Question

1. The City of Blueville has implemented Red Light Camera Program for one year and the City Manager wants to evaluate the effectiveness of program. After reviewing the traffic accident reports in the past several decades years, Assistant City Manager found that there were average of 115 accidents each year, normally distributed with standard deviation of 25. There were 100 accidents during the first year.

1. Test the (alternative) hypothesis that the number of accidents was reduced significantly at 0.05 significance level. (10 points)

Calculate 95% confidence interval, and what will be your conclusion? Explain why you reach or do not reach the same conclusion as in question (a).

Answer 1

Ho :   µ =   115                  
Ha :   µ <   115       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
population std dev ,    σ =    25.0000                  
Sample Size ,   n =    1                  
Sample Mean,    x̅ =   100.0000                  
                          
'   '   '                  
Standard Error , SE = σ/√n =   25.0000   / √    1   =   25.0000      
Z-test statistic= (x̅ - µ )/SE = (   100.000   -   115   ) /    25.0000   =   -0.60
                          
critical z value, z* =       -1.6449   [Excel formula =NORMSINV(α/no. of tails) ]              
                          
p-Value   =   0.2743   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value>α, Do not reject null hypothesis                       

there is no evidance that  the number of accidents was reduced significantly

..........

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   25.0000   / √   1   =   25.000000
margin of error, E=Z*SE =   1.9600   *   25.00000   =   48.999100
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    100.00   -   48.999100   =   51.000900
Interval Upper Limit = x̅ + E =    100.00   -   48.999100   =   148.99910
95%   confidence interval is (   51.00   < µ <   149.00   )

CI contains 115 , so results are not siginificant

result is same as part (a) , because sample size is small leading the std error to larger

thanks

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