SOLUTION: I am having trouble getting to the point of a final answer. I am able to solve the Margin of Error easy enough I cannot get to the 90% confidence interval. Here is the problem:

Question 1095615: I am having trouble getting to the point of a final answer. I am able to solve the Margin of Error easy enough I cannot get to the 90% confidence interval.
Here is the problem: In a study of the accuracy of fast food drive-through orders, Restaurant A had 355 accurate orders and 76 that were not accurate.
Construct a 90% confidence interval estimate of the percentage of orders that are not accurate?
What I have so far is CI 90% Za is 1.64 My MoE is 0.036 I cannot get to the

Any help would be appreciated. Thank you
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The sample size is 355 + 76 = 431
:
The sample proportion is 76 / 431 = 0.1763 approximately 0.18
:
Standard Error(SE) = sqrt[ p * ( 1 - p ) / n ] = sqrt[0.18*(1-0.18)/431] = 0.0185
:
alpha(a) = 1 - (90/100) = 0.10
:
critical probability(p*) = 1 - α/2 = 0.95
:
We express the critical value(CV) as a z score(sample size > 30, we can use the normal distribution), find the z-score having a cumulative probability equal to our p* value of 0.95. The associated CV is 1.645
:
ME = CV * SE = 1.645 * 0.0185 = 0.0304 approximately 0.03
:
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90% confidence interval = 0.18 + or - 0.03 which is (0.15, 0.21)
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