# Evaluating Research Findings Using StatisticsEssay

Pages: 8 (2176 words)  ·  Bibliography Sources: 0  ·  File: .docx  ·  Topic: Criminal Justice

Probability and Statistics

Does familiarity with neighbors lead to lower crime rates? A researcher has theorized that if neighbors know each other, then the crime rates will be lower than in communities in which neighbors do not have a relationship. When communities grow quickly the closeness of the community degrades. A study was done to see if there is a relationship between the crime rate and the percentage change in neighborhood population. 10 neighborhoods in a large metropolis were studied. The results are as follows.

Percentage Change in Neighborhood Population

Crime Rate (Crime per 1000 population)

Identify the Explanatory (X) and Response (Y) variable.

The explanatory variable (X) is the percent change in neighborhood population, because the hypothesis states that increased turnover of residents will breed isolation and encourage crime. The response variable is crime rate (Y), because resident turnover rates theoretically explain a corresponding response in crime rates.

2. Find Persons Correlation Coefficient and determine if there exists a significant linear relationship.

X = 29+2+11+17+7+6+51+13+19+7 = 162

Y = 173+35+132+127+69+53+223+104+99+45 = 1060

X2 = 292+22+112+172+72+62+512+132+192+72 = 4520

Y2 = 1732+352+1322+1272+692+532+2232+1042+992+452 = 144648

XY = 29*173+2*35+11*132+17*127+7*69+6*53+51*223+13*104+19*99+7*45 = 24420

n = 10

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r = {?XY -- [(?X*?Y)/n]}/?{[?X2 -- (?X2/n)][ ?Y2 -- (?Y2/n)]} = {24420 -- [(162*1060)/10]}/?{[4520 -- (1622/10)][144648 -- (10602/10)]} = 0.92645

df = 10-2 = 8

Assuming an alpha of .05, the critical value of r is 0.632 given 8 degrees of freedom. Since rdet ? rcrit, there is a significant linear relationship between percent change in neighborhood population and crime rate.

3. Make a scatter plot of the data. Does the scatter plot support the decision from problem 2?

## Essay on Evaluating Research Findings Using Statistics Assignment

The scatter plot reveals what appears to be a significant linear relationship between the two variables, although there may be a ceiling effect above a 50% change in the neighborhood population. The scatter plot does support the decision from problem 2.

4. Find the Least Squares Regression Line for the data.

MX = ?X/10 = 16.2

MY = ?Y/10 = 106.0

SDX = ?{[?(X -- MX) 2]/n-1} = ?{[(29 -- 16.2)2+(2 -- 16.2)2+(11 -- 16.2)2+(17 -- 16.2)2+(7 -- 16.2)2+(6 -- 16.2)2+(51 -- 16.2)2+(13 -- 16.2)2+(19 -- 16.2)2+(7 -- 16.2)2]/9} = 14.513

SDY = ?{[?(Y -- MY) 2]/n-1} = ?{[(173 -- 106.0)2+(35 -- 106.0)2+(132 -- 106.0)2+(127 -- 106.0)2+(69 -- 106.0)2+(53 -- 106.0)2+(223 -- 106.0)2+(104 -- 106.0)2+(99 -- 106.0)2+(45 -- 106.0)2]/9} = 59.896

For Y = a + bX

b = r (SDY/SDX) = 0.92645(59.896/14.513) = 3.2836

a = MY -- bMX = 106.0 -- (3.2836)16.2 = 44.0578

Therefore, Y = 44.0578 + (3.2836*X) and X = (Y -- 44.0578)/3.2836

5. What percentage of variation in the crime rate that is explained by the linear relationship between the crime rate and the percentage of change in neighborhood population? (Hint: Coefficient of Determination)

Coefficient of Determination = r2 = (0.92645)2 = 0.8583

Close to 86% of the change in crime rate can be explained by the change in neighborhood population.

Use the least squares regression line to predict the crime rate for a neighborhood with a change in population of 45%

If X = 45, then Y = 44.0578 + (3.2836*45) = 192 (after rounding off)

Part 2

A study was done to explore the number of Chocolate bars consumed by 16-year-old girls in a months' time. The results are shown below.

Number of Chocolate Bars Consumed

56

46

12

62

39

24

59

51

39

52

28

41

10

64

27

0

34

5

55

32

42

24

14

63

1

63

52

58

52

26

1. Identify the level of measurement used in this study.

The above data represents ratio data, because it has a natural zero point, order, and equal intervals.

2. Create a Frequency Table with 6 classes.

Values

Classes

Number of Students

0-10

Very Low

4

11-21

Low

2

22-32

Low Medium

6

33-43

High medium

5

44-54

High

5

55-65

Very High

8

3. Create a Histogram based on the Frequency table in problem 2.

4. Find the mean, median, and mode of the number of chocolate bars consumed by 16-year-old girls in a month.

Mean (M) = (56+46+12+62+39+24+59+51+39+52+28+41+10+64+27+0+34+5+55+ 32+42+ 24+14+63+1+63+52+58+52+26)/30 = 1131/30 = 37.7

Median = 0, 1, 5, 10, 12, 14, 24, 24, 26, 27, 28, 32, 34, 39, 39, 41, 42, 46, 51, 52, 52, 52, 55, 56, 58, 59, 62, 63, 63, 64 = (39+41)/2 = 40

Mode = 0, 1, 5, 10, 12, 14, 24, 24, 26, 27, 28, 32, 34, 39, 39, 41, 42, 46, 51, 52, 52, 52, 55, 56, 58, 59, 62, 63, 63, 64 = 52

5. Find the Variance and Standard Deviation of the number of chocolate bars consumed by 16-year-old girls in a month.

M = 37.7

Variance (s2) = [?(X -- MX) 2]/n-1 = [(56 -- 37.7)2+(46 -- 37.7)2+(12 -- 37.7)2+(62 -- 37.7)2+(39 -- 37.7)2+(24 -- 37.7)2+(59 -- 37.7)2+(51 -- 37.7)2+(39 -- 37.7)2+(52 -- 37.7)2+(28 -- 37.7)2+(41 -- 37.7)2+(10 -- 37.7)2+(64 -- 37.7)2+(27 -- 37.7)2+(0 -- 37.7)2+(34 -- 37.7)2+(5 -- 37.7)2+(55 -- 37.7)2+(32 -- 37.7)2+(42 -- 37.7)2+(24 -- 37.7)2+(14 -- 37.7)2+(63 -- 37.7)2+(1 -- 37.7)2+(63 -- 37.7)2+(52 -- 37.7)2+(58 -- 37.7)2+(52 -- 37.7)2+(26 -- 37.7)2]/29 = 398.22

SD = ? s2 = 19.9554

Part 3

1. A researcher is interested to learn if there is a linear relationship between the hours in a week spent exercising and a person's life satisfaction. The researchers collected the following data from a random sample, which included the number of hours spent exercising in a week and a ranking of life satisfaction from 1 to 10 ( 1 being the lowest and 10 the highest).

Participant

Hours of Exercise

Life Satisfaction

1

3

1

2

14

2

3

14

4

4

14

4

5

3

10

6

5

5

7

10

3

8

11

4

9

8

8

10

7

4

11

6

9

12

11

5

13

6

4

14

11

10

15

8

4

16

15

7

17

8

4

18

8

5

19

10

4

20

5

4

a. Find the mean hours of exercise per week by the participants.

M = (3+14+14+14+3+5+10+11+8+7+6+11+6+11+8+15+8+8+10+5)/20 = 177/20 = 8.85

b. Find the variance of the hours of exercise per week by the participants.

s2 = [?(X -- MX) 2]/n-1 = [(3 -- 8.85)2+(14 -- 8.85)2+(14 -- 8.85)2+(14 -- 8.85)2+(3 -- 8.85)2+(5 -- 8.85)2+(10 -- 8.85)2+(11 -- 8.85)2+(8 -- 8.85)2+(7 -- 8.85)2+(6 -- 8.85)2+(11 -- 8.85)2+(6 -- 8.85)2+(11 -- 8.85)2+(8 -- 8.85)2+(15 -- 8.85)2+(8 -- 8.85)2+(8 -- 8.85)2+(10 -- 8.85)2+(5 -- 8.85)2]/19 = 13.397

c. Determine if there is a linear relationship between the hours of exercise per week and the life satisfaction by using the correlation coefficient.

X = hours of exercise per week

Y = life satisfaction

X = 3+14+14+14+3+5+10+11+8+7+6+11+6+11+8+15+8+8+10+5 = 177

Y = 1+2+4+4+10+5+3+4+8+4+9+5+4+10+4+7+4+5+4+4 = 101

X2 = 32+142+142+142+32+52+102+112+82+72+62+112+62+112+82+152+82+82+102+52 = 1821

Y2 = 12+22+42+42+102+52+32+42+82+42+92+52+42+102+42+72+42+52+42+42 = 627

XY = 3*1+14*2+14*4+14*4+3*10+5*5+10*3+11*4+8*8+7*4+6*9+11*5+6*4+ 11*10+8*4+15*7+8*4+8*5+10*4+5*4 = 876

n = 20

r = {?XY -- [(?X*?Y)/n]}/?{[?X2 -- (?X2/n)][ ?Y2 -- (?Y2/n)]} = {876 -- [(177*101)/20]}/?{[1821 -- (1772/20)][627 -- (1012/20)]} = -0.103346

df = 20-2 = 18

Assuming an alpha of .05, the critical value of r is 0.444 given 18 degrees of freedom. Since rdet < rcrit, there is not a significant linear relationship between hours of exercise and life satisfaction.

d. Describe the amount of variation in the life satisfaction ranking that is due to the relationship between the hours of exercise per week and the life satisfaction.

Coefficient of Determination = r2 = (-0.103346)2 = 0.010703

Close to 1% of the change in life satisfaction can be explained by the hours of exercise per week.

e. Develop a model of the linear relationship using the regression line formula.

MX = ?X/n = 177/20 = 8.85

MY = ?Y/n = 101/20 = 5.05

SDX = ?s2 = ?13.397 = 3.6602

SDY = ?{[?(X -- MX) 2]/n-1} = ?{[(1 -- 5.05)2+(2 -- 5.05)2+(4 -- 5.05)2+(4 -- 5.05)2+(10 -- 5.05)2+(5 -- 5.05)2+(3 -- 5.05)2+(4 -- 5.05)2+(8 -- 5.05)2+(4 -- 5.05)2+(9 -- 5.05)2+(5 -- 5.05)2+(4 -- 5.05)2+(10 -- 5.05)2+(4 -- 5.05)2+(7 -- 5.05)2+(4 -- 5.05)2+(5 -- 5.05)2+(4 -- 5.05)2+(4 -- 5.05)2]/19} = 2.4810

For Y = a + bX

b = r (SDY/SDX) = -0.103346(2.4810/3.6602) = -0.07005

a = MY -- bMX = 5.05 -- (-0.07005*8.85) = 5.6699

Therefore, Y = 5.6699 + (-0.07005*X)

2. Insomnia has become an epidemic in the United States. Much research has been done in the development of new pharmaceuticals to aide those who suffer from insomnia. Alternatives to the pharmaceuticals are being sought by sufferers. A new relaxation technique has been tested to see if it is effective in treating the disorder. Sixty insomnia sufferers between the ages of 18 to 40 with no underlying health conditions volunteered to participate in a clinical trial. They were randomly assigned to either receive the relaxation treatment or a proven pharmaceutical treatment. Thirty were assigned to each group. The amount of time it took each of them to fall asleep was measured and recorded. The data is shown below. Use the appropriate t-test to determine if the relaxation treatment is more… [END OF PREVIEW] . . . READ MORE

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