Variables
Using SPSS to Create A Chart or Graph
Variables | Graph/Chart |
Year | Bar Chart |
p_unemployment | Histogram |
p_cashassistance | Histogram |
p_fstamps | Histogram |
PERBLACK | Histogram |
PERHIS | Histogram |
UnderDeg | Pie Chart |
GradDeg | Pie Chart |
Veterans | Histogram |
p_foreign_born | Histogram |
2.
3. The table below gives the variable, the suitable representation and interpretation of the representations
Variables | Graph | Interpretation |
Year | Bar Chart | The bar plot shows that majority of the data was collected in the year 2013 followed by year 2012 and 2014 |
p_unemployment | Histogram | The Histogram shows that the percentage of the population unemployed for the period of interest is normally distributed |
p_cashassistance | Histogram | The Histogram of the percentage of population that uses cash assistance program also followed a normal distribution |
p_fstamps | Histogram | The Histogram of the percentage of the population that used food stamp programs is rightly skewed which reveals that the mean is greater than the median. |
PERBLACK | Histogram | The Histogram of the percentage of the population that identifies as black or African American is rightly skewed which reveals that the mean is greater than the median. |
PERHIS | Histogram | The Histogram of the percentage of the population that identifies as Hispanic is also rightly skewed which reveals that the mean is greater than the median. |
UnderDeg | Pie Chart | The pie plot reveals that the majority of the sampled location has 10-19.9 percent of the population with a bachelor’s degree followed by location with 20-20.9 percent of the population with a bachelor’s degree. |
GradDeg | Pie Chart | The pie plot for GradDeg reveals that the majority of the sampled location has 0-9.9 percent of the population with a graduate degree followed by location with 10-19.9 percent of the population with a graduate degree. The least of the location have 20 percent or more of the population with graduate degree |
Veterans | Histogram | The Histogram of the number of veterans in the community is rightly skewed which reveals that the mean is greater than the median. |
p_foreign_born | Histogram | The Histogram of the percentage of foreign-born members of the community is rightly skewed which reveals that the mean is greater than the median. |
Identifying an Appropriate Chart or Graph
Variables | Appropriate Visualization |
SOUTH | Bar Plot |
FE.Cleaned | Line plot |
vc_rate | Histogram |
FTSWORN | Histogram |
FTCIV | Histogram |
ISSU_SPEC | Line plot |
CP_TrnRec | Bar plot |
CP_TrnIns | Bar plot |
CP_PTNR | Pie plot |
Swat | Pie plot |
3. The table below gives the interpretation of the various variables visualization
Variables | Visualization | Interpretation |
SOUTH | Bar Plot | The bar plot shows that the majority of the sampled community are in the Southern US |
FE.Cleaned | Line plot | The line plot gives the visualization the number of officer-involved fatal encounters in the community. The visualization revealed that majority of the community have no officer-involved in fatal encounter. |
vc_rate | Histogram | The Histogram is rightly skewed which shows that the mean is greater than the median for the violent crime rate |
FTSWORN | Histogram | The Histogram is rightly skewed which shows that the mean is greater than the median for number of full-time sworn officer |
FTCIV | Histogram | The FTCIV also have a rightly skewed histogram with mean greater than the median. |
ISSU_SPEC | Line plot | The line plot revealed that majority of the community has 3 numbers of specialized units in each department. |
CP_TrnRec | Bar plot | The bar plot revealed that majority of the sample community requires the recruits to undergo at least 8 hours of community policing training |
CP_TrnIns | Bar plot | The bar plot revealed that majority of the sample community agency does not require currently employed officers to undergo at least 8 hours of community policing training |
CP_PTNR | Pie plot | The pie plot shows that majority of the sampled community agency does participate in community problem-solving partnerships |
Swat | Pie plot | The pie plot shows that the majority of the sampled community agency does not have their own SWAT team |
Creating New Variables
Statistics | ||||||
ORDBLACK | ORDCASH | FE2 | LRGAGENCY | ORDFOR | ||
N | Valid | 463 | 463 | 463 | 463 | 463 |
Missing | 0 | 0 | 0 | 0 | 0 | |
Median | 1.0000 | 2.0000 | .0000 | .0000 | 2.0000 | |
Mode | 1.00 | 2.00 | .00 | .00 | 1.00 |
The table above give the median and mode value for each of the variables.
Statistics | ||||||
ORDBLACK | ORDCASH | FE2 | LRGAGENCY | ORDFOR | ||
N | Valid | 463 | 463 | 463 | 463 | 463 |
Missing | 0 | 0 | 0 | 0 | 0 | |
Range | 2.00 | 4.00 | 1.00 | 1.00 | 3.00 | |
Percentiles | 25 | 1.0000 | 1.0000 | .0000 | .0000 | 1.0000 |
50 | 1.0000 | 2.0000 | .0000 | .0000 | 2.0000 | |
75 | 1.0000 | 3.0000 | 1.0000 | .0000 | 2.0000 |
Interpreting Confidence Intervals
Variables | 95% Confidence Interval | |
Lower | Upper | |
vc_rate | 441.6012 | 521.4842 |
p_fstamps | 13.969 | 15.545 |
p_unemployment | 6.424 | 6.837 |
FTCIV | 59.64 | 90.83 |
FTSWORN | 224.19 | 363.01 |
Variables | 90% Confidence Interval | |
Lower | Upper | |
vc_rate | 448.0433 | 515.0420 |
p_fstamps | 14.096 | 15.418 |
p_unemployment | 6.457 | 6.803 |
FTCIV | 62.16 | 88.31 |
FTSWORN | 235.39 | 351.82 |
Hypothesis Testing
Chi-Square Tests |
|||||
Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |
Pearson Chi-Square | 57.900^{a} | 1 | .000 | ||
Continuity Correction^{b} | 55.821 | 1 | .000 | ||
Likelihood Ratio | 54.983 | 1 | .000 | ||
Fisher's Exact Test | .000 | .000 | |||
Linear-by-Linear Association | 57.775 | 1 | .000 | ||
N of Valid Cases | 463 | ||||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 23.40. | |||||
b. Computed only for a 2x2 table |
Chi-Square Tests | |||
Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 8.635^{a} | 2 | .013 |
Likelihood Ratio | 8.335 | 2 | .015 |
Linear-by-Linear Association | .735 | 1 | .391 |
N of Valid Cases | 463 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 8.48. |
Chi-Square Tests | |||
Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 1.400^{a} | 3 | .706 |
Likelihood Ratio | 1.390 | 3 | .708 |
Linear-by-Linear Association | .448 | 1 | .503 |
N of Valid Cases | 451 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 11.67. |
X2(3, 451) = 1.40, p=0.706
Step 4: Conclusion
The p-value is greater than 0.05, we do not reject the null hypothesis and conclude that departments in communities with larger proportions of foreign born individuals are not likely to have SWAT teams.
Question 8
Are police departments in communities with larger proportions of citizens with undergraduate degrees more likely to engage in community partnerships? Using UnderDeg and CP_PTNR assess this research question using the hypothesis testing process. You MUST specify each step of the hypothesis testing process.
Question 8 Solution
Step 1: Hypothesis
Null Hypothesis: Police departments in communities with larger proportions of citizens with undergraduate degrees is independence of the engagement in community partnerships
Alternative Hypothesis: Police departments in communities with larger proportions of citizens with undergraduate degrees is not independence of the engagement in community partnerships
Step 2: Rejection Rule
Reject the null hypothesis if the p-value is less than 0.05
Step 3: Test Statistic
The chi square test of independence is obtain and the result display below
Chi-Square Tests | |||
Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 4.458^{a} | 3 | .216 |
Likelihood Ratio | 4.470 | 3 | .215 |
Linear-by-Linear Association | .995 | 1 | .318 |
N of Valid Cases | 455 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 16.89. |
Chi-Square Tests | |||
Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 12.366^{a} | 2 | .002 |
Likelihood Ratio | 12.531 | 2 | .002 |
Linear-by-Linear Association | .134 | 1 | .715 |
N of Valid Cases | 455 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 20.35. |
X2(2, 455) = 12.366, p=0.002
Step 4: Conclusion
The test p-value is less than 0.05, we reject the null hypothesis police departments in communities with larger proportions of citizens with graduate degrees are more likely to engage in community partnerships
Question 10
Do agencies that require recruits to undergo significant community policing training also require current officers to undergo extensive community policing training? Using CP_TrnRec and CP_TrnIns assess this question using the hypothesis testing process. You MUST specify each step of the hypothesis testing process.
Question 10 Solution
Step 1: Hypothesis
Null Hypothesis: Agencies that require recruits to undergo significant community policing training is independence of the current officers undergoing extensive community policing training
Alternative Hypothesis: Agencies that require recruits to undergo significant community policing training is not independence of the current officers undergoing extensive community policing training
Step 2: Rejection Rule
Reject the null hypothesis if the p-value is less than 0.05
Step 3: Test Statistic
The chi square test of independence is obtain and the result display below
Chi-Square Tests | |||||
Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |
Pearson Chi-Square | 50.958^{a} | 1 | .000 | ||
Continuity Correction^{b} | 49.303 | 1 | .000 | ||
Likelihood Ratio | 56.200 | 1 | .000 | ||
Fisher's Exact Test | .000 | .000 | |||
Linear-by-Linear Association | 50.825 | 1 | .000 | ||
N of Valid Cases | 381 | ||||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 44.54. | |||||
b. Computed only for a 2x2 table |