3 The type of household for the U S population and for a ran

3. The type of household for the U. S. population and for a random sample of 755 households in Plainville, CT is shown in the table below. Use a chi square test to determine if Plainville fits the distribution of U. S. households. Use a = .01

Solution

Doing an observed/expected value table,          
O   E   (O - E)^2/E  
189   196.3   0.271472236  
213   218.95   0.161692167  
77   67.95   1.205334805  
191   188.75   0.026821192  
85   83.05   0.045785671  
          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    1.711106072      
          
As df = a - 1,           
          
a =    5      
df = a - 1 =    4      
          
Then, the critical chi^2 value is          
          
significance level =    0.01      
chi^2(crit) =    13.27670414      
          
Also, the p value is          
          
p =    0.788699217      
          
Thus, comparing chi^2 and chi^2(crit) [or, p and significance level], we   FAIL TO REJECT THE NULL HYPOTHESIS.      
          
Thus, there is significant evidence that the distribution stated is followed. [ANSWER]