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Robbery Rate 2.24
A criminologist studying the relationship between population
density and
robbery rate in medium sized U.S. cities collected the following
data from a random sample of sixteen
cities; X is the population density of the city( number of
people per unit area) and Y is the robbery rate
last year number of robberies per 100000 people). Assume that
the first order regression model (2.1)
is appropriate.
a) Obtain the estimated regression function. Plot the estimated
regression function and the data
does the linear regression appear to give a good fit here?
Discuss.
sol. we will find first Y=bo+b1X
Therefore
Y= 182.972+.262X_________________________Equ.(1)
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b) Obtain point estimates of the following. (1). the difference
in the mean robbery rate in cities
that differ by one unit in population density. (2). the mean
robbery rate in last year in cities
with population density X=60. (3). 10 (4). 2
Sol. (1). If there is one unit chane in X.then there will be
b1=.262 change in Y.
(2) now we will put X=60 in Equation (1)
then estimated value of Y will be 198.692
(3). 10 = -3.68316
(4). 2=114.067
Q 3.30. Refer to Robbery rate problem 2.24
a) Test whether or not there is a linear association between
robbery rate and population density
using t test using a= .01. state the alternatives; decesion
rules and conclusions. what is the
P-Value of the test.
b) test wheather or not bo=0 Control the risk of type I Error at
.01; State the alternatives. decesion
the rules and conclusions. why might there be interest in
resting whether or not bo=0
c) Estimate b1 with a 99 percent confidence interval. inerpret
your interval estimate.
sol: (1). Ho:There is linear association between robbery rate
and populaion density.
H1:There is no linear association between robbery rate and
population density.
level of significance =.01
test statistics: t =1.467
critical region: p value > , .164 > .01
conclusion: Accept Ho because our P value is greater than our a
value
so There is linear association between robbery rate and
population density.
P-value of the test = .164
