You wish to estimate a population proportion with a confidence interval with a margin of error no larger than 0.03. You have a budget of $5000 for collecting sample data with which to determine the interval estimate. Collection c sample data in similar situations has cost $10 per item. A plot sample yielded a sample proportion of 0.75. Using the largest possible sample (given the budget) and p^dot = 0.75, the confidence level that you can have in your estimate is: 0.9394. 0.4394. 0.8788. 0.1212.

The answer is 0.8788

Step-by-step explanation:

From question given, let us recall the following:

We know that  Ƶα /2  * √p (1-p)/n

when we use  n≤  5000/10 =500

P = 0.75

The Margin of error = 0.03

Putting this values together we arrive at

Ƶα/2 = 0.03/√0.75 * 0.25/500

= 1.549

Now,

Ф (1.549) = 0.9394

Therefore the confidence level becomes:

1- (1-∝)/2 = 0.9394

∝ = 0.8788

The answer is 0.8788

the answer is simply just 6n add the two together as they have an n

answer: what is the question you're trying to ask?

step-by-step explanation: