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a. There is a positive linear relationship between population density and the rate of violent crime. As population density increases, the rate of violent crime increases.
b. There is a positive linear relationship between prison overcrowding and the number of assaults on correctional officers. As prison overcrowding increases, the number of assaults on correctional officers increases.
c. There is a negative linear relationship between the number of officers on foot patrol, and rates of property crime. As the number of officers on foot patrol increases, the rate of property crime decreases.
d. There is no relationship between the income level of parents and the frequency of alcohol use by teenage children.
e. There is a negative linear relationship between the probability of being arrested for armed robbery and the rate of armed robbery. As the probability of being arrested increases, the rate of armed robbery decreases.
f. There is a curvilinear relationship between age and the number of criminal acts. Criminal acts are infrequent both in early and later ages.
a. There is a moderately strong negative linear relationship between the median income level in a neighborhood and its rate of crime. As the median income level in a community increases, its rate of crime decreases.
b. There is a weak positive linear relationship between the number of hours spent working after school and self-reported delinquency. As the number of hours spent working after school increases, the number of self-reported delinquent acts increases.
c. There is a strong positive linear relationship between the number of prior arrests and the length of current sentence. As the number of prior arrests increases, the length of the sentence received for the last offense increases.
d. There is a weak negative linear relationship between the number of jobs held between the ages of 15-17 and the number of arrests as an adult
e. There is no linear relationship between the divorce rate and a state's rate of violent crime.
a. (-.55)2 = .30. Thirty percent of the variance in neighborhood crime rates is explained by the median income level of the neighborhood.
b. (.17)2 = .03. Three percent of the variance in self-reported delinquency is explained by the number of hours a youth works after school.
c. (.74)2 = .55. Fifty-five percent of the variance in sentence length is explained by the number of prior arrests.
d. (-.12)2 = .01. One percent of the variance in the number of arrests as an adult is explained by the number of jobs held when 15-17 years of age.
e. (-.03)2 = .0009. Less than one percent of the variance in a state's violent crime rate is explained by its divorce rate.
a. An increase in the fine imposed by one dollar decreases the number of price fixing citations by .017.
b. An increase of one percent in unemployment increases the rate of property crime by .715.
c. an increase in one year's education increases a police officer's salary by $1,444.53.
a.
b. In order to calculate correlation and regression statistics, there is some preliminary work we can do, like determine the sum of the x and y scores, and the squares of those scores. We do this for the data below:
Self Control Self-Reported Delinquency x x2 y y2 xy 45 2025 5 25 225 63 3969 10 100 630 38 1444 2 4 76 77 5929 23 529 1771 82 6724 19 361 1558 59 3481 7 49 413 61 3721 17 289 1037 88 7744 24 576 2112 52 2704 14 196 728 67 4489 20 400 1340 S= 632 42230 141 2529 9890
Using our computational formula, we can calculate the value of the regression coefficient, b:
The value of the slope coefficient is .428. This tells us that a one score increase on the low self-control scale increases the number of self-reported criminal acts by .428.
c. We will determine the y intercept from the regression equation:
. We first need to find the mean values of x and y (we already know that the value of b is .428). The mean value of y is equal to 141/10 or 14.1. The mean value of x is equal to 632/20 or 63.2. Now we simply substitute the known values into the equation, and solve for the unknown a:
The value of the y intercept or a, then, is equal to -12.95.
d. The predicted number of self-reported offenses when the self-control scale is equal to 70 can now be determined from our regression prediction equation:
The predicted number of offenses, therefore, is 17.01.
e. We can use our computational formula to determine the value of r:
From our calculations in part b above, we have all the information we need to solve for r:
There is a strong positive correlation between low self-control and the number of self-reported criminal offenses.
We now want to conduct a hypothesis test about r. Our null hypothesis is that r = 0, and our alternative hypothesis is that r > 0. We predict direction because we have reason to believe that there is a positive correlation between low self-control and the number of self-reported crimes. To determine if this estimated r value is significantly different from zero with an alpha level of .01, we calculate a t statistic, with n - 2 degrees of freedom. We go to the t table to find our critical value of t with 10 - 2 = 8 degrees of freedom, an alpha level of .01, and a one-tailed test. The critical value of t is 2.896. Our decision rule is to reject the null hypothesis if tobt ³ 2.896. Now, we calculate our t statistic:
We have a tobt of 5.19. Since 5.19 > 2.896, we decide to reject the null hypothesis. There is a significant positive correlation between low self-control and self-reported crime.
f. to determine the amount of variance explained in y by x we simply determine the coefficient of determination, which is the value of r squared. Our r was .88, (.88)2 = .77, so 77% of the variance in self-reported crime is explained by low self-control.
g. based on our results, we would conclude that there is a significant positive linear relationship between low self-control and self-reported offending. Our results are consistent with the earlier findings of Grasmick and colleagues (1993) and the theory of low self-control developed by Gottfredson and Hirschi (1990).
b. In order to calculate correlation and regression statistics, there is some preliminary work we can do, like determine the sum of the x and y scores, and the squares of those scores. We do this for the data below:
Police Response Time Community Rate of
In Minutes Crime per 1,000
x x2 y y2 xy
14 196 82.9 6872.41 1160.6
3 9 23.6 556.96 70.8
5 25 42.5 1806.25 212.5
6 36 39.7 1576.09 238.2
5 25 63.2 3994.24 316.0
8 64 51.3 2631.69 410.4
7 49 58.7 3445.69 410.9
4 16 44.5 1980.25 178.0
10 100 61.2 3745.44 612.0
12 144 73.5 5402.25 882.0
S = 74 664 541.1 32011.27 4491.4
Using our computational formula, we can calculate the value of the regression coefficient, b:
The value of the slope coefficient is 4.19. This tells us that a one minute increase in police response time increases the crime rate by 4.19 per 1,000. The longer the response time, the higher the crime rate. Stated conversely, the shorter or faster the response time, the lower the crime rate.
c. We will determine the y intercept from the regression equation:
. We first need to find the mean values of x and y (we already know that the value of b is 4.19). The mean value of y is equal to 541.1/10 or 54.11. The mean value of x is equal to 74/10 or 7.4. Now we simply substitute the known values into the equation, and solve for the unknown a:
The value of the y intercept or a, then, is equal to 23.1.
d. The predicted community rate of crime when the police response time is 11 minutes can now be determined from our regression prediction equation:
The predicted crime rate, therefore, is 69.19 crimes per 1,000 population.
e. We can again use our computational formula to determine the value of r:
From our calculations in part b above, we have all the information we need to solve for r:
There is a strong positive correlation between low self-control and the number of self-reported criminal offenses.
We now want to conduct a hypothesis test about r. Our null hypothesis is that r = 0, and our alternative hypothesis is that r > 0. We predict direction because we have reason to believe that there is a positive correlation between the number of minutes it takes the police to respond and the community's rate of crime (the longer the response time, the higher the crime rate). To determine if this estimated r value is significantly different from zero with an alpha level of .05, we calculate a t statistic, with n - 2 degrees of freedom. We go to the t table to find our critical value of t with 10 - 2 = 8 degrees of freedom, an alpha level of .05, and a one-tailed test. The critical value of t is 1.86. Our decision rule is to reject the null hypothesis if tobt ³ 1.86. Now we calculate our tobt:
We have a tobt of 4.77. Since 4.77 > 1.86, we decide to reject the null hypothesis. There is a significant positive correlation between the length of police response time and community crime rates.
f. to determine the amount of variance explained in y by x we simply determine the coefficient of determination, which is the value of r squared. Our r was .86, (.86)2 = .74, so 74% of the variance in community crime rates is explained by police response time.
g. based on our results, we would conclude that there is a significant positive linear relationship between police response time and community crime rates.
b. In order to calculate correlation and regression statistics, there is some preliminary work we can do, like determine the sum of the x and y scores, and the squares of those scores. We do this for the data below:
Community Percent On Hours of Daily
Number Welfare Police Patrol
x x2 y y2 xy
1 40 1600 20 400 800
2 37 1369 15 225 555
3 32 1024 20 400 640
4 29 841 20 400 580
5 25 625 15 225 375
6 24 576 20 400 480
7 17 289 15 225 255
8 15 225 20 400 300
9 12 144 10 100 120
10 8 64 20 400 160
11 4 16 40 1600 160
12 2 4 50 2500 100
S = 245 6777 265 7275 4525
Using our computational formula, we can calculate the value of the regression coefficient, b:
The value of the slope coefficient is -.499. This tells us that a one percent increase in the percent of the population that is on welfare decreases the hours of daily police patrol by -.499 (about one-half hour). The greater the percent of the population receiving public assistance in a neighborhood, the fewer the number of hours of police patrol.
c. We will determine the y intercept from the regression equation:
. We first need to find the mean values of x and y (we already know that the value of b is -.499). The mean value of y is equal to 265/12 or 22.1. The mean value of x is equal to 245/12 or 20.4. Now, as in our other problems we simply substitute the known values into the equation, and solve for the unknown a:
The value of the y intercept or a, then, is equal to 32.28.
d. The predicted number of daily hours of police patrol when the percent receiving public assistance in the neighborhood is 30% can now be determined from our regression prediction equation:
The predicted number of hours of police patrol, therefore, is 17.31 hours.
e. From our calculations in part b above, we have all the information we need to solve for r:
There is a moderately strong negative correlation between the percent of the community on welfare and the number of daily hours of police patrol.
We now want to conduct a hypothesis test about r. Our null hypothesis is that r = 0, and our alternative hypothesis is that r < 0. We predict direction because we have reason to believe that there is a negative correlation between the affluence of the neighborhood and the number of hours of daily police patrol. To determine if this estimated r value is significantly different from zero with an alpha level of .05, we calculate a t statistic, with n - 2 degrees of freedom. We go to the t table to find our critical value of t with 12 - 2 = 10 degrees of freedom, an alpha level of .05, and a one-tailed test. The critical value of t is -1.812. The critical value is negative because in our alternative hypothesis we predicted that the population value of r was less than zero. Our decision rule is to reject the null hypothesis if tobt £ 1.812.
We have a tobt of -2.13. Since -2.13 £ -1.812, we decide to reject the null hypothesis. There is a significant negative correlation between the percent of a neighborhood that is receiving public assistance and the number of hours of daily police patrol.
f. to determine the amount of variance explained in y by x we simply determine the coefficient of determination, which is the value of r squared. Our r was -.56, (-.56)2 = .31, so 31% of the variance in the number of hours of police patrol is explained by the percent of the neighborhood that is on welfare.
g. based on our results, we would conclude that there is a significant negative linear relationship between the affluence of a neighborhood and the number of hours that the police patrol it per day.
h. what are the values of b and r when the 11th and 12th neighborhoods in the data set are left out? Here are the new data:
Community Percent On Hours of Daily
Number Welfare Police Patrol
x x2 y y2 xy
1 40 1600 20 400 800
2 37 1369 15 225 555
3 32 1024 20 400 640
4 29 841 20 400 580
5 25 625 15 225 375
6 24 576 20 400 480
7 17 289 15 225 255
8 15 225 20 400 300
9 12 144 10 100 120
10 8 64 20 400 160
S = 239 6757 175 3175 4265
First, we will estimate b and then r:
Instead of a b of -.499 and a moderately strong inverse r of -.56, when the last two observations are deleted, the value of b becomes .079, and the correlation coefficient indicates a weak positive correlation between the affluence of a neighborhood and the number of hours of daily police patrols. Why this difference? Look at the scatterplot for the new data.
We would conclude from this new data that there is not a very strong relationship between the percent on welfare and the number of hours of police patrol. You can see that without the last two observations the slope of the data points changes. These last two data points are very unusual. They are unusually affluent neighborhoods, with 4 and 2% of the persons receiving public assistance, respectively. Moreover, they receive an unusually high number of hours of police patrols. These two neighborhoods are, then, outliers, and as outliers they can distort your conclusions.
feedback form |
permissions |
international |
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Copyright ©2001 The McGraw-Hill Companies.
digital solutions |
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customer service |
mhhe home
Any use is subject to the
Terms of Use and Privacy Policy.
McGraw-Hill Higher Education is one of the many fine businesses of the
The McGraw-Hill Companies.