Question

2. Express the confidence interval 82.7% ± 6.9% in interval form. Express the answer in decimal format (do not enter as percents).

3. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 394 with 20.1% successes at a confidence level of 99.8%.

M.E. =_____%

Answer 1

2)

A 95% confidence interval for population proportion p is ,

- E < p < + E

82.7% - 6.9% < p < 82.7% + 6.9%

75.8% < p < 89.6%

0.758 < p < 0.896

3)

Given that,

n = 394

Point estimate = sample proportion = = 0.201

1 - = 0.799

Z_/2 = 3.09

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 3.09 * (((0.201 * 0.799) / 394)

= 0.062

M. E. = 6.2%