Ccj/4701 Final Exam
Name: Stacey-Ann Rubio PID: 3062746 Date: April 16, 2014
CCJ 4701 Final Exam Spring 2014
Florida International University is committed to maintaining academic integrity. As a result, students who engage in academic dishonesty are subject to a series of sanctions from failure on an assignment to failure for the course to a referral to the appropriate administrator. Your submission of the exam indicates that you are aware of the university’s policies and agree to follow them.
Directions: Please complete the exam using MS Word. All of your work must be shown and typed to receive full credit. Exams that are written by hand and scanned or photographed will not be graded. The exam is due to Blackboard Monday, April 14, 2014 …show more content…
by 6 pm for CCJ 4701-U01 or Wednesday, April 16, 2014 by 6 pm for CCJ 4701-U04. There will be no exceptions. Any students who do not submit the exam to Blackboard by the respective due date will have to come to the university to take an exam with me. Good luck!
1. A researcher wants to know if there was a difference between urban and rural jurisdictions in the hiring of female police officers last year. In urban jurisdictions (N = 77), the mean number of females hired was 9.65 (s = 4.17). Police departments in nonurban areas (N = 84) hired an average of 6.80 females (s = 4.16). Using an alpha of .05, test the null hypothesis that there is no difference between the means.
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between urban and rural jurisdictions in the hiring of female police officers last year. ii. The research hypothesis is that there is a difference between urban and rural jurisdictions in the hiring of female police officers last year.
b. What are the degrees of freedom?
Formula: df=N1+N2-2 Given: N1=77, N2=84 df= 77 + 84 -2 df= 159
c. What is the critical value?
i. Due to the fact that the degrees of freedom is 159 and the table goes from 100 to infinity, infinity is the closest value that can give the closest critical value being 1.96 in the two-tailed test category. Since the critical value decreases as the obtained value increases, the critical value has to be less than 1.96 which is the critical value of 100. (Page 372)
d. Do you reject or fail to reject the null hypothesis?
Given:
Urban (Group 1): Nonurban (Group 2): N1=77 N2=84 Mean (): 9.65 Mean (): 6.80 Standard Deviation (s1): 4.17 Standard Deviation (s2): 4.16 Variance (s12): 17.39 Variance (s22): 17.31
Formula:
Due to the critical value being 1.96 and the obtained value being 4.34, we must reject the null. This is because the obtained value is larger than the critical value.
e. What does this mean?
i. This means that there is an actual difference between the two groups. There is in fact a difference between urban and rural jurisdictions in the hiring of female police officers last year.
2. According to the UCR, Southern states that had the death penalty (N = 15) executed an average of 6.00 people per state in 2009 (s = 6.07). Non-Southern states that authorized capital punishment (N = 22) averaged 1.32 executions per state (s = 0.69). Using an alpha level of .05, test the null hypothesis of no difference.
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between the death penalty of Southern states and Non-Southern states. ii. The research hypothesis is that there is a difference between the death penalty of Southern states and Non-Southern states.
b. What are the degrees of freedom?
Formula: df=N1+N2-2 Given: N1=15, N2=22 df= 15 + 22 -2 df= 35
c. What is the critical value?
i. Due to the fact that the degree of freedom is 35, the critical value is 2.03. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 372)
d. Do you reject or fail to reject the null hypothesis?
Given:
Urban (Group 1): Nonurban (Group 2): N1=15 N2=22 Mean (): 6.00 Mean (): 1.32 Standard Deviation (s1): 6.07 Standard Deviation (s2): 0.69 Variance (s12): 36.84 Variance (s22): 0.48
Formula:
Due to the critical value being 2.03 and the obtained value being 3.61, we must reject the null. This is because the obtained value is larger than the critical value.
e. What does this mean?
i. This means that there is an actual difference between the two groups. There is in fact a difference between the death penalty of Southern states and Non-Southern states.
3. In 2004, the high court in the state of New York declared the state’s death penalty legislation unconstitutional. The table before contains UCR homicide data for New York’s six largest cities before and after the death penalty was abolished. Using alpha = .05, test the null hypothesis of no difference.
City
Homicides per 100,000 citizens in 2004
Homicides per 100,000 citizens in 2005
Amherst Town
0.00
0.90
Buffalo
17.86
19.77
New York
27.04
24.64
Rochester
16.70
24.91
Syracuse
11.09
13.26
Yonkers
7.58
4.56
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between the homicide rate before and after the death penalty was abolished. ii. The research hypothesis is that there is a difference between the homicide rate before and after the death penalty was abolished.
b. What are the degrees of freedom?
Formula: df=N-1 Given: N1=6 df= 6-1 df= 5
c. What is the critical value?
i. Due to the fact that the degree of freedom is 5, the critical value is 2.571. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 371)
d. Do you reject or fail to reject the null hypothesis?
Before (2004)
After (2005)
0.00
0.90
-0.90
0.81
17.86
19.77
-1.91
3.65
27.04
24.64
2.4
5.76
16.70
24.91
-8.21
67.40
11.09
13.26
-2.17
4.71
7.58
4.56
3.02
9.12
-7.77
91.45
Formula:
Due to the critical value being 2.571 and the obtained value being -0.79, we must accept the null. This is because the obtained value is smaller than the critical value.
e. What does this mean?
i. This means that there is no difference between the two groups. There is no difference between the homicide rates before and after the death penalty was abolished.
4. A professor is interested in whether or not bullying decreases after a school-based intervention. He assesses the students’ perceptions of bullying before and after showing them an anti-bullying film using an anti-bullying scale where higher numbers indicate a greater tolerance for bullying. Test the null hypothesis of no difference using alpha = .05.
Student
Before
After
Alan
7
5
Robert
4
4
Gloria
4
1
Norma
3
4
Michael
9
8
Megan
5
5
Lisa
8
7
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between bullying before and after the anti-bullying film is shown. ii. The research hypothesis is that there is a difference between bullying before and after the anti-bullying film is shown.
b. What are the degrees of freedom?
Formula: df=N-1 Given: N1=7 df= 7-1 df= 6
c. What is the critical value?
i. Due to the fact that the degree of freedom is 6, the critical value is 2.447. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 371)
d. Do you reject or fail to reject the null hypothesis?
Before
After
7
5
2
4
4
4
0
0
4
1
3
9
3
4
-1
1
9
8
1
1
5
5
0
0
8
7
1
1
6
16
Formula:
Due to the critical value being 2.447 and the obtained value being 1.69, we must accept the null.
This is because the obtained value is smaller than the critical value.
e. What does this mean?
i. This means that there is no difference between the two groups. There is no difference between bullying before and after the anti-bullying film is shown
5. The SPSS output below contains data on the number of prior arrests among three types of felony defendants: those facing at least 2 murder charges; those facing one murder charge; and those facing rape charges. Using the output, answer the following questions:
a. What is the average number of prior arrests for the sample?
i. 3.4286
b. What is the average number of prior arrests for multiple-murder defendants?
i. 5.5000
c. Do the number of prior arrests differ significantly as a result of the charge types? How do you know?
i. Yes, because the significance between groups is 0.021, which is less than the significant level of 0.05. Due to the significance between the groups being smaller, it is statistically significant.
d. Do multiple-murder defendants have a significantly different number of prior arrests compared to other types of defendants? How do you
know?
i. Yes, between rape arrests. This occurs because the significance between the two groups is 0.027, which is less than the significance level of 0.05. Due to that, there is a significant difference.
e. What are the differences between multiple murder defendants and other defendants?
i. 3.33167
ANOVA
Arrests
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
60.790
2
30.395
4.521
.021
Within Groups
168.067
25
6.723
Total
228.857
27
Multiple Comparisons
Dependent Variable: Arrests
Bonferroni
(I) Group
(J) Group
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Multiple Murder
Single Murder
1.40000
1.33892
.917
-2.0356
4.8356
Rape
3.66667*
1.29641
.027
.3401
6.9932
Single Murder
Multiple Murder
-1.40000
1.33892
.917
-4.8356
2.0356
Rape
2.26667
1.11018
.156
-.5820
5.1154
Rape
Multiple Murder
-3.66667*
1.29641
.027
-6.9932
-.3401
Single Murder
-2.26667
1.11018
.156
-5.1154
.5820
*. The mean difference is significant at the 0.05 level.
ANOVA
Arrests
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
104.734
2
52.367
.238
.790
Within Groups
4622.600
21
220.124
Total
4727.333
23
Multiple Comparisons
Dependent Variable: Arrests Bonferroni
(I) Group
(J) Group
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Public Defender
Assigned Counsel
-2.30952
8.25431
1.000
-23.7819
19.1628
Private Attorney
2.76623
7.17339
1.000
-15.8943
21.4267
Assigned Counsel
Public Defender
2.30952
8.25431
1.000
-19.1628
23.7819
Private Attorney
5.07576
7.52984
1.000
-14.5120
24.6635
Private Attorney
Public Defender
-2.76623
7.17339
1.000
-21.4267
15.8943
Assigned Counsel
-5.07576
7.52984
1.000
-24.6635
14.5120
6. A researcher is interested in the public’s perception that stricter handgun regulations would help curb the murder rate in the U.S. He admits that some people think that tougher gun laws will not reduce murder because people who are motivated to kill but who cannot obtain a handgun will just find a different weapon. This idea, called the substitution effect, would result in a negative correlation between handgun and knife murders. Using data from 10 states, the researcher calculates the murder rates per 100,000 state residents. Using the output below, answer the following questions:
a. On average, how many murders using knives occur per 100,000 residents?
i. 1.8510
b. Is the correlation between the two variables statistically significant? How do you know?
i. Yes it is because the significance of the correlation is 0.026, which is smaller than the significance level of 0.05 making it statistically significant.
c. Is the correlation positive or negative? What does this mean?
i. The correlation is positive because it has a direct relationship. This means both are increasing shown by the fact that it is statistically significant.
Descriptive Statistics
Mean
Std. Deviation
N
Handgun
1.8510
.68663
10
Knife
.5610
.12749
10
Correlations
handgun
Knife
Handgun
Pearson Correlation
1
.695*
Sig. (2-tailed)
.026
N
10
10
Knife
Pearson Correlation
.695*
1
Sig. (2-tailed)
.026
N
10
10
*. Correlation is significant at the 0.05 level (2-tailed).
7. A researcher is interested in whether it takes defendants who face multiple charges longer to get through the adjudication process compared to defendants who only have one charge? Using alpha = .01, answer the following questions:
a. What is the mean number of charges faced by the defendants?
i. 2.5000
b. How many individuals are in the sample?
i. 10
c. Is the correlation statistically significant at the .01 level? How do you know?
i. No it is not because the significance of the correlation is 0.035, which is larger than the significance level of 0.01 making it not statistically significant.
d. What is the strength of the correlation?
i. The strength of the correlation is a strong relationship.
e. How much of the variance in the months until release is explained by the variance in the number of charges?
i.
Descriptive Statistics
Mean
Std. Deviation
N
Numberofcharges
2.5000
1.35401
10
Monthsuntilrelease
1.9240
2.44471
10
Correlations
numberofcharges monthsuntilrelease Numberofcharges
Pearson Correlation
1
.667*
Sig. (2-tailed)
.035
N
10
10
Monthsuntilrelease
Pearson Correlation
.667*
1
Sig. (2-tailed)
.035
N
10
10
*. Correlation is significant at the 0.05 level (2-tailed).
CCJ 4701 Final Exam Spring 2014
Florida International University is committed to maintaining academic integrity. As a result, students who engage in academic dishonesty are subject to a series of sanctions from failure on an assignment to failure for the course to a referral to the appropriate administrator. Your submission of the exam indicates that you are aware of the university’s policies and agree to follow them.
Directions: Please complete the exam using MS Word. All of your work must be shown and typed to receive full credit. Exams that are written by hand and scanned or photographed will not be graded. The exam is due to Blackboard Monday, April 14, 2014 …show more content…
by 6 pm for CCJ 4701-U01 or Wednesday, April 16, 2014 by 6 pm for CCJ 4701-U04. There will be no exceptions. Any students who do not submit the exam to Blackboard by the respective due date will have to come to the university to take an exam with me. Good luck!
1. A researcher wants to know if there was a difference between urban and rural jurisdictions in the hiring of female police officers last year. In urban jurisdictions (N = 77), the mean number of females hired was 9.65 (s = 4.17). Police departments in nonurban areas (N = 84) hired an average of 6.80 females (s = 4.16). Using an alpha of .05, test the null hypothesis that there is no difference between the means.
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between urban and rural jurisdictions in the hiring of female police officers last year. ii. The research hypothesis is that there is a difference between urban and rural jurisdictions in the hiring of female police officers last year.
b. What are the degrees of freedom?
Formula: df=N1+N2-2 Given: N1=77, N2=84 df= 77 + 84 -2 df= 159
c. What is the critical value?
i. Due to the fact that the degrees of freedom is 159 and the table goes from 100 to infinity, infinity is the closest value that can give the closest critical value being 1.96 in the two-tailed test category. Since the critical value decreases as the obtained value increases, the critical value has to be less than 1.96 which is the critical value of 100. (Page 372)
d. Do you reject or fail to reject the null hypothesis?
Given:
Urban (Group 1): Nonurban (Group 2): N1=77 N2=84 Mean (): 9.65 Mean (): 6.80 Standard Deviation (s1): 4.17 Standard Deviation (s2): 4.16 Variance (s12): 17.39 Variance (s22): 17.31
Formula:
Due to the critical value being 1.96 and the obtained value being 4.34, we must reject the null. This is because the obtained value is larger than the critical value.
e. What does this mean?
i. This means that there is an actual difference between the two groups. There is in fact a difference between urban and rural jurisdictions in the hiring of female police officers last year.
2. According to the UCR, Southern states that had the death penalty (N = 15) executed an average of 6.00 people per state in 2009 (s = 6.07). Non-Southern states that authorized capital punishment (N = 22) averaged 1.32 executions per state (s = 0.69). Using an alpha level of .05, test the null hypothesis of no difference.
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between the death penalty of Southern states and Non-Southern states. ii. The research hypothesis is that there is a difference between the death penalty of Southern states and Non-Southern states.
b. What are the degrees of freedom?
Formula: df=N1+N2-2 Given: N1=15, N2=22 df= 15 + 22 -2 df= 35
c. What is the critical value?
i. Due to the fact that the degree of freedom is 35, the critical value is 2.03. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 372)
d. Do you reject or fail to reject the null hypothesis?
Given:
Urban (Group 1): Nonurban (Group 2): N1=15 N2=22 Mean (): 6.00 Mean (): 1.32 Standard Deviation (s1): 6.07 Standard Deviation (s2): 0.69 Variance (s12): 36.84 Variance (s22): 0.48
Formula:
Due to the critical value being 2.03 and the obtained value being 3.61, we must reject the null. This is because the obtained value is larger than the critical value.
e. What does this mean?
i. This means that there is an actual difference between the two groups. There is in fact a difference between the death penalty of Southern states and Non-Southern states.
3. In 2004, the high court in the state of New York declared the state’s death penalty legislation unconstitutional. The table before contains UCR homicide data for New York’s six largest cities before and after the death penalty was abolished. Using alpha = .05, test the null hypothesis of no difference.
City
Homicides per 100,000 citizens in 2004
Homicides per 100,000 citizens in 2005
Amherst Town
0.00
0.90
Buffalo
17.86
19.77
New York
27.04
24.64
Rochester
16.70
24.91
Syracuse
11.09
13.26
Yonkers
7.58
4.56
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between the homicide rate before and after the death penalty was abolished. ii. The research hypothesis is that there is a difference between the homicide rate before and after the death penalty was abolished.
b. What are the degrees of freedom?
Formula: df=N-1 Given: N1=6 df= 6-1 df= 5
c. What is the critical value?
i. Due to the fact that the degree of freedom is 5, the critical value is 2.571. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 371)
d. Do you reject or fail to reject the null hypothesis?
Before (2004)
After (2005)
0.00
0.90
-0.90
0.81
17.86
19.77
-1.91
3.65
27.04
24.64
2.4
5.76
16.70
24.91
-8.21
67.40
11.09
13.26
-2.17
4.71
7.58
4.56
3.02
9.12
-7.77
91.45
Formula:
Due to the critical value being 2.571 and the obtained value being -0.79, we must accept the null. This is because the obtained value is smaller than the critical value.
e. What does this mean?
i. This means that there is no difference between the two groups. There is no difference between the homicide rates before and after the death penalty was abolished.
4. A professor is interested in whether or not bullying decreases after a school-based intervention. He assesses the students’ perceptions of bullying before and after showing them an anti-bullying film using an anti-bullying scale where higher numbers indicate a greater tolerance for bullying. Test the null hypothesis of no difference using alpha = .05.
Student
Before
After
Alan
7
5
Robert
4
4
Gloria
4
1
Norma
3
4
Michael
9
8
Megan
5
5
Lisa
8
7
a. What is the null hypothesis? What is the research hypothesis?
i. The null hypothesis is that there is no difference between bullying before and after the anti-bullying film is shown. ii. The research hypothesis is that there is a difference between bullying before and after the anti-bullying film is shown.
b. What are the degrees of freedom?
Formula: df=N-1 Given: N1=7 df= 7-1 df= 6
c. What is the critical value?
i. Due to the fact that the degree of freedom is 6, the critical value is 2.447. This is the case because it is a two-tailed test and after referring to the table in the back, one can find out so. (Page 371)
d. Do you reject or fail to reject the null hypothesis?
Before
After
7
5
2
4
4
4
0
0
4
1
3
9
3
4
-1
1
9
8
1
1
5
5
0
0
8
7
1
1
6
16
Formula:
Due to the critical value being 2.447 and the obtained value being 1.69, we must accept the null.
This is because the obtained value is smaller than the critical value.
e. What does this mean?
i. This means that there is no difference between the two groups. There is no difference between bullying before and after the anti-bullying film is shown
5. The SPSS output below contains data on the number of prior arrests among three types of felony defendants: those facing at least 2 murder charges; those facing one murder charge; and those facing rape charges. Using the output, answer the following questions:
a. What is the average number of prior arrests for the sample?
i. 3.4286
b. What is the average number of prior arrests for multiple-murder defendants?
i. 5.5000
c. Do the number of prior arrests differ significantly as a result of the charge types? How do you know?
i. Yes, because the significance between groups is 0.021, which is less than the significant level of 0.05. Due to the significance between the groups being smaller, it is statistically significant.
d. Do multiple-murder defendants have a significantly different number of prior arrests compared to other types of defendants? How do you
know?
i. Yes, between rape arrests. This occurs because the significance between the two groups is 0.027, which is less than the significance level of 0.05. Due to that, there is a significant difference.
e. What are the differences between multiple murder defendants and other defendants?
i. 3.33167
ANOVA
Arrests
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
60.790
2
30.395
4.521
.021
Within Groups
168.067
25
6.723
Total
228.857
27
Multiple Comparisons
Dependent Variable: Arrests
Bonferroni
(I) Group
(J) Group
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Multiple Murder
Single Murder
1.40000
1.33892
.917
-2.0356
4.8356
Rape
3.66667*
1.29641
.027
.3401
6.9932
Single Murder
Multiple Murder
-1.40000
1.33892
.917
-4.8356
2.0356
Rape
2.26667
1.11018
.156
-.5820
5.1154
Rape
Multiple Murder
-3.66667*
1.29641
.027
-6.9932
-.3401
Single Murder
-2.26667
1.11018
.156
-5.1154
.5820
*. The mean difference is significant at the 0.05 level.
ANOVA
Arrests
Sum of Squares
Df
Mean Square
F
Sig.
Between Groups
104.734
2
52.367
.238
.790
Within Groups
4622.600
21
220.124
Total
4727.333
23
Multiple Comparisons
Dependent Variable: Arrests Bonferroni
(I) Group
(J) Group
Mean Difference (I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
Public Defender
Assigned Counsel
-2.30952
8.25431
1.000
-23.7819
19.1628
Private Attorney
2.76623
7.17339
1.000
-15.8943
21.4267
Assigned Counsel
Public Defender
2.30952
8.25431
1.000
-19.1628
23.7819
Private Attorney
5.07576
7.52984
1.000
-14.5120
24.6635
Private Attorney
Public Defender
-2.76623
7.17339
1.000
-21.4267
15.8943
Assigned Counsel
-5.07576
7.52984
1.000
-24.6635
14.5120
6. A researcher is interested in the public’s perception that stricter handgun regulations would help curb the murder rate in the U.S. He admits that some people think that tougher gun laws will not reduce murder because people who are motivated to kill but who cannot obtain a handgun will just find a different weapon. This idea, called the substitution effect, would result in a negative correlation between handgun and knife murders. Using data from 10 states, the researcher calculates the murder rates per 100,000 state residents. Using the output below, answer the following questions:
a. On average, how many murders using knives occur per 100,000 residents?
i. 1.8510
b. Is the correlation between the two variables statistically significant? How do you know?
i. Yes it is because the significance of the correlation is 0.026, which is smaller than the significance level of 0.05 making it statistically significant.
c. Is the correlation positive or negative? What does this mean?
i. The correlation is positive because it has a direct relationship. This means both are increasing shown by the fact that it is statistically significant.
Descriptive Statistics
Mean
Std. Deviation
N
Handgun
1.8510
.68663
10
Knife
.5610
.12749
10
Correlations
handgun
Knife
Handgun
Pearson Correlation
1
.695*
Sig. (2-tailed)
.026
N
10
10
Knife
Pearson Correlation
.695*
1
Sig. (2-tailed)
.026
N
10
10
*. Correlation is significant at the 0.05 level (2-tailed).
7. A researcher is interested in whether it takes defendants who face multiple charges longer to get through the adjudication process compared to defendants who only have one charge? Using alpha = .01, answer the following questions:
a. What is the mean number of charges faced by the defendants?
i. 2.5000
b. How many individuals are in the sample?
i. 10
c. Is the correlation statistically significant at the .01 level? How do you know?
i. No it is not because the significance of the correlation is 0.035, which is larger than the significance level of 0.01 making it not statistically significant.
d. What is the strength of the correlation?
i. The strength of the correlation is a strong relationship.
e. How much of the variance in the months until release is explained by the variance in the number of charges?
i.
Descriptive Statistics
Mean
Std. Deviation
N
Numberofcharges
2.5000
1.35401
10
Monthsuntilrelease
1.9240
2.44471
10
Correlations
numberofcharges monthsuntilrelease Numberofcharges
Pearson Correlation
1
.667*
Sig. (2-tailed)
.035
N
10
10
Monthsuntilrelease
Pearson Correlation
.667*
1
Sig. (2-tailed)
.035
N
10
10
*. Correlation is significant at the 0.05 level (2-tailed).