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]
