A number of minor automobile accidents occur at various high-risk intersections.

Question

A number of minor automobile accidents occur at various high-risk intersections. Eight intersections were chosen at random, and the traffic lights of those intersections were modified. The numbers of minor accidents during a six-month period before and after the modifications were:

NUMBER OF ACCIDENTS

A - B - C - D - E - F - G - H

5 - 7 - 6 - 4 - 8 - 9 - 8 - 10 (BEFORE MODIFICATION)

3 - 7 - 7 - 0 - 4 - 6 - 8 - 2 (AFTER MODIFICATION)

Sorry, the number above are supposed to be in a chart but I wasn't able to draw one..

Question: At the .01 significance level is it reasonable to conclude that the modification reduced the number of traffic accident?

Explanation / Answer

If we assume that the driving habits of the drivers before and after are the same, the samples may still include people who are new, and people who have moved may not be included. You check with your instructor. I have done two tests. One assumes that it is a paired sample (paired t-test) while the other assumes independent samples (independent t-test for 2 samples)

H0: No difference in accidents after modification.
HA: Modification reduced accident

Paired t-test:
This is a one-tailed test. The critical value of t (from t table with 7 degrees of freedom) with 0.01 significance is 2.998
H0: dbar = 0
HA: dbar not = 0
Mean of differences -2.5
Standard deviation of differences = 2.9277
Standard error of difference-mean = sd / n
SE = 2.9277002 / 2.8284271
Standard error of difference-mean 1.0350983
t = dbar /se
t = -2.5 / 1.035098
t = -2.4152
This is not less than -2.998. This means do not reject H0. Modification has not reduced the number of accidents. The fact that the sample sizes are small should be noted.

Independent t-test
____________________________________________________________________________________________
The critical t-value with 0.01 level of significance and 15 degrees of freedom is 2.602
Sample 1 size 8
Sample 2 size 8
Sample 1 mean 7.125
Sample 2 mean 4.625
Sample 1 S.D. 2.031
Sample 2 S.D. 2.8253
Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)
Pooled variance s = [(7)(4.125)+(7)(7.982142857142857))] / (14) = 6.053571428571429
Multiply s by sqrt(1/8+1/8) = 0.5
Denominator of t = 1.2302003321178454
t = (7.13 - 4.63) = 2.5 / 1.2302003321178454
t =2.0322
Degree of freedom 14

___________________________________________________________________________________________
The computed t is again less than the critical t and the conclusion is the same as the paired test.

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