A report from 2004 classified 718 fatal bicycle accidents according to the month

Question

A report from 2004 classified 718 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table.

(a) Use the given data to test the null hypothesis H0: 1 = 1/12, 2 = 1/12, . . . , 12= 1/12, where 1 is the proportion of fatal bicycle accidents that occur in January,2 is the proportion for February, and so on. Use a significance level of 0.01. (Use 2 decimal places.)
2 =

(b) The null hypothesis in Part (a) specifies that fatal accidents were equally likely to occur in any of the 12 months. But not all months have the same number of days. Test the hypotheses proposed in H0: 4 =  4 = 6 = 9 = 11 = 30/366 0.082, 2 = 29/366 0.079, 1 = 3 = 5 = 7 = 8 = 10 = 12 = 31/366 0.085 using a 0.05 significance level. (Use 2 decimal places.)

2 =

Month Number of Month Accidents January 36 February 32 March 42 April 60 May 79 June 74 July 98 August 83 September 65 October 65 November 43 December 41

Explanation / Answer

a)

Doing an observed/expected value table,          

          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    82.12256267   [ANSWER]  
          
As df = a - 1,           
          
a =    12      
df = a - 1 =    11      
          
Then, the critical chi^2 value is          
          
significance level =    0.01      
chi^2(crit) =    24.72497031      
          
Also, the p value is          
          
p =    5.7271E-13      
          
As chi^2 > 24.725, and P < 0.01, we   REJECT THE NULL HYPOTHESIS.      
          
Thus, there is significant evidence that the distribution of month accidents is not uniform among the months.

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b)

Doing an observed/expected value table,          
  

          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    78.68259378   [ANSWER]  
          
As df = a - 1,           
          
a =    12      
df = a - 1 =    11      
          
Then, the critical chi^2 value is          
          
significance level =    0.01      
chi^2(crit) =    24.72497031      
          
Also, the p value is          
          
p =    2.65153E-12      
          
As chi^2 > 24.725, and P < 0.01, we   REJECT THE NULL HYPOTHESIS.      
          
Thus, there is significant evidence that the distribution of month accidents is not as stated. [CONCLUSION]   

O E (O - E)^2/E 36 59.83333 9.4935 32 59.83333 12.94754 42 59.83333 5.315227 60 59.83333 0.000464 79 59.83333 6.13974 74 59.83333 3.354225 98 59.83333 24.34587 83 59.83333 8.969824 65 59.83333 0.446147 65 59.83333 0.446147 43 59.83333 4.73584 41 59.83333 5.928041

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