference.Maryland prisoners have significantly more tattoos than the

general population of incarcerated individuals.



b. Suppose your sample size was only 40 prisoners. Can you still conclude that there is a significant difference in the mean number of tattoos? Conduct the appropriate hypothesis test to answer this question. Interpret your results.n = 40

Step 1: H0: μMD tattoos = 2.9H1: μMD tattoos ≠ 2.9

Step 2: t distribution two-tailed

Step 3: Alpha = .05

tcrit = ± 2.021 Reject if tobt > 2.021 or < -2.021[or reject of |tobt| > 2.021]

Step 4:

Step 5: 5.00 > 2.021, so we still reject the null hypothesis of no difference.Even with a sample of only 42 prisoners, we still conclude that Maryland prisoners have significantly more tattoos than the

general population of incarcerated individuals.

c. You take a new sample of 100 prisoners who have tattoos and find that 42% of their tattoos are located on their arms. Tattoo Me magazine reports that 38% of all individuals with tattoos have them on their arms. Is the proportion of tattoos on prisoners’ arms significantly greater than the proportion in the general public? Using an alpha of .01, conduct a hypothesis test to answer this question. Interpret your results.

Step 1: H0: pMD arm tattoos = .38H1: pMD arm tattoos > .38

Step 2: Np, Nq > 5 (100*.38 = 38; 100*.62 = 62)

z distribution one-tailed

Step 3: Alpha = .01

zcrit = 2.33 Reject if zobt > 2.33

Step 4:



zobt=p̂−p

(√ p(1−p )n )

=. 42−.38

(√( . 38)( . 62)100 )

=. 04.049

=. 82

Step 5: .82 < 2.33, so we fail to reject the null hypothesis. There is no evidence that Maryland prisoners are more likely to have tattoos on their arms than the general public.

13. You are interested in examining if the gender of the offender is related to whether or not armed robbers are arrested. You know that the overall probability of arrest for armed robbers is .7. You obtain a random sample of 10 female armed robbers and observe that only two of them were arrested. Assuming we can use a binomial to answer this question, calculate the probability of observing exactly two arrests out of 10 robberies in your data.

P(r )=( n!r ! (n−r )! ) pr qn−r

= P(2)=(10 !

2 !(10−2)! ) .72 .310−2

= (45)(.49)(.00006561)

= .0014

14. There has been a lot of talk about the “gang gene” in criminology and the national news lately. The gang gene is a particular gene called MAOA and males with it are supposed to be at high risk of being violent, including members of a gang. You take a random sample of 1,000 residents of a neighborhood in Compton, California, and test for the presence of both the MAOA gene and whether or not there is evidence that the person has been or is currently a gang member.

Here are the data you get as your Observed Frequencies:

Not in a Gang In a Gang TotalHas MAOA Gene 416 384 800No MAOA Gene 110 90 200

Total 526 474 1,000

Here are your Expected Frequencies under the assumption of independence:

Not in a Gang In a Gang TotalHas MAOA Gene 421 379 800No MAOA Gene 105 95 200

Total 526 474 1,000



a. What is the independent variable here and what is the dependent variable? Presence or absence of the MAOA gene is the independent variable. Being in a gang is the dependent variable.

b. What is the probability that someone is a member of a gang? p =(474/1000)= .474

c. What is the relative risk of gang membership for those without and with the MAOA gene? with MAOA p = (379/800) = .473without MAOA p = (95/200) =.475

d. Test the null hypothesis that the MAOA gene and gang membership are independent from each other against the alternative that they are dependent or that there is an association between them. Use an alpha of .05 and state each step of your hypothesis test.

Step 1: H0: χ2 = 0 H1: χ2 > 0

Step 2: Chi-square test of independence which has a chi-square distribution

Step 3: With 1 df and an alpha of .05, the critical value of chi-square is 3.841, so we will reject the null hypothesis if χ2 obtained ≥ 3.841.

Step 4:

fo fo2 fe fo2/fe416 173,056421 411.06384 147,456379 389.06110 12,100 105 115.24 90 8,100 95 85.26

Σ=1000.62

χ2 = 1000.62 – 1000 = .62

Step 5: Since chi-square obtained is less than our critical value of 3.841, we will fail to reject the null hypothesis and conclude that there is no relationship between having the MAOA gene and being a gang member. Those with the gene are no more likely to be gang members than those without the MAOA gene to be a gang member.



e. Calculate and interpret the strength of the relationship between the MAOA gene and gang membership? Should we conclude that all the hype about the gang gene is supported by the facts?

There is only a very weak relationship between having the MAOA gene and being a member of a gang.

15. Do first time offenders who are sent to prison earn less money from legal work after release than first time offenders who were given probation? To test this idea, you take a random sample of 125 first time offenders who were sent to prison after conviction. In the year after they were released, the average amount of money they earned in legal work was $12,800, with a standard deviation of $1,100. You take a second independent sample of 90 first time offenders who received probation after conviction. In the year after they were released, the average amount of money they earned in legal work was $14,600 with a standard deviation of $500. Test the null hypothesis that those sent to prison earn the same amount of money after release as those given probation against the alternative that they earn less. Use an alpha of .05 and state each step of your hypothesis test. What is your conclusion? In your hypothesis test, you cannot assume that the population standard deviations are equal (σ1 ≠ σ2). Assume that you have 120 degrees of freedom.

a. What is the independent variable? The independent variable is whether one was sent to prison or given probation.

b. What is the dependent variable?



The dependent variable is the amount of money earned in legal work (a continuous variable).

c. Test the null hypothesis that those sent to prison earn the same amount of money after release as those given probation against the alternative that they earn less. Use an alpha of .05 and state each step of your hypothesis test. What is your conclusion? In your hypothesis test, you cannot assume that the population standard deviations are equal (σ1 ≠ σ2). Assume that you have 120 degrees of freedom.

STEP 1: H0: μ wages after prison = μ wages after probation

H1: μ wages after prison < μ wages after probation

STEP 2: This is a two population mean, separate variance t test. The test statistic has a student’s t probability distribution.

STEP 3: There are 120 degrees of freedom. With 120 degrees of freedom, and a one-tailed alpha of .05, the critical value of t is -1.658. So our decision rule will be to reject the null hypothesis if the t we obtain is ≤ -1.658.

STEP 4:



STEP 5: Since the obtained t is less that the critical t (it lies in the critical region on the left side), I would reject the null hypothesis.

Convicted offenders who are sentenced to prison earn significantly less than those who are given probation.

16. It has always been claimed that being unemployed puts you at higher risk of committing a crime. To investigate this possibility you examine the relationship between being unemployed in the past year and being arrested. Here’s the data that you get:

OBSERVED FREQUENCIES

Empl

oyed

Unem

ployed

Total

No

Arrest

65 25 90

One

or More

Arres

ts

35 45 80

Total 100 70 170

a. What is the independent variable? Whether or not the person was employed or unemployed

b. What is the dependent variable? Whether the person had no arrests or one or more arrests

c. What is the unconditional risk of having one or more arrests? p(one or more arrests) = (80 / 170) = .47

d. What is the relative risk of having one or more arrests for those who were employed in the past year? p(one or more arrests | employed) = 35 / 100 = .35

e. What is the relative risk of having one or more arrests for those who were unemployed in the past year? p(one or more arrests | unemployed) = 45 / 70 = .64



f. Are the events of employment and arrest independent events? Explain. No, they are not independent events because the conditional probabilities are not equal to the unconditional probability. This tells us that knowing someone’s employment status tells us something about whether they have been arrested.

g. Test the null hypothesis that employment/unemployment in the past year and arrest are independent against the alternative hypothesis that they are dependent or are related to one another. Use an alpha of .01 and state each step of your hypothesis test. What is your decision and what do you conclude? To help you out, I have provided you with the table of expected frequencies below.

STEP 1: H0: employment status and arrest are unrelated. χ2 = 0.H1: employment status and arrest are related. χ2 > 0.

STEP 2: This is a chi-square test of independence with a chi-square test. The probability distribution is the chi-square.

STEP 3: With an alpha of .01 and 1 degree of freedom, our critical value of chi-square is 6.635. Our decision rule is to reject the null hypothesis if Χ2 obtained is ≥ 6.635.

STEP 4:

EXPECTED FREQUENCIES

Empl

oyed

Unem

ployed

Total

No

Arrest

53 37

One

or More

Arres

ts

47 33

Total 170



fo fe fo-fe (fo-fe)2 (fo-fe)2/fe

65 53 12 144 144/53= 2.72

25 37 -12 144 144/37=3.89

35 47 -12 144 144/47=3.06

45 33 12 144 144/33=4.36

Σ=14.03

STEP 5: Our obtained chi-square is 14.03, which is greater than the critical chi-square of 6.635; so I would reject the null hypothesis and assume that being unemployed is related to having one or more arrests.

h. Calculate an appropriate measure of association and describe the strength of the relationship between (if any) between unemployment and crime.

Yule’s Q (gamma for a 2x2 table) =

So, there is a moderate positive relationship between employment status and arrest. Those who are not employed have a higher risk of being arrested.

17. One of the complaints about immigration in the United States is that immigrants are more criminal than those who are natives. This, of course, is an empirical question, so you go to Arizona and take a random sample of 61 illegal immigrants from Mexico (from a group of immigrants detained by INS) and a second independent random sample of 61 persons in the same area who are not immigrants. For each person in both samples, you do a records check to see how many arrests they have had in the past (both in the U.S. and Mexico) and calculate the average number of arrests in each group. Here’s the data you obtained:



a. What is the independent variable? The independent variable is one’s immigration status—immigrant or non-immigrant.

b. What is the dependent variable? The dependent variable is the NUMBER of arrests (a continuous variable).

c. Test the null hypothesis that the mean number of arrests for illegal immigrants is no different from that for non-immigrants, against the alternative hypothesis that the average number of arrests for immigrants is greater than that for non-immigrants. Use an alpha of .01. You may assume that the population standard deviations are equal (σ1 = σ2). State each step of your hypothesis test. What is the conclusion that you would draw about crime between immigrants and non-immigrants?

STEP 1: H0: μ immigrant arrest = μ non-immigrant arrest

H1: μ immigrant arrest > μ non-immigrant arrest

STEP 2: This is a two-population mean, pooled-sample t test. The test statistic has a

Student’s t probability distribution.

STEP 3: There are 120 degrees of freedom (61 + 61 - 2) = 120, with 120 degrees of

freedom, and a one-tailed alpha of .01, the critical value of t is 2.358. So our decision

rule will be to reject the null hypothesis if the t we obtain is ≥ 2.358.

STEP 4:



STEP 5: Since t obtained is not greater than the critical t of 2.358, I would fail to reject the null hypothesis. There is no relationship between immigrant status and the number of arrests.

18. As the instructor, I am interested in whether or not students who fail this statistics course one semester do better on their exams the second time they take the class. To examine this, I record the exam scores for a sample of 10 students who took the class last semester and are repeating the class this semester. The exam scores for these 10 students are reported below for last semester and this semester. Alpha = .05.

a. What are the independent and dependent variables and their levels of measurement? IV First- or second-timer (nominal)DV Exam scores (interval or ratio)

b. Conduct the appropriate hypothesis test to determine whether or not student scores are significantly greater for students taking the class the second semester compared to their scores the first semester. State each step of the test and interpret your results.

Step 1: H0: μD = 0 or μsecond time = μfirst time

H1: μD > 0 or μsecond time > μfirst time

Step 2: t test for Dependent Samples



Step 3: Alpha = .05, one-tail test df = (n – 1) = (10 – 1) = 9 [n is the number of pairs of observations]

tcrit = 1.833 Reject the null if tobt > 1.833

Step 4:

StudentSemester 1

Semester 2 Difference Deviation Squared Dev.

x1 x2 xD = x2 – x1 ( xD−X̄ D) ( xD−X̄ D)2

A 65 78 13 7 49B 70 72 2 -4 16C 54 66 12 6 36D 66 57 -9 -15 225E 42 50 8 2 4F 69 82 13 7 49G 70 70 0 -6 36H 64 62 -2 -8 64I 39 55 16 10 100J 53 60 7 1 1

Σ ∑ x D=60 ∑ ( xD−X̄ D)2=580

X̄ D=∑ x D

n=60

10=6

sD=√ Σ (x D − X̄ D)2

n−1=√580

9=8 . 028

tobt=X̄ D

sD

√n =

68 .028

√10

=2 .363

Step 5: 2.363 > 1.833, so reject the null hypothesis. The exam scores for students taking CCJS200 the second time are significantly greater than their exam scores from the first semester they took the class.



19. You want to investigate whether or not children who have parents with criminal records are more likely to get in trouble with the law than children who have parents with no criminal record. You take a sample of 50 children who have at least one parent who has been convicted of a crime, and you take a separate sample of 100 children whose parents have never been convicted of a crime. You find that 44% of the sample of children with criminal parents have been in trouble with the law whereas 39% of the sample of children with non-criminal parents have been in trouble with the law.

a. What are the independent and dependent variables and their levels of measurement? IV Criminal or non-criminal parent (nominal)DV Proportion of kids in trouble w/law (ratio)

b. Conduct the appropriate hypothesis test to see if there is a significant difference in the proportion of children who have been in trouble with the law between the two groups. State the steps of your test and interpret your results. Alpha = .05.

Criminal Parents Non-Criminal Parents n1 = 50 n2= 100p̂1 = .44 p̂2 = .39

Step 1: H0: pcrim parents = pnon-crim parents

H1: pcrim parents ≠ pnon-crim parents

Step 2: z Statistic for Difference of Proportions [you don’t need to test p̂1 n1>5 , etc. . . . ]

Step 3: Alpha = .05, two-tail test zcrit = ±1.96 Reject the null if zobt > 1.96 or zobt < –1.96[or reject if |zobt| > 1.96]

Step 4:

p̂=n1 p̂1+n2 p̂2

n1+n2=

(50 )( . 44 )+(100 )( .39 )150

=. 407

q̂=(1− p̂)=1−. 407=.593

zobt=p̂1− p̂2

√ p̂ q̂√ n1+n2

n1 n2

=. 44−. 39

√( . 407 )( .593 )√1505000

=. 05( . 491 )( .173 )

=.589



Step 5: .589 < 1.96, so we fail to reject the null hypothesis. There is insufficient evidence to conclude that children with criminal parents are more likely to get in trouble with the law.

20. As the head of research for the Probation Department of Bikini Bottom, you want to know if reducing the caseload size of your probation officers will result in better care and therefore fewer new arrests for those on probation. You compare the number of arrests for probationers who are in one of four groups:

1. Low-case load (less than 25 probationers)2. Middle-low case load (between 25 and 49 probationers)3. Middle-high case load (between 50 and 74 probationers)4. Large case load (between 75 and 100 probationers)

You have eight probationers in each group.a. What is the independent variable and what are the values of that variable?The independent variable is the caseload size of a probation officer. The values are low, middle-low, middle-high, and large.

b. What is the dependent variable?The dependent variable is the number of arrests for each person on probation.

c. Test the null hypothesis that the mean number of offenses in the four groups is the same (µ1 = µ2 = µ3 = µ4) against the alternative that they are different (µ1 ≠ µ2 ≠ µ3 ≠ µ4), use an alpha of .05. State each step of your hypothesis test. What would you conclude and more importantly? The data are below:

Step 1: H0: µ1 = µ2 = µ3 = µ4 H1: µ1 ≠ µ2 ≠ µ3 ≠ µ4

Step 2: This is an analysis of variance F test and the F test has an F probability distribution.

Step 3: With an alpha of .05 and 3 between and 28 within degrees of freedom, the critical value of F is 2.95. My decision rule is to reject the null hypothesis if Fobt ≥ 2.95.

Step 4: Calculate the various sums of squares and degrees of freedom to get my obtained F.



Group (k)

# of Arrests (x)

1 2 1.375 3.84375

0.625 0.390625 -1.84375

3.39941 -2.46875

6.09473

1 1 1.375 3.84375

-0.375 0.140625 -2.84375

8.08691 -2.46875

6.09473

1 2 1.375 3.84375

0.625 0.390625 -1.84375

3.39941 -2.46875

6.09473

1 0 1.375 3.84375

-1.375 1.890625 -3.84375

14.77441 -2.46875

6.09473

1 2 1.375 3.84375

0.625 0.390625 -1.84375

3.39941 -2.46875

6.09473

1 1 1.375 3.84375

-0.375 0.140625 -2.84375

8.08691 -2.46875

6.09473

1 1 1.375 3.84375

-0.375 0.140625 -2.84375

8.08691 -2.46875

6.09473

1 2 1.375 3.84375

0.625 0.390625 -1.84375

3.39941 -2.46875

6.09473

2 3 1.625 3.84375

1.375 1.890625 -0.84375

0.71191 -2.21875

4.92285

2 2 1.625 3.84375

0.375 0.140625 -1.84375

3.39941 -2.21875

4.92285

2 3 1.625 3.84375

1.375 1.890625 -0.84375

0.71191 -2.21875

4.92285

2 1 1.625 3.84375

-0.625 0.390625 -2.84375

8.08691 -2.21875

4.92285

2 1 1.625 3.84375

-0.625 0.390625 -2.84375

8.08691 -2.21875

4.92285

2 0 1.625 3.84375

-1.625 2.640625 -3.84375

14.77441 -2.21875

4.92285

2 0 1.625 3.84375

-1.625 2.640625 -3.84375

14.77441 -2.21875

4.92285

2 3 1.625 3.84375

1.375 1.890625 -0.84375

0.71191 -2.21875

4.92285

3 4 6 3.84375

-2 4 0.15625 0.02441 2.15625 4.64941

3 5 6 3.84375

-1 1 1.15625 1.33691 2.15625 4.64941

3 6 6 3.84375

0 0 2.15625 4.64941 2.15625 4.64941

3 5 6 3.84375

-1 1 1.15625 1.33691 2.15625 4.64941

3 7 6 3.84375

1 1 3.15625 9.96191 2.15625 4.64941

3 7 6 3.84375

1 1 3.15625 9.96191 2.15625 4.64941

3 8 6 3.84375

2 4 4.15625 17.27441 2.15625 4.64941



3 6 6 3.84375

0 0 2.15625 4.64941 2.15625 4.64941

4 6 6.375 3.84375

-0.375 0.140625 2.15625 4.64941 2.53125 6.40723

4 5 6.375 3.84375

-1.375 1.890625 1.15625 1.33691 2.53125 6.40723

4 7 6.375 3.84375

0.625 0.390625 3.15625 9.96191 2.53125 6.40723

4 8 6.375 3.84375

1.625 2.640625 4.15625 17.27441 2.53125 6.40723

4 5 6.375 3.84375

-1.375 1.890625 1.15625 1.33691 2.53125 6.40723

4 6 6.375 3.84375

-0.375 0.140625 2.15625 4.64941 2.53125 6.40723

4 9 6.375 3.84375

2.625 6.890625 5.15625 26.58691 2.53125 6.40723

4 5 6.375 3.84375

-1.375 1.890625 1.15625 1.33691 2.53125 6.40723

Σ 123 Σ =43.63 Σ =220.22

Σ =176.59

SStotal = 220.22SSbetween = 176.59SSwithin = 43.63

df between = 3df within = 28df total = 31

Source SS df Variance F

Between 176.59 3 58.86 37.73

Within 43.63 28 1.56

Since my obtained F is greater than the critical F, my decision is to reject the null hypothesis. Some of the groups have different numbers of arrests. Probation case load size does make a difference.

d. If the critical difference from Tukey’s HSD test was 2.5, which pair of means would be different from each other? Based upon this, what is the optimal number of probationers you would assign to each probation officer based on your analysis?



Difference between the low and middle low group is |.25|; these groups are not different.

Difference between the low and middle high group is |4.625|; these groups are different.

Difference between the low and large group is |5.000|; these groups are different.

Difference between the middle low and middle high group is |4.375|; these groups are different.

Difference between the middle low and large group is |4.75|; these groups are different.

Difference between the middle high and large group is |.375|; these groups are not different.

It looks like the low and middle-low groups are different from the middle-high and large groups but not different from each other.

Once a caseload size reaches 50 or more, the number of arrests is higher but less than 49 cases does not make a difference. So I would have my probation officers have no more and no less than 49 probationers.

e. What is the strength of the relationship between caseload size and number of arrests? Explain and calculate the relevant statistic.

η2 =

There is a strong relationship; more specifically, about 80% of the variance in arrests is due to variation in the caseload size of the probation officer.



21. You think that a Head Start program for kids will reduce their risk of later getting involved in delinquency. To test this idea you take a sample of 95 homes where the child had been enrolled in Head Start and 125 homes in the same neighborhood where the child had not been enrolled in Head Start. You find that by age 18, 25 of the kids who had been in Head Start had at some time been arrested and sent to the juvenile court while 55 of the kids who had not been in Head Start had an arrest and juvenile court record by age 18. Test the null hypothesis that being in Head Start has no effect on the proportion of kids who end up being delinquent compared with kids not in Head Start, against the alternative that those in Head Start are less likely to be arrested and sent to court as juveniles. Use an alpha of .05 in your hypothesis test and state each step in your hypothesis test. What is your interpretation about the value of Head Start?

For your convenience, here is your data:

Head Start Not in Head Startn = 95 n = 125

Step 1: H0: PHeadStart = PNoHeadStart

H1: PHeadStart < PNoHeadStart

Step 2: Two-population difference of proportions test; z test with a standard normalprobability (z) distribution.

Step 3: One-tailed alpha of .05. zcrit = -1.64 (or -1.65); reject the null hypothesis if zobt ≤ -1.64 (or ≤ -1.65).

Step 4:



Step 5: Since zobt > our critical value of -1.64 or (-1.65), we fail to reject the null hypothesis. Participation in Head Start has no effect on the risk of being a delinquent as a juvenile.

22. You have a problem on your police force of a group of 10 identified officers who have had an unusually high number of citizen complaints filed against them. Rather than fire these officers, you send them to a training session dealing with how to interact better with citizens. You have a record of the number of citizen complaints against each of these 10 officers in the year both before and after the training session. Test the null hypothesis that the training session had no effect on the number of citizen complaints against the alternative that the session reduced the number of complaints the officer had filed against them. Use an alpha of .01 in your hypothesis test and explicitly state each step. What is your conclusion about the training session?

Here is your data:

Officer Number of Number ofComplaints Before Complaints After



1 7 32 9 63 14 114 8 95 12 56 7 47 10 48 9 69 11 1010 16 15

Here’s a hint: The standard deviation of the difference scores is 2.357.

a. Calculate and interpret the average difference score.

This tells us that on average there were three fewer citizen complaints per officer after taking the training session.

b. Test the null hypothesis that the training session had no effect on the number of citizen complaints against the alternative that the session reduced the number of complaints the officer had filed against them. Use an alpha of .01 in your hypothesis test and explicitly state each step. What is your conclusion about the training session?

Step 1: H0: µD = 0

H1: µD < 0

Step 2: Dependent samples t test; t distribution

Step 3: n - 1 or 9 df; one-tailed test, alpha of .01; tcrit = -2.821Decision rule is to reject H0 if tobt ≤ -2.821.

Step 4:

Officer Number of Number ofComplaints Before Complaints After D

1 7 3 -4 -112 9 6 -3 00



3 14 11 -3 004 8 9 1 4165 12 5 -7 -4166 7 3 -3 007 10 4 -6 -398 9 6 -3 009 11 10 -1 2410 16 15 -1 24

Σ=-30 Σ=50

Step 5: Since our decision rule was to reject H0 if tobt ≤ -2.821 and the obtained t is -4.021, we would reject the null hypothesis. Going to the training session did lower the number of citizen complaints afterward.

23. Does getting married lead criminal offenders to quit crime (desist)? To test this you take a sample of 10 men who as juveniles spent time in a reform school, and a matched group of 10 men (matched no age, social class, and race) who also spent time in a juvenile reform school but did not get married. You collect data on the number of offenses they committed between 17 and 40 and here are your results for the two groups:



Mean N

Std. Deviation

Std. Error Mean

Pair 1

Married

4.4000

10

1.64655

.52068

Not Married

2.6000

10

1.50555

.47610

Other helpful information:

Test the null hypothesis that being married as an adult is unrelated to desistance or quitting crime against the alternative that those who get married as adults commit less crime between the ages 18 and 32 than those who stay single. Use an alpha of .05 and state each step of your hypothesis test. What do you conclude?

Step 1: H0: µD = 0 H1: µD < 0

Step 2: This is a dependent samples t test, which has a t probability distribution.

Step 3: With an alpha of .05 and 9df with a one-tailed test, the critical value of t is -1.833, so we will reject the null hypothesis if tobt ≤ -1.833.

Step 4:

Step 5: Since our t obtained of -2.17 is less than the critical value of -1.833, we would reject the null hypothesis and conclude that those offenders who get married do commit fewer crimes than those who do not get married.

24. Do long prison sentences result in a lower crime rate? You divide 24 states where you have violent crime rate data into how severe their punishments are for violent felonies into three groups:



(1) Low severity (sentences on average for violent offenses are less than 5 years)(2) Medium severity (sentences on average for violent offenses are between 5 and 10 years)(3) High severity (sentences on average for violent offenses are more than 10 years)

Then, using the FBI’s Uniform Crime Reports, you get the violent crime rate for these 24 states.

Here are your data where each entry is the rate of violent crime in that state in 2014:

Low Severity Medium Severity High Severity

8.3 9.2 10.57.6 11.1 12.78.0 10.7 10.46.9 9.7 8.98.0 12.5 11.97.3 10.8 11.17.9 8.7 9.28.6 13.7 14.2

n1 = 8 n2 = 8 n3 = 8

Hint: Total sum of squares = 96.446

a. With this data, test the null hypothesis that µlow = µmedium = µhigh against the alternative that µlow = µmedium = µhigh, use an alpha of .01 and state each step of your hypothesis test. What do you conclude about sentence severity and violent crime rates?

Step 1: H0: µlow = µmedium = µhigh

H1: µlow ≠ µmedium ≠ µhigh

Step 2: Analysis of variance F test, which has an F probability distribution.

Step 3: alpha = .01, with 2 and 21 degrees of freedom, so Fcrit = 5.78, so our decision is to reject Ho if Fobt ≥ 5.78.

Step 4:

ANOVACrime Rate

Sum of Squares

df

Mean Square F



Between Groups

52.680 2 26.341

2.633Within

Groups43.776

21

2.085

Total96.446

23

Step 5: Since the obtained F of 12.633 is greater than the critical F of 3.96, we would reject the null hypothesis. Some of the means are different from some of the others.

b. Using a critical difference score of |2.32|, which pair of means are different, and what would you conclude now about the relationship between sentence severity and violent crime rates?

The low severity is different from both medium and high severity in that low-severity states have lower rates of violent crimes than both medium- and high-severity states. Medium- and high-severity states are not different from each other.

c. Calculate and interpret the strength of the relationship between sentence length severity and the rate of violent crime.



25. You work in the research section of the Maryland Department of Probation and Parole and your department ran an experiment in which four different types of probation treatment were tried on its probation clients:

a. probation onlyb. probation and finec. probation and short jail termd. probation and long jail term

After a year, you collected information on the number of new arrests each person had during their year of probation. The mean number of mean arrests for each group are reported below.

Your boss comes to you and asks, “Based on the results of the experiment, which treatment should we use in the future and why.” You have to answer this question based on the data provided:

SSTotal = 335.10SSBetween = 125.70

a. What is your independent variable here and what is your dependent variable? The independent variable is the type of probation practice and the dependent variable is the number of new arrests during the year of probation.

b. Conduct a hypothesis test that different probation treatments have no effect on the number of new arrests against the alternative that some treatment is better than some other. Use an alpha of .01.

Step 1: H0: μprobation = μprobation&fine = μprobation&shortjail = μprobation&longjail

H1: μprobation ≠ μprobation&fine ≠ μprobation&shortjail ≠ μprobation&longjail

Step 2: Analysis of variance F test, F probability distribution



Step 3: α = .01 3 and 36 degrees of freedom (really 3 and 40 df for the F table)Fcritical = 4.31, reject the null hypothesis if Fobt ≥ 4.31.

Step 4:

ANOVANew Arrests

Sum of Squares df

Mean Square F

Sig.

Between Groups

125.700

341.90

07.

203.001

Within Groups

209.400

36

5.817

Total 335.100

39

Step 5: Based on the above ANOVA, I would reject the null hypothesis and conclude that at least some of the group means are significantly different from each other.

c. Suppose your critical difference (CD) score was |3.58|, which pair of means are significantly different?

DifferenceProbation only vs. probation and fine 3.90Probation only vs. probation and short jail 3.90 All these are significantly different from each other.Probation only vs. probation & long jail 4.40

Probation & fine vs. probation & short jail .000Vs. probation & long jail .5000 None of these are

significantly different from each other.Probation & short jail vs. probation & long jail .500

Probation only is worse than the other three, but the other three are not significantly different from each other.

d. How much variation in new arrests are you explaining by knowing what treatment group someone was in?



so we are explaining about 37.5% of the variance.

e. What course of action will you tell your boss to take regarding the best probation treatment program and why?Do probation and anything else (fine, short- or long-term jail) and it will reduce new arrests better than probation alone. Why? Because the only difference in the observed means (which caused us to reject the null) was between probation only and everything else.

26. You work in a police department that is being sued by the American Civil Liberties Union. They claim that there are significantly more complaints against the police in poor neighborhoods than the rich neighborhoods you serve. You look at all complaints made against police officers in your department for the last year and classify them as either being made by citizens in poor or more affluent neighborhoods. You then calculate the mean number of citizen complaints in each of the two neighborhoods. Here are your data:

Test the null hypothesis that the two population means are equal against the alternative that they are different from each other. Use an alpha of .01, with 23 df, and you cannot assume that the two population standard deviations are equal (σ1 ≠ σ2). State each step of your hypothesis. Will you win or lose the law suit? Explain.

Step 1: H0: µpoor = µaffluent

H1: µpoor ≠ µaffluent

Step 2: Separate variance t test for the difference between two-population means. Probability distribution is the t distribution.

Step 3: Two-tailed alpha of .01 with 23 df = tcritical = ±2.807, so our decision is to reject the null hypothesis if tobt ≤ -2.807 or if tobt ≥ 2.807.



Independent Samples TestLevene's

Test for Equality of Variances t test for Equality of Means

FS

ig. t df

Sig. (two-tailed)

Mean

Difference

Std. Error

Difference

Number of Complaints Against the Police

Equal variances not assumed

-6.405

23.358

.000

-8.35000

1.26984

Step 4:

Group StatisticsWealth of

Community NM

eanStd.

DeviationStd.

Error Mean

Number of Complaints Against the Police

.00 20

2.1000

1.83246

.40975

Affluent 20

10.4500

5.37514

1.20192

Step 5:The tobt is -6.405, so we would reject the null hypothesis. There are significantly more complaints about the police in poorer than more affluent neighborhoods. You would lose the law suit.

27. You are interested in understanding the property crime rates in different neighborhoods in your state. You take a sample of 20 neighborhoods and collect information on three variables:

their property crime rate (y)whether the police use foot patrol (coded as “1”) or car patrol (coded as “0”) (x1)the divorce rate in the neighborhood (x2)

Here is your data:



Prop.Crime DivorceRate Patrol Rate

(y) (x1) (x2)23.40 .00 4.5044.00 .00 6.4036.70 .00 5.0043.00 .00 8.7018.00 .00 2.9045.00 .00 7.9038.60 .00 6.5044.20 .00 6.407.00 .00 1.7051.00 .00 9.2011.00 1.00 2.3013.20 1.00 9.2016.70 1.00 8.7019.50 1.00 3.0018.40 1.00 4.2021.00 1.00 7.9020.00 1.00 8.5014.40 1.00 4.3017.10 1.00 3.8011.00 1.00 2.50

And some supporting information:

ryx1 = -.69ryx2 = .55rx1x2 = -.096

a. You ran a bivariate regression with crime rate as the dependent variable and the type of neighborhood patrol practice as the independent variable, here are your results:

Coefficientsa

Model

Unstandardized Coefficients

Standardized

Coefficients

tSi

g.BStd.

Error Beta

1 (Constant)

35.090

3.274

10.717

.000



FootPatrol

-18.860

4.630

-.693-

4.073.001

a. Dependent Variable: Property Crime Rate

R2 = .48

Interpret the intercept, the slope coefficient, and whether your failed to reject or rejected the null hypothesis that βfootpatrol = 0 (with an alpha of .05). How much variance in property crime rates does type of police patrol explain?

The intercept shows that the predicted mean crime rate for neighborhoods with car patrol (coded as “0”) is 35.09.

The slope coefficient of -18.860 shows that the difference in the crime rate between neighborhoods with foot patrols and those with car patrols is 18.860. More specifically, the mean crime rate for neighborhoods with foot patrols is 18.860 less than for neighborhoods with car patrols.

With a t obtained of -4.073, (and an alpha of .001), I would reject the null hypothesis that βfootpatrol = 0, foot patrols reduce crime compared with car patrols.

Type of police patrol explains 48% of the variance in crime rates.

b. You ran a bivariate regression with crime rate as the dependent variable and the neighborhood divorce rate as the independent variable, here are your results:

Coefficientsa

Model


Standardized

Coefficients

tSi

g.BStd.

Error Beta

1 (Constant)

8.744

6.694

1.306

.208

Divorce Rate

2.978

1.079

.5452.

760.0

13


R2 = .30

Do a scattergram of the relationship between the divorce rate and crime. Interpret the intercept, the slope coefficient, and whether your failed to reject or rejected the null hypothesis that βdivorce rate = 0 (with an alpha of .05). How much variance in property crime rates does foot patrol explain?





The intercept shows that the property crime rate in a neighborhood is predicted to be 8.744 when the divorce rate is equal to 0. The slope coefficient shows that for each 1 unit change in the divorce rate the property crime rate increases by 2.978.With an alpha of .05, I would reject the null hypothesis that βdivorcerate = 0 because the probability of a t greater than or equal to 2.760 is .013, which is less than .05.The divorce rate explains 30% of the variance in property crime rates.

c. You then ran a multivariate regression with neighborhood crime rates as the dependent variable and both type of police patrol and neighborhood divorce rate as the independent variables. Here’s what you got:

Coefficientsa

Model


Standardized

Coefficients

tSi

g.BStd.

Error Beta

1 (Constant)

19.470

4.925

3.954

.001

Divorce Rate

2.638

.716 .4833.

687.0

02

Foot Patrol

-17.594

3.568

-.646-

4.931.0

00


R2 = .68

Interpret the intercept, the slope coefficient, and whether your failed to reject or rejected the null hypothesis that βdivorce rate = 0 and βfootpatrol = 0 (with an alpha of .05). Which variable has the stronger effect on property crime rates, and why do you say this? How much variance in property crime rates does foot patrol and divorce rates explain? Finally, was the neighborhood divorce rate a good second variable to add to your original regression equation that had only foot patrol in it? Why do you say this?

The intercept shows that when a neighborhood has both car patrol (x1 = 0) and a divorce rate of 0, the predicted rate of neighborhood property crime is 19.470.

The slope coefficient for the divorce rate shows that a 1 unit increase in the divorce rate increases neighborhood property crime by 2.638, holding constant the type of police patrol.

The slope coefficient for foot patrol shows that neighborhoods with foot patrol have an average of 17.594 fewer crimes, holding constant the divorce rate in the neighborhood.

We would reject or the null hypothesis that βdivorce rate = 0 and βfootpatrol = 0 with an alpha of .05, because the p value for each is less than .05.



The presence of foot patrol is more strongly related to neighborhood crime rates than the divorce rate. The standardized beta for divorce is greater |-.646| vs. |.482|, and the absolute value of the t test is greater for foot patrol.

The divorce rate was a good variable to add to the equation because it is related to the dependent variable (r = .55), but unrelated to the other independent variable (r = -.096).

28. Here is a graph that shows the relationship between murder rates and the percentage of the population that has a handgun for 20 cities in the United States.

The correlation here is r = .21

Here is a second graph of the same data where we added only one new data point.



The correlation now is r = .47.

How do you explain this?

The new data point is an outlier and it is making the relationship linear. The new data point is an influential observation.

New Data

Point

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EXAM ANSWERS · Web viewEXAM ANSWERS 1. For each of the following statements, write which one is the independent and which is the dependent variable. a. You think there is a relationship