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MrMuchimi
3 years ago
6

Two different red-light-running signal systems were installed at various intersection locations with the goal of reducing angle-

type crashes. Red-Light-Running System A resulted in 60% angle crashes over a sample of 720 total crashes. Red-Light-Running System B resulted in 52% angle crashes over a sample of 680 total crashes. Was there a difference between the proportions of angle crashes between
Mathematics
1 answer:
nekit [7.7K]3 years ago
3 0

Complete Question

Two different red-light-running signal systems were installed at various intersection locations with the goal of reducing angle-type crashes. Red-Light-Running System A resulted in 60% angle crashes over a sample of 720 total crashes. Red-Light-Running System B resulted in 52% angle crashes over a sample of 680 total crashes. Was there a difference between the proportions of angle crashes between the two red-light-running systems installed? Use an alpha of 0.10.

Answer:

Yes there is a difference between the proportions of angle crashes between the two red-light-running systems installed

Step-by-step explanation:

From the question we are told that

   The first sample  proportion  is  \r  p_ 1 =  0.60

   The  second sample proportion is  p_2 =  0.52

    The first sample size is  n_1 =  720

     The second sample size is  n_2  =  680

     The  level of significance is  \alpha = 0.10

     

The null hypothesis is H_o  :  \r p_1 -  \r p_2 =  0

The  alternative hypothesis is  H_a :  \r p_1 - \r p_2 \ne  0

Generally the pooled proportion is mathematically represented as

       p_p =  \frac{(\r p_1 *  n_1 ) + (\r p_2 * n_2)}{n_1 + n_2 }

=>     p_p =  \frac{(0.6 *  720) + ( 0.52 * 680)}{720 +680 }

=>    p_p  = 0.56

 Generally the test statistics is evaluated as        

       t  =  \frac{ ( \r p_1  - \r p_2 )  - 0 }{ \sqrt{ (p_p  (1- p_p)  *  [ \frac{1}{n_1 } +  \frac{1}{n_2 }  ])} }

        t  =  \frac{ (0.60   - 0.52 )  - 0 }{ \sqrt{ (0.56  (1- 0.56)  *  [ \frac{1}{720} +  \frac{1}{680 }  ])} }    

       t  =  3.0

The  p-value obtained from the z-table is  

      p-value  =  P(Z> t ) =  0.0013499

From the question we see that p-value  <  \alpha so the null hypothesis is rejected

 Hence we can conclude that there is a difference between the proportions

         

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