Question

In a test of the quality of two television commercials, each commercial was shown in a separate test area six times over a one-week period. The following week a telephone survey was conducted to identify individuals who had seen the commercials. Those individuals were asked to state the primary message in the commercials. The following results were recorded.

Commercial A Commercial B

Number who saw the commercial: 145 Number who saw the commercial: 195

Number who recalled the message: 61 Number who recalled the message: 57

Use a=.05 and test the hypothesis that there is no difference in the recall proportions for the two commercials.

Formulate the null and the alternative hypotheses.

What is the value of the test statistic?

What is the p-value( round to 4 decimals)

Compute a 95% confidence interval for the difference between the recall proportions for the two populations (to 4 decimals).

( , )

Answer 1

Ho:   p1 - p2 =   0          
Ha:   p1 - p2 ╪   0          
                  
sample #1   ----->   A          
first sample size,     n1=   145          
number of successes, sample 1 =     x1=   61          
proportion success of sample 1 , p̂1=   x1/n1=   0.4207          
                  
sample #2   ----->   B          
second sample size,     n2 =    195          
number of successes, sample 2 =     x2 =    57          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.292          
                  
difference in sample proportions, p̂1 - p̂2 =     0.4207   -   0.2923   =   0.1284
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.3471          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0522          
Z-statistic = (p̂1 - p̂2)/SE = (   0.128   /   0.0522   ) =   2.4594
                  
p-value =        0.0139 [excel formula =2*NORMSDIST(z)]      
decision :    p-value<α,Reject null hypothesis               
                  
Conclusion:   There is enough evidence to conclude that   there is a difference in the recall proportions for the two commercials

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level of significance, α =   0.05              
Z critical value =   Z α/2 =    1.960   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0524          
margin of error , E = Z*SE =    1.960   *   0.0524   =   0.1026
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.128   -   0.1026   =   0.0258
upper limit = (p̂1 - p̂2) + E =    0.128   +   0.1026   =   0.2310
                  
so, confidence interval is (   0.0258   < p1 - p2 <   0.2310   )