Question

1) The following is a 90% confidence interval for a population proportion, p: CI = (0.26, 0.54). What margin of error, ME, was used to construct this interval? Give your answer to 2 decimal places.

2)The following is a 90% confidence interval for a population proportion, p: CI = (0.26, 0.54). Find the sample proportion used in the construction of this interval. Give your answer to 2 decimal places.

3)The following is a 90% confidence interval for a population proportion, p: CI = (0.26, 0.54). How large was the sample, n, used to construct this interval?  Give your answer as a whole integer. Hint: Use your answers to question 1 and question 2. Otherwise, solve the problem from scratch.

Answer 1

Solution :

A 90% confidence interval = (0.26,0.54)

1) Margin of error = E = 0.54 - 0.26 / 2

Margin of error = E = 0.14

2) Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = (0.26 + 0.54) / 2

Point estimate = = 0.40

1 - = 1 - 0.40 = 0.60

3) At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645 / 0.14)² * 0.40 * 0.60

= 33.135

sample size = n = 34