91 of the 1860 Americans said that they would rather own a h

91% of the 1860 Americans said that they would rather own a home than drive an expensive car. Please use the result of this survey to estimate with 98% confidence interval of percentage of all Americans who prefer homeownership to a more expensive automobile.

Solution

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.91          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.006635681          
              
Now, for the critical z,              
alpha/2 =   0.01          
Thus, z(alpha/2) =    2.326347874          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.894563098          
upper bound = p^ + z(alpha/2) * sp =    0.925436902          
              
Thus, the confidence interval is              
              
(   0.894563098   ,   0.925436902   ) [ANSWER]

91% of the 1860 Americans said that they would rather own a home than drive an expensive car. Please use the result of this survey to estimate with 98% confiden

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