A manufacturer is planning on putting out an advertisement c

A manufacturer is planning on putting out an advertisement claiming that over x percent of the users of his product are satisfied with it. To determine x, a random sample of 500 users was questioned. If 92 percent of these people indicated satisfaction and the manufacturer wants to be 95 percent confident about the validity of the advertisement, what value of x should be used in the advertisement? What value should be used if the manufacturer was willing to be only 90 percent confident about the accuracy of the advertisement?

Solution

For 95% confidence:

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.92          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.012132601          
              
Now, for the critical z,              

alpha/2 =   0.025          

Thus, z(alpha/2) =    1.959963985          

Thus,              

      
lower bound = p^ - z(alpha/2) * sp =   0.89622054 or 89.62% [ANSWER]

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For 90% confidence:
          
Note that              
              
p^ = point estimate of the population proportion = x / n =    0.92          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.012132601          
              
Now, for the critical z,              

alpha/2 =   0.05          

Thus, z(alpha/2) =    1.644853627          
Thus,              
      
lower bound = p^ - z(alpha/2) * sp =   0.900043648 or 90.00% [ANSWER]