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Data Analysis in Criminal Justice

by: Tasfia Kamal

Data Analysis in Criminal Justice

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Tasfia Kamal
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Data Analysis in Criminal Justice
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Date Created: 04/10/16
Class Activity Chapter 2 1 Class Activities Chapter 2 1.  Dr. Smith has concluded the initial portion of data collection on her study into rural  crime in America. She has placed the task of identifying the various types of data for a  random selection of a few of her variables to her new research assistant.  Help Dr.  Smith’s new assistant by identifying the level of measurement of each of the variables  listed in the table below. Income # Homicides Sex Age Temperature Suspect °F $20,215 17 F 21 56 Wendy $150,987 22 M 31 89 Daniel $78,002 49 M 37 102 Albus $67,897 82 M 29 98 Sydney $45,612 12 F 54 45 Lily $35,815 3 F 42 36 Joanne $52,270 8 F 35 72 Cynthia $80,903 12 M 23 67 Ronald Answer:   The level of measurements would be as follows: If the Suspect column counts, I believe  it would be categorical and nominal.  The temperature would count as continuous and  interval.  The age is continuous and interval.  The sex column would be categorical and  nominal.  Homicides would be continuous and ratio.  Income would also be continuous  and ratio.     Class Activity Chapter 2 2 2.  Take the following data set containing a continuous variable and generate  categorical data by creating categories into which each of the numbers listed below  could fall. Income $20,215  (1) $150,987 (4)  $78,002 (2)  $67,897 (2) $45,612 (1) $35,815 (1) $52,270 (1)  $80,903 (2) As far as turning the continuous ratio income into a categorical ordinal one, we’d want to identify the ratio and section off the individuals.  Turning the income into level of  incomes is the goal.  Rounding them off, you’d get $20,000 and $160,000.  160­20=140  and 140/4=35.  The four groups would be 20­55 (1), 55­90 (2), 90­125 (3) and 125­160  (4) respectfully.  The only concern would be if someone were to actually have $55,000,  because they would fall into two categories.  By doing sectioning off the individuals  incomes, you are able to turn it into an ordinal category. The numbers next to their  income show which group they would fall into.  Notice that none of the individuals fall  into the third category of the range sets, 90­125.    Chapter 3 Class Activities 1.  A new research assistant has been tasked with the chore of completing a table to  assist in the organization of the data.  He has been given the following incomplete table. Help the Professor’s assistant and complete the table cells for cumulative frequency,  percent, proportion, cumulative proportion, and cumulative percent. Homicides f cf p cp pct cpct 0 5 5 .07 .07 6.67 6.67 1 15 20 .2 .27 20 26.67 2 11 31 .15 .42 14.67 41.34 3 7 38 .09 .51 9.33 50.67 4 4 42 .05 .56 5.33 56 5 33 75 .44 1.00 44 100 6 N=75 Σ=1.00 Σ=100 Cumulative columns are constructed by summing the f, p, and pct columns successively from  row to row. Here, the column labeled cf, which stands for cumulative frequency. The cp and cpct columns contain cumulative proportions and percentages, respectively. As shown on the table,  the sum of f is 75, the sum of p is 1.00 and the sum of the pct is 100.  Gau Chapter 4 Class Activities  1 Chapter 4 Class Activities INSTRUCTIONS 1) Download the document named "In-Class 3 Chapter 4" 2) Run SPSS analyses and past the outputs in this document 3) Answer the questions in this document 4) Upload all your work on the link "In-Class 3" SPSS PRACTICE Practice # 1  A simple frequency analysis of the variable securitylevel  Dataset: Census of State and Federal Adult Correctional Facilities  SPSS Commands Analyze > Descriptive > Statistics > Frequencies, Select the variable you want from the list on the leftside and either drag it to the right or click the arrow to move it over. Place the output on this word document, Minimum annual salary for chief Cumulative Frequency Percent Valid Percent Percent Valid 30000 1 1.4 1.4 1.4 33258 1 1.4 1.4 2.8 34000 1 1.4 1.4 4.2 Gau Chapter 4 Class Activities  2 36000 1 1.4 1.4 5.6 39578 1 1.4 1.4 7.0 39790 1 1.4 1.4 8.5 40000 1 1.4 1.4 9.9 42597 1 1.4 1.4 11.3 46197 1 1.4 1.4 12.7 46301 1 1.4 1.4 14.1 46633 1 1.4 1.4 15.5 47024 1 1.4 1.4 16.9 47238 1 1.4 1.4 18.3 47810 1 1.4 1.4 19.7 50000 2 2.8 2.8 22.5 50088 1 1.4 1.4 23.9 53000 1 1.4 1.4 25.4 53136 1 1.4 1.4 26.8 54935 1 1.4 1.4 28.2 55000 2 2.8 2.8 31.0 55931 1 1.4 1.4 32.4 56545 1 1.4 1.4 33.8 56597 1 1.4 1.4 35.2 57138 1 1.4 1.4 36.6 57449 1 1.4 1.4 38.0 57500 1 1.4 1.4 39.4 57541 1 1.4 1.4 40.8 58000 1 1.4 1.4 42.3 58278 1 1.4 1.4 43.7 59932 1 1.4 1.4 45.1 60000 2 2.8 2.8 47.9 60278 1 1.4 1.4 49.3 61256 1 1.4 1.4 50.7 62000 1 1.4 1.4 52.1 62774 1 1.4 1.4 53.5 64810 1 1.4 1.4 54.9 65000 3 4.2 4.2 59.2 65104 1 1.4 1.4 60.6 Gau Chapter 4 Class Activities  3 66210 1 1.4 1.4 62.0 66980 1 1.4 1.4 63.4 69354 1 1.4 1.4 64.8 69888 1 1.4 1.4 66.2 71850 1 1.4 1.4 67.6 73813 1 1.4 1.4 69.0 75268 1 1.4 1.4 70.4 75496 1 1.4 1.4 71.8 75986 1 1.4 1.4 73.2 76897 1 1.4 1.4 74.6 77000 1 1.4 1.4 76.1 77334 1 1.4 1.4 77.5 80000 3 4.2 4.2 81.7 81453 1 1.4 1.4 83.1 82745 1 1.4 1.4 84.5 83678 1 1.4 1.4 85.9 85000 1 1.4 1.4 87.3 85155 1 1.4 1.4 88.7 89500 1 1.4 1.4 90.1 90299 1 1.4 1.4 91.5 95000 1 1.4 1.4 93.0 96600 1 1.4 1.4 94.4 101039 1 1.4 1.4 95.8 108150 1 1.4 1.4 97.2 110000 1 1.4 1.4 98.6 128051 1 1.4 1.4 100.0 Total 71 100.0 100.0 Interpretation: Frequency: According to the chart, only 1 officer earns $30,000 as his salary and 3 officers earn  $80,000 as their salary. The frequency symbolized the head count per salary. Cumulative Percent: The cumulative percentage symbolizes what percentage of people earn a  certain amount of money. For example, 81.7% of the officers earn $80,000 or less.  Gau Chapter 4 Class Activities  4 Practice # 2 The SPSS Chart Builder Pie Chart of the variable: securitylevel  Before constructing graphs or charts, visitthe Measure column in the Variable View  SPSS Commands Graphs > Chartbuilder > Element Properties > change COUNT to PERCENT > apply > ok Bar Graph of the variable: securitylevel  Select a graph from the "Bar" menu on the left  Drag itto the preview area this time we will ask for counts (i.e., frequencies), which is the Chart Builder's default Gau Chapter 4 Class Activities  5 Practice # 3 There are two different ways to obtain central tendency output. Analyze > Descriptive Statistics> Descriptives and Analyze > Descriptive Statistics> Frequencies . Both ofthese functions will produce central tendency analyses, the Frequencies option offers a broader array of descriptive statistics and Gau Chapter 4 Class Activities  6 even some charts and graphs. Forthisreason, Use Frequencies rather than Descriptives. Once you have opened the Frequencies box, click on the Statistics button to open a menu ofoptions for measures of central tendency. Select Mean, Median, and Mode, Copy and your outputs in this document. Statistics Minimum annual salary for chief   N Valid 71 Missing 0 Mean 65020.62 Median 61256.00 a Mode 65000 a. Multiple modes exist. The  smallest value is shown Median: Here 61256.00 cuts the distribution exactly in half such that 50% of the scores are  above that value and 50% is below that. Mean: Here 65020.62 is the arithmetic average of the entire set data. QUESTIONS 1. An additional research assistant has been hired to help with basic grading and statistical analyses. The new assistant is very eager, but is perplexed by the mean calculated below. Help the new assistant debug these calculations to determine what, if anything, went awry. Grades f fx 80 5 400 75 9 300 70 4 280 65 5 325 Gau Chapter 4 Class Activities  7 55 2 275 90 4 360 95 3 285 100 2 200 n = 8 Σ = 2425 The equation that should be used is X-bar = Σfx/N This gives X-bar = 2425/8...for a mean of 303.125; this is incorrect. This answer is wrong because 75*9 is 675, not 300. And, 55*2 is 110, not 275.  2. Use the following data set, calculate the mean, identify the mode, and spot the median. # Homicides 2001 Cork........... 5 Cornwall.....3 York............4 Middleton..... 11 Bloomington 6 Normal...... 2 Riverton... 7 Manchac... 1 Rooston..... 9 Mean: 10.67 Mode: Middleton Median: 5 3. Take the data set below and calculate the mean, median, and mode. # Homicides 2001 Cork........... 5 Cornwall.....3 York............4 Middleton..... 11 Bloomington 6 Normal...... 2 Riverton... 7 Manchac... 1 Colston.......8 Rooston..... 9 Gau Chapter 4 Class Activities  8 Mean: 5.6 Mode: Middleton Median: 5.5 4. Using the data below, calculate the deviation scores. What is a good trick to checking mathematical calculations at this point? Or, in other words, what should be the sum of the deviation scores? Grades 89 80 75 65 92 82 90 95 Mean: 83.5 Raw Score                d 89                    89­83.5= 5.5 80                    80­83.5= ­3.5 75                    75­83.5= ­8.5 65                    65­83.5= ­18.5 92                    92­83.5= 8.5 82                    82­83.5= ­1.5 90                    90­83.5= 6.5 95                    95­83.5= 11.5  The sum of the deviation score should be 1.  Gau Chapter 5 Class Activities 1 Chapter 5 Class Activities SPSS PRACTICE  SPSS offers ranges, variances, and standard deviations. It will not provide you with  variation ratios, but those are easy to calculate by hand. # 1 ) Measures of Dispersion with Juvenile Arrest Data Analyze Descriptive Statistics   Frequencies  Statistics  Measures of Dispersion Statistics Number of juveniles arrested for  homicide, 2012   N Valid 8 Missing 0 Mean 5.25 Median 1.50 Mode 0 Std. Deviation 7.797 Variance 60.786 Range 20 Minimum 0 Maximum 20 Mean: 5.25 is the arithmetic average of this data set. Median: 1.50 cuts this distribution exactly in half.  Mode: There is 0 frequently occurring category or value in this set of scores.  Standard Deviation: 7.797 is the mean of all the deviation scores. Variance: 60.786 is the average squared deviation from the mean in this data set. #2) Trends With Hate Crimes data  Graphs   Gau Chapter 5 Class Activities 2 According to this data, the number of the hate crime is increasing every year. The numbers  peaked in 2002 and went down once in 2005.  Gau Chapter 5 Class Activities 3 QUESTIONS  1.  Using the data below calculate the Variance. Show your calculations below 2 Grades                     1    x ­ẋ                             1ẋ)               (x   89                              89­83.5= 5.5                                    30.25 80                              80­83.5= ­3.5                                   12.25 75                              75­83.5= ­8.5                                   72.25 65                              65­83.5= ­18.5                                 342.25 92                              92­83.5= 8.5                                    72.25 82                              82­83.5= ­1.5                                   2.25 90                              90­83.5=6.5                                     42.25 95                              95­83.5= 11.5                                  132.25 N=8 ẋ= 83.5 2 ∑(x 1ẋ) = 706 S = ∑(x ­ẋ)  / N­1 1     = 706/ (8­1)     =706/7     = 100.86 2.  Use the data above and calculate the Standard Deviation. Show your calculations  below Grades 89 80 75 65 92 82 90 95 S= √100.86= 10.04 Gau Chapter 5 Class Activities 4 3. What do you think the above standard deviation indicates about the data in the  previous question? ­ It means, 10.04 is the mean of all the deviation scores.  Gau Chapter 6 Class Activities   1 Chapter 6 Class Activities 1.  Take the scores below and convert them into Z scores. 2  Final Exam Grades                x ­ẋ1                                        1­ẋ)  (x   89                                        89­85.42857=3.571429                  12.7551 90                                        90­85.42857=4.571429                  20.89796                     93                                        93­85.42857=7.571429                  57.32653 67                                        67­85.42857= ­18.4286                  339.6122 74                                        74­85.42857= ­11.42866                130.6122 85                                        85­85.42857= ­.42857                    0.183673 100                                     100­85.42857=14.57143                 212.3265 Answer: a) Calculate the mean, X­bar: ẋ= 85.42857 b) Calculate the standard deviation (square root of variance)   S  =   2 Σ(X­X­bar)     N­1 2 Don’t forget to take the square root!  √ S 2  ∑(x 1ẋ) = 773.7143 2 2 S = ∑(x ­1)  / N­1     =773.7143/ 6     =128.9524 S= √128.9524 = 11.35572 c) Use the formula:  Z  = x ­ X­bar   S Z 1 89­85.42857/11.35572= 0.3145                      Z = 74­ 55.42857/11.35572= ­1.0064   Z 2 90­85.42857/11.35572= 0.4026                      Z = 85­ 65.42857/11.35572= ­0.0377   Z 3 93­85.42857/11.35572= 0.6669                      Z = 100­785.42857/11.35572= 1.2831  Z 4 67­85.42857/11.35572= ­1.6223 Gau Chapter 6 Class Activities   2 2.  Take the Z­scores from problem #1 and fine the corresponding areas under the  Standardized Normal curve. Final Exam Grades                      Z                                   Area 89      0.32                                       0.1255 90      0.40                                       0.1554 93      0.67                                       0.2486 67     ­1.62                                       0.4474 74     ­1.01                                       0.3438 85     ­0.04                               0.0160 100                                      1.                                     0.4032  Each Z score should be rounded to the closest decimal... Gau Chapter 7 Class Activities  1 Chapter 7 Class Activities 1. Professor Trelawney’s research assistant is hard at work again.  This time the  Professor has asked the assistant to begin some basic inferential analysis.  The  assistant has been given a data set consisting of 45 observations taken by the  professor for a study into neighborhood design and crime.  The assistant is now at the  point of trying to apply one of the theoretical distributions to the results obtained in the  first round of analysis.  Should the assistant try to apply the principles of the Z  distribution or those associated with the t distribution?  Why? The assistant should try applying the principles of the t distribution because the sample is too small. To use the z distribution, the sample size has to be either equal to or bigger than 100. As in this problem, N is 45, which is smaller than 100, the criterion is not met. Hence, the assistant cannot use z distribution and has to use t distribution instead. 2.  Identify the characteristics of the t distribution.  How is this distribution similar to that  of the Z?  How does the t distribution differ from the Z distribution? The t distribution is similar to z distribution because they both are symmetrical, unimodal, and has a constant area of 1.00. The only different between these two is that t is a family of several different curves rather than one fixed, single curve like z is. The t distribution changes shape depending on the size of the sample. When the sample is small, the curve is wide and flattish; as the sample size increases, the t curve becomes more and more normal until it looks identical to the z curve. 3.  Explain what is meant by the term ‘sampling error’.  How is this important in  statistical analysis? Sampling error is the uncertainty introduced into a sample statistic by the fact that any given sample is only one of an infinite number of samples that could have been drawn from that population. It is important in every statistical analysis researchers would have assumed that any given sample’s mean must be equal to the population mean if they were not aware of the sampling error. 4.  Explain the difference between a statistic and a parameter. Without peeking, what  are the symbols for common statistics and their corresponding parameters? Statistic and parameter are two different concepts. Statistics are numbers derived from samples, while parameters are numbers based on populations.  Gau Chapter 7 Class Activities  2 Statistics are estimates of parameters. Population parameter has only one mean and standard deviation. Sample statistics, by contrast, vary from sample to sample. In any given sample, these statistics might be exactly equal to, roughly equal to, or completely different from their corresponding parameters. Gau Chapter 8 Class Activities 1 Chapter 8 Class Activities 1. A researcher has collected data on the number of homicides in major Southern  cities in the U.S.  With a sample of 208 cities she has found a mean homicide  rate of 46.9, with a standard deviation of 12.35.   Help out our researcher and calculate the confidence interval for some of the  professor’s data.  The α = .05. To answer this question:   First write down what information is provided by the problem. N = 208 X­bar = 46.9 S = 12.35 α = .05; since the confidence level is 95% (because α is .05), z = a .96 Now, simply plug the numbers into the formula. X­bar± Z α /√N­1) = CI 46.9± 1.96 (12.35/√208­1) = 46.9± 1.96 (12.35/14.39)= 46.9± 1.96 (.86)= 46.9± 1.69 LL= 46.9­ 1.69= 45.21 UL= 46.9+ 1.69= 48.59 the 95% CI:  45.21 ≤ μ ≥ 48.59 Interpret this CI:  There is a 95% chance (or .95 probability) that the true population mean is 45.21, 48.59, or some number in between those values. This also mean there is a 5% chance that the true population mean is not in this range.  Gau Chapter 8 Class Activities 2 2.  The researcher has collected some additional data for her study into homicide in  Southern U.S. cities.  However, before she adds the new data to the general data base  she wishes to check the confidence levels.  She has a new sample size of 31 homicdes, with a standard deviation of 7.89, and a mean of 22.34.  The α = .05.  Calculate a confidence level based on this new data. Answer:   Write down all of the information provided in the question first. N = 31 S = 7.89 X­bar = 22.34 α = .05 This time the t distribution must be used due to the small sample size. df = N ­ 1 = 30 two­tailed, α = .05 t = 2.042 X­bar± tα S/√N­1) = CI 22.34±2.042 (7.89/√30) = 22.34±2.042 (1.44)= 22.34±2.94 LL= 22.34­ 2.94= 19.40 UL= 22.34+2.94= 25.28 the 95% CI:  19.40 ≤ μ ≥ 25.28 Interpret this CI:  Gau Chapter 8 Class Activities 3 There is a 95% chance (or .95 probability) that the true population mean is 19.40, 25.28, or some number in between those values. This also mean there is a 5% chance that the true population mean is not in this range.  3.  The researcher has determined that unlawful homicides account for a proportion of  0.86 in a larger data set of 208 homicides in Southern U.S. cities.   Calculate a confidence interval for α = .05. Answer: As in the cases above write down the information provided in the question. N = 208 P­hat = .86 α = .05 .86± 1.96√. 86(1­.86)/208= .86± 0.047 LL= .86­0.047= .813 UL= .86+0.047= .907 the 95% CI:   .813 ≤ P ≥ .907 Interpret this CI:  There is a 95% chance (or .95 probability) that the true population mean is .813, .907,  or some number in between those values. This also mean there is a 5% chance that the true population mean is not in this range.  Gau Chapter 9 Class Activities 1 Chapter 9 Class Activities 1. Describe the possible explanations for the situation when a sample statistic fails to equal a population parameter? - Anytime a sample statistics is not equal to population parameter, there are two possible explanation of this difference. First, it can be just random fluctuation; which means it is just a meaningless fluke. Second it can be genuine discrepancy which mean there must be a bona fide effect statistical effect. Statistics are estimates of population parameter which might contain errors. However, population parameter is fixed. Infinite of random samples can be drawn from the population. Each of those random sets might have different mean and standard deviation. That is why sample statistics may vary from sample to sample. Hence, a sample statistic can fail to equal a population parameter. 2. A researcher has the results of her initial analysis on two variables of interest in her study on regional homicide differences in the United States. She has tested an Independent Variable denoted as ‘REGION’ with labels of ‘Southeast’, ‘Northeast’, Midwest’, ‘Great Plains’, and ‘West’. The Dependent Variable tested in this analysis was ‘Homicide’ indicating the actual number of homicides that have occurred in the calendar year. 2. a) What type of hypothesis test might she have used? What are the levels of measurement for those two variables? Here, the independent variable (region) is categorical and the dependent variable (homicide) is continuous. Hence, the researcher should use ANOVA for her analysis. 2. b) What are two potential explanations to account for these results? There are two potential explanation of the researcher’s disparity. First this can be just a fluke; which means the inequality might be just meaningless. Secondly, there must be a genuine discrepancy in the statistics. For example, crimes vary from place to place. So not every place in one area contain the same rate of crime. Depending on which area the sample was taken from, the samples can vary from the population mean and standard deviation. 3. Based on your background in this program, a) State a problem in the field of criminal justice, Gau Chapter 9 Class Activities 2 The correction department of criminal justice system is less successful in rehabilitating the prisoners. That is why people that goes to prisons are more inclined to return after certain years. The correction officers are supposed help the prisoners to get better and go back to the society to lead a better life. But often some of the program and the officers fail to work and help the whole prison system. That is why every year the prison population rate goes higher than the last year. b) Then propose a potential cause of the problem, Often people have negative outlook that no matter what they do, prisoners always going to come back to prisons. Hence, the officers genuinely don’t help the prisoners to get better. c) Transform it into a hypothesis Because of these negative outlooks towards the prisoners, correction officers are less successful in rehabilitating the prisoners. That is what causes the prison population rate go up. d) State Null hypothesis and alternative hypothesis Here, prison population rate is the dependent variable and the negative outlook is the independent variable. Null hypothesis: There is no relationship between the prison population and the negative outlook of the correction officers. Alternative hypothesis: There is a positive correlation between the prison population and the negative outlook of the correction officers. When the negative outlook of the correction officers go up, the number of the prison population goes up as well. Chapter 10 & 11 Class Activity 9 (Hypothesis Testing Using Two Categorical Variables) – Chi2  QUESTION 1. Describe when we should use Chi2 test by providing an example?  (Think about levels of measurement) Answer - We should use the Chi2 test when the dependent variable and the independent variable are both categorical. Following the chart from the book can point this out. Gender and the feelings towards capital punishment are an example we could use, much like the ones provided in the next section. QUESTION 2. Download the dataset named “10 GSS 2012 Chi2” on Blackboard.  Practice what you learned in Chapter 10 about hypothesis testing. Using  variables “SEX” and “CAPPUN” in SPSS file; (a) set Ho and H , (b)1run the  analysis, (c) interpret the results. The following information is to help you answer the question 2: The Chi Square analysis is found via the sequence: Analyze à Descriptive Statistics à  Crosstabs à Statistics Check Chi­Square;  2 The obtained value of the X obtstatistic is located on the line labeled Pearson Chi­ Square.  – By default, SPSS provides only observed frequencies in the crosstabs table. If you  want expected frequencies or percentages, you must go into Cells and request  them. P value – In SPSS output, the probability associated with the obtained value of the test  statistic. When p < a, the null hypothesis is rejected. Answer - Independent variable to the rows dependent variable to the columns. It’s important to find the relationship and find how strong they are. So you need to check if it’s statistically significant. Larger means you reject it. P value and significance. .05 If it’s less than they are statically significant in the relationship towards the two variables a) The null hypothesis is that there is no relationship between the gender and attitudes towards capital punishment. The alternative is that there Gau Chapter 11 Class Activities  2 is a relationship between gender and the attitudes towards capital punishment. With p being less than .05 we can reject the null with the significant relationship. The systematic measures chart allows the viewer to see the relationship with more ease. Chi­Square Tests Asymptotic Significance (2­ Exact Sig. (2­ Exact Sig. (1­ Value df sided) sided) sided) Pearson Chi­Square 31.461a 1 .000 Continuity Correction 31.110 1 .000 Likelihood Ratio 31.623 1 .000 Fisher's Exact Test .000 .000 Linear­by­Linear Association 31.454 1 .000 N of Valid Cases 4518 a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 711.40. b. Computed only for a 2x2 table Symmetric Measures Approximate Value Significance Nominal by Nominal Phi .083 .000 Cramer's V .083 .000 N of Valid Cases 4518 c) Looking at the pearson chi2 line and significance, there is a staticaiclly significant relationship between gender and attitudes on capital punishment. It is significant because the p <.05. so we will reject the null hypothesis. Symmetric Measures Approximate Value Significance Nominal by Nominal Phi .083 .000 Cramer's V .083 .000 N of Valid Cases 4518 The significance is at .083, so it’s a weak relationship. It’s in the bottom third bracket for scoring, therefore it doesn’t work in the manner we were hoping for in the experiment. The next step is to look at the magnitudes of the relationship between the two variables (gender and attitude) and check if they are weak and or poor. Cramer’s V is for magnitude shows us that .083 is weak since it’s small, though it is still important. Class Activity 10 (Hypothesis Testing Using T­Test) QUESTION 3. Describe the differences between a dependent samples test and an  independent samples test. Answer - The t-test makes the difference. The characteristics can make or break the case. Continuous for the dependent and the independent is binomial. Seeing that under most circumstance, the individuals of the case can’t be in both categories during the test. Breaking up the inmates from that info helps the person in charge of the test differentiate the subsamples further along in the study. Inmates in prison vs the guards. The characterization of the cases. Just because you picked one case, that doesn’t mean you can use it as another case. You’re limited to one case in the guidelines of the study. The independent variable is limited. Cases can’t be both for IV. The variable is independent, but the case isn’t. Newark precincts are at 40-60. If you hot spot in half and not in the other, they can’t count as both. The chance for being picked is equal, but after that, you can’t switch. Experimental design has the guidelines to follow, straying from those could be costly in the long run. QUESTION 4.  Practice what you learned in Chapter 11 about hypothesis testing.  Using variables “offense” and “charges” in SPSS file; (a) set Ho and H ,1(b) run  the analysis, (c) interpret the results. Independent Samples T Test.  Dataset: Juvenile Defendants in Criminal Courts (juveniles charged with felonies in 40  states) The following information is to help you answer the question 2: a) The sample crimes can’t both be labelled violent and property as they are independent. The null hypothesis is that there is no relationship Gau Chapter 11 Class Activities  4 between the offense type and the number of juveniles. The alternative is that there is a relationship between the type of crime and the number of juveniles charged. b) We are comparing the means here. Based off the relationship from the means chart (with the help of Levene’s test), P = 0.0 and it’s less than .05, that means it’s statistically significant. The mean difference between violence and property means it’s statistically significant. So we will reject the null hypothesis. Independent Samples Test Levene's Test for Equality of Variances t­test for Equality of Means 95% Confidence Sig. Interval of the Difference (2­ Mean Std. Error F Sig. t df tailed) Difference DifferenceLower Upper Total  Equal  number  variances  34.081 .000 4.761 1293 .000 1.498 .315 .881 2.116 of  assumed charges Equal  variances not  7.041 1287.925 .000 1.498 .213 1.081 1.916 assumed Analyze > Compare Means > Independent Sample T Test IV is designated as the Grouping Variable   DV à Test Variable(s) Specify the value of IV Levene`s test for equality of variance: it is a hypothesis test. The null is that equality of  variance is equal.  “F” column displays the obtained value of F statistics, and Sig. column shows the p  value for F.  If p<.05, then the null is rejected. QUESTION 5. Practice what you learned in Chapter 11 about hypothesis testing.  Using variables “before” and “after” in SPSS file; (a) set Ho and H , (b) run t1e  analysis, (c) interpret the results. Test  : Dependent­Sample T Test  Dataset : Installation of CCTV in 64 public locations  Command : Analyze > Compare Means > Paired­Samples T Test >  a) The null hypothesis is that the introduction will not make an impact on the crime. There is not a mean difference between before and after the CCTVs are introduced. The alternative is that there will be a significant mean difference between the before and after the introduction of the CCTVs. c) Since p is equal to .094 and since it’s bigger than .05, that means it’s larger and it’s over the alpha. P=.094 the mean difference is not statistically significant and the null hypothesis is retained and we reject the alternative. Paired Samples Test Paired Differences 95% Confidence Interval of the Std. Std. Error Difference Sig. (2­ Mean Deviation Mean Lower Upper t df tailed) Pair  Number of Crimes  1 BEFORE CCTV ­  Number of Crimes  3.985 18.882 2.342 ­.694 8.663 1.701 64 .094 AFTER CCTV Chapter 12 Class Activities QUESTION 2.  ANalysis Of VAriance There are 4 variables in the SPSS datafile. Using JDCC dataset, Decide what variable  you will use, and then; (a) set Ho and H , 1 (b) run the analysis,  (c) interpret the results. Dataset: Juvenile Defendants in Criminal Courts (juveniles charged with felonies in 40  states) Analyze > Compare Means > One­Way ANOVA  PostHoc> select both Bonferoni & Tukey    a) H =0There is no difference in mean in groups of the independent variable. H = There is statistically significant difference in mean in groups of the 1 independent variables. ANOVA Days to release   Sum of Squares df Mean Square F Sig. Between Groups 49125.707 2 24562.854 9.631 .000 Within Groups 1550662.129 608 2550.431 Total 1599787.836 610 The means are significantly different from each other in the independent variable. If it is less than .05 it rejects the null hypothesis and retains the alternative hypothesis. Here, because the p value is .000 which is less than . 05 that means there is statistically significant difference in mean in groups of the independent variables. Multiple Comparisons Dependent Variable:   Days to release   95% Confidence Interval Mean Difference (I) Race (J) Race (I­J) Std. Error Sig. Lower Bound Upper Bound Tukey HSD Black Hispanic 10.066 6.012 .216 ­4.06 24.19 * White 19.561 4.462 .000 9.08 30.04 Hispanic Black ­10.066 6.012 .216 ­24.19 4.06 White 9.494 6.167 .273 ­4.99 23.98 * White Black ­19.561 4.462 .000 ­30.04 ­9.08 Hispanic ­9.494 6.167 .273 ­23.98 4.99 Bonferroni Black Hispanic 10.066 6.012 .284 ­4.37 24.50 White 19.561* 4.462 .000 8.85 30.27 Hispanic Black ­10.066 6.012 .284 ­24.50 4.37 White 9.494 6.167 .373 ­5.31 24.30 White Black ­19.561* 4.462 .000 ­30.27 ­8.85 Hispanic ­9.494 6.167 .373 ­24.30 5.31 *. The mean difference is significant at the 0.05 level. The pots hog test result indicate that the mean days to relate among juveniles are significantly different for whites and blacks. The means difference of each group is equal to each other. ANOVA Jail sentence (months)   Sum of Squares df Mean Square F Sig. Between Groups 44.033 2 22.016 2.147 .119 Within Groups 2830.677 276 10.256 Total 2874.710 278 The relationship between jail sentence and race is not significant. The Anova test results indicated that jail sentence and race is not significantly related. The p value is around .12 and bigger than .05. So we retain the null hypothesis and reject the alternative hypothesis. If one assumes that jail sentence varies by race, he or she is not right. Gau Chapter 13 Class Activities  1 Chapter 13 Class Activities  QUESTION 1.  Bivariate Correlation There are 4 variables in the SPSS datafile. Using PPCS dataset, Decide what variable  you will use, and then; (a) set Ho and H , 1 (b) run the analysis,  (c) interpret the results. a) H0=There is no correlation between the age of the respondent and the length of the traffic stop. H1= There is correlation between the age of the respondent and the length of the traffic stop. When the age of the respondent is increasing, the length of the traffic stop is also increasing. Before running the statistical analysis make sure to check out whether the relationship is linear. The line is straight, so its linear. It tells us , there is a negative linear relationship between . now its time to look at whether it is statistically significant. Correlations Length of traffic Age stop Age Pearson Correlation 1 ­.113 Sig. (1­tailed) .154 N 179 83 Length of traffic stPearson Correlation ­.113 1 Sig. (1­tailed) .154 N 83 83 The pierson correlation is negative. Its ­.113. because its close to 0, the correlation is  very weak. its not significantly correlated because its bigger than the p value which is . 05. we retain the null and reject the alternate.  Correlations Gau Chapter 13 Class Activities  2 Number of face­ Length of traffic to­face police stop contacts ** Length of traffic stop Pearson Correlation 1 .312 Sig. (2­tailed) .004 N 83 83 Number of face­to­face  Pearson Correlation .312** 1 police contacts Sig. (2­tailed) .004 N 83 179 **. Correlation is significant at the 0.01 level (2­tailed). Significant because .004 is less than the p value. That’s why its statistically significant.  Pearson is .31 which is moderate correlation.  QUESTION 2.  Bivariate Correlation There are 4 variables in the SPSS datafile. Using GSS dataset, Decide what variable  you will use, and then; (a) set Ho and H ,  1 (b) run the analysis,  (c) interpret the results. Gau Chapter 13 Class Activities  1 Chapter 13 Class Activities  QUESTION 1.  Bivariate Correlation There are 4 variables in the SPSS datafile. Using PPCS dataset, Decide what variable  you will use, and then; (a) set Ho and H , 1 (b) run the analysis,  (c) interpret the results. a) H 0There is no correlation between the age of the respondent and the length of the traffic stop. H 1 There is a correlation between the age of the respondent and the length of the traffic stop. When the age of the respondent is increasing, the length of the traffic stop is also increasing. Before running the statistical analysis make sure to check out whether the relationship is linear. The line is straight, so it is linear. It tells us, there is a negative linear relationship between them. Now it is time to look at whether it is statistically significant. b) Correlations Length of traffic Age stop Age Pearson Correlation 1 ­.113 Sig. (1­tailed) .154 N 179 83 Length of traffic stPearson Correlation ­.113 1 Sig. (1­tailed) .154 N 83 83 c) The Pierson correlation is negative and it is -.113. Because it is close to 0, the correlation is very weak. It is not significantly correlated because it is bigger than the p value which is .05. As a result, we retain the null and reject the alternate. Gau Chapter 13 Class Activities  2 QUESTION 2.  Bivariate Correlation There are 4 variables in the SPSS datafile. Using GSS dataset, Decide what variable  you will use, and then; (a) set Ho and H ,  1 (b) run the analysis,  (c) interpret the results. a) H = There is no correlation between the number of face- to face police 0 contacts and the length of traffic stop. H = There is a correlation between the number of face- to face police 1 contacts and the length of traffic stop. b) Correlations Number of face­ Length of trafficto­face police stop contacts ** Length of traffic stop Pearson Correlation 1 .312 Sig. (2­tailed) .004 N 83 83 Number of face­to­face  Pearson Correlation .312* 1 police contacts Sig. (2­tailed) .004 N 83 179 **. Correlation is significant at the 0.01 level (2­tailed). c) The Pierson Correlation is positive and it is .312. Because of it being . 312, the correlation is moderate. It is significant because .004 is less than the p value which is .05 That is why it is statistically significant. As a result, we retain the alternate hypothesis and reject the null hypothesis. Tasfia Kamal 12/13/15 Professor Onat Data Analysis in Criminal Justice Data Analysis Final Paper Chosen article for this paper: The Death Penalty: A Multi-Level Analysis of Public Opinion: Burgason, Kyle & Pazzani, Lynn Introduction: For this study, the researchers used a southern university’s 16000 students as a population. Among those students, they asked 150 students as a random sample to ask to participate voluntarily to fill out the survey about whether the favor or oppose the death penalty. 145 students among those 150 students successfully submitted their surveys which gives them a 95% successful response rate. In that survey they were simply asked whether they favor or oppose death penalty along with several background questions. The researchers examined several variable which includes, gender, race, political party, education level etc. Their initial hypothesis: There is a negative correlation between the level of education and opinion on death penalty. 1 Research result: The found that Education level is inversely related to support for the death penalty. So, there is a negative relationship between the level of education and opinion on death penalty. The Null Hypothesis and the Alternative Hypothesis:  H0 = There is no relationship between the level of education and the opinion on death penalty.  H1 = There is a relationship between the level of education and the opinion on death penalty. Here, the level of education is the independent variable and the opinion on death penalty is the dependent variable. They both are categorical as that is one of the requirements to use the Chi square experiment. We will be using chi2 to analyze to see if there is any significant relationship between these two variable. The p value for this study is .05. We can also determinate whether the relationship is weak or strong looking at the Phi or Cramer’s V significance. Chi­Square Tests Asymptotic Significance (2­ Value df sided) Pearson Chi­Square 97.883a 4 .000 Likelihood Ratio 95.633 4 .000 Linear­by­Linear Association 42.777 1 .000 N of Valid Cases 4518 a. 0 cells (0.0%) have expected count less than 5. The minimum  expected count is 119.84. Symmetric Measures Approximate Value Significance Nominal by Nominal Phi .147 .000 Cramer's V .147 .000 N of Valid Cases 4518 Analysis:  Looking at the Pearson chi2 line and significance, there is a statistically significant relationship between the level of education and the opinion on capital punishment. It is significant because the significance is .000 2 which is less than the p value .05. Hence, we will reject the null hypothesis and retain the alternative hypothesis.  The Phi and Cramer’s V value is .147 which shows a weak relationship between the variables because it is too small. But it is still important. Work Cited: • Burgason, Kyle & Pazzani, Lynn. The Death Penalty: A Multi-level th Analysis of Public Opinion. December 24 , 2014. 3


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