Red Light Cameras To combat red-light-running crashes, many states are installin

Question

Red Light Cameras

To combat red-light-running crashes, many states are installing red light cameras at dangerous intersections. These cameras photograph the license plates of vehicles that run red lights and automatically issue tickets. How effective are these photo enforcement programs? The Virginia Department of Transportation (VDOT) conducted a comprehensive study of its newly adopted program and published the results in a 2012 study. In one portion of the study, VDOT provided crash data both before and after installation of the cameras at several intersections. The data, measured as the crash rate caused by red light running per intersection per year, for 13 randomly selected intersections in Fairfax County are given in the data file called “Red Light Cameras.” At the 0.10 significance level, test the claim that the installation of the cameras decreased the mean number of crashes at intersections where the cameras were installed.

Intersection:

1
2
3
4
5
6
7
8
9
10
11
12
13

Before Camera:

3.6
0.27
0.29
4.55
3.09
2.29
2.4
1.09
3.15
3.21
1.88
1.35
7.35

After Camera:

1.87
0
0
1.79
2.04
3.14
2.72
0.24
1.57
0.43
0.28
1.09
4.92

a) Conduct a full hypothesis test by following the steps below. Enter an answer for each of these steps in your document.

i) State the null and alternative hypotheses using correct notation.

ii) State the significance level for this problem.

iii) Calculate the test statistic “by-hand.” Show the work necessary to obtain the value by typing your work and provide the resulting test statistic.

iv) Calculate the p-value.

v) State whether you reject or do not reject the null hypothesis and your reason for your answer in one sentence.

vi) State your conclusion in context of the problem (i.e. interpret your results and/or answer the question being posed) in one or two complete sentences.

b) Construct a 90% confidence interval using the above data. Please do this “by hand” using the formula and showing your work (please type your work, no images accepted here).  

Explanation / Answer

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.356
since our test is right-tailed
reject Ho, if to > 1.356
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 1.11
We have d = 1.11
pooled variance = calculate value of Sd= S^2 = sqrt [ 32.175-(14.43^2/13 ] / 12 = 1.16
to = d/ (S/n) = 3.449
critical Value
the value of |t | with n-1 = 12 d.f is 1.356
we got |t o| = 3.449 & |t | =1.356
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.4491 ) = 0.00241
hence value of p0.1 > 0.00241,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
test statistic: 3.449
critical value: reject Ho, if to > 1.356
decision: Reject Ho
p-value: 0.00241

that the installation of the cameras decreased the mean number of crashes at intersections where the cameras were installed

Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = di/n
Sd = Sqrt( di^2 – ( di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( di/n ) =14.43/13=1.11
Pooled Sd( Sd )= Sqrt [ 32.175- (14.43^2/13 ] / 12 = 1.16
Confidence Interval = [ 1.11 ± t a/2 ( 0.67/ Sqrt ( 13) ) ]
= [ 1.11 - 1.782 * (0.322) , 1.11 + 1.782 * (0.322) ]
= [ 0.537 , 1.683 ]

X Y X-Y (X-Y)^2 3.6 1.87 1.73 2.993 0.27 0 0.27 0.073 0.29 0 0.29 0.084 4.55 1.79 2.76 7.618 3.09 2.04 1.05 1.103 2.29 3.14 -0.85 0.723 2.4 2.72 -0.32 0.102 1.09 0.24 0.85 0.723 3.15 1.57 1.58 2.496 3.21 0.43 2.78 7.728 1.88 0.28 1.6 2.56 1.35 1.09 0.26 0.068 7.35 4.92 2.43 5.905 14.4 32.17

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