(S 9.1) Recall the formula for a proportion confidence interval is p ^? z p ^(1?

Question

(S 9.1) Recall the formula for a proportion confidence interval is

p^?zp^(1?p^)n????????<p<p^+zp^(1?p^)n????????

Thus, the margin of error is E=zp^(1?p^)n???????? .

NOTE: the margin of error can be recovered after constructing a confidence interval on the calculator using algebra (that is, subtracting p^ from the right endpoint.)

In a simple random sample of size 56, taken from a population, 21 of the individuals met a specified criteria.

a) What is the margin of error for a 90% confidence interval for p, the population proportion?

Round your response to at least 4 decimal places.

b) What is the margin of error for a 95% confidence interval for p?

Round your response to at least 4 decimal places.

NOTE: These margin of errors are greater than .10 or 10%.

c) How big of a sample is needed to be certain that we have a margin of error less than .10 (or 10%) at 90% confidence?

Explanation / Answer

for estimate proportion p=21/56=0.375

here std error =sqrt(p(1-p)/n) =0.0647

a) for 90% CI ; critical z =1.645

therefore margin of error for a 90% confidence interval =z*std error=0.1064

b)

for 95% CI ; critical z =1.96

therefore margin of error for a 95% confidence interval =z*std error=0.1268

c)

here margin of error E = 0.1 for90% CI crtiical Z          = 1.645 estimated proportion=p= 0.375 required sample size n =         p*(1-p)*(z/E)2= 64.00

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