A study was conducted to determine if there was a difference in the humor conten

Question

A study was conducted to determine if there was a difference in the humor content in the British and American trade magazine advertisements. In an independent random sample of 250 American trade magazine advertisements, 55 were humorous. An independent random sample of 200 British trade magazines contained 52 humorous ads. Does this data provide evidence at the 0.05 level of significance that there is a difference in the proportion of the humorous ads in British versus American trade magazines? Make sure to provide the null and the alternative hypotheses and the decision rule. Also, provide the p-value and interpret it.

Explanation / Answer

Formulating the hypotheses          
Ho: p1 - p2   =   0  
Ha: p1 - p2   =/=   0  
Here, we see that pdo =    0   , the hypothesized population proportion difference.  
          
Getting p1^ and p2^,          
          
p1^ = x1/n1 =    0.22      
p2 = x2/n2 =    0.26      
          
Also, the standard error of the difference is          
          
sd = sqrt[ p1 (1 - p1) / n1 + p2 (1 - p2) / n2] =    0.040600493      
          
Thus,          
          
z = [p1 - p2 - pdo]/sd =    -0.985209721      
          
As significance level =    0.05   , then the critical z is  
          
zcrit =    1.959963985      

Hence, reject Ho when |z| > 1.96.
          
Also, the p value is          
          
P =    0.324521067  

Hence, if Ho is true, there is a 0.3245 probability of obtaining a sample at least this extreme.  
          
As |z| < 1.96, and P > 0.05, then we    FAIL TO REJECT THE NULL HYPOTHESIS.

There is no significant difference in the proportion of the humorous ads in British versus American trade magazines. [CONCLUSION]      

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