A national public television network has found that 35% of its contributions are

Question

A national public television network has found that 35% of its contributions are for less than $20, with 45% for $20-$50, and 20% for more than $50. In a random sample of 200 of the contributions to a local station, 42% were for less than $20, 43% were for $20-$50, and 15% were over $50. At the 0.05 level, does the local stations distribution of contributions differ significantly from that experienced nationally? Also what is the most accurate statement you can make about the p-value for this test?

Explanation / Answer

Doing an observed/expected value table,          
O   E   (O - E)^2/E  
70   84   2.333333333  
90   86   0.186046512  
40   30   3.333333333  
          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    5.852713178      
          
As df = a - 1,           
          
a =    3      
df = a - 1 =    2      
          
Then, the critical chi^2 value is          
          
significance level =    0.05      
chi^2(crit) =    5.991464547      
          
Also, the p value is          
          
p =    0.05359194      
          
As chi^2 < 5.991, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.      
          
Thus, there is no significant evidence that the local stations distribution of contributions differ from that experienced nationally at 0.05 level. [CONCLUSION]  

If the national distribution of contributions are indeed as reported by the national public television network, there is 0.05359 probability of obtaining a sample at least this extremely far from the expected distribution. [STATEMENT ABOUT P VALUE]

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