Question

In: Statistics and Probability

The Vic Government is considering increasing the number of police employed in a region in Victoria...

The Vic Government is considering increasing the number of police employed in a region in Victoria in an effort to reduce crime. Before making the final decision on the number of police to be employed, the Ministry of Police asked that various regions of similar size throughout Vic to be surveyed to determine the relationship between the number of police employed and the number of crimes reported per day. Data collected is shown in the table.

Region Number of Police Number of Crimes per day
1 34 28
2 44 14
3 36 12
4 48 9
5 49 15
6 24 36
7 32 28
8 20 42
9 25 30
10 32 31

a. Calculate the intercept coefficient of the sample linear regression equation. Display working.

b. Provide an interpretation of the intercept coefficient you calculated in terms of the relation between number of police and number of crimes.

c. State the estimated sample linear regression equation.

d. Predict the number of crimes per day if 45 polices are employed. Display working. Comment on the validity of this prediction.

e. Conduct a test on the slope coefficient to see if a negative relation exists between the two variables. Use a 1% level of significance. Display working of the six steps hypothesis test.   The t test-statistic has been calculated. It equals -6.06.

f. Calculate the coefficient of determination for the regression line. Display working

g. Provide an interpretation of the calculated coefficient of determination in terms of the relation between number of police and number of crimes.

Solutions

Expert Solution

X Y XY
total sum 344.000 245.000 7509.00 12742.000 7135
mean 34.4000 24.5000

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   -0.9061

there is STRONF, LINEAR, AND NEGATIVE relation between two

...............

B)

sample size ,   n =   10          
here, x̅ =Σx/n =   34.4000   ,   ȳ = Σy/n =   24.5  
                  
SSxx =    Σx² - (Σx)²/n =   908.400          
SSxy=   Σxy - (Σx*Σy)/n =   -919.000          
SSyy =    Σy²-(Σy)²/n =   1132.500          
estimated slope , ß1 = SSxy/SSxx =   -919.000   /   908.400   =   -1.0117
                  
intercept,   ß0 = y̅-ß1* x̄ =   59.3014          
                  

..............

C)

so, regression line is   Ŷ =   59.30   +   -1.01   *x

...............

D)

Predicted Y at X=   45   is                  
Ŷ =   59.301   +   -1.012   *   45   =   13.776
.........

E)

slope hypothesis test
Ho:   ß1=   0          
H1:   ß1< 0          
n=   10              
alpha=   0.01              
estimated std error of slope =Se(ß1) = Se/√Sxx =    5.035   /√   908   =   0.1670
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    -1.0117   /   0.1670   =   -6.06
                  
Degree of freedom ,df = n-2=   8              
p-value =    0.0003              
decision :    p-value<α , reject Ho              

...................


R² =    (Sxy)²/(Sx.Sy) =    0.821

..........

G)

82.1% OF VARIATION IS EXPAINED BY NUMBER OF POLICE OF NUMBER OF CRIMES

.................

THANKS

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