Help me with this stats problems please!

Question

1.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 237 with 65.8% successes at a confidence level of 99.8%.

M.E. = %

2.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 247 with 114 successes at a confidence level of 98%.

M.E. = %

3.Out of 300 people sampled, 285 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

 ___< p < ___

Answer 1

margin of error = z-critical value * sqrt(p*(1-p)/n), where n = sample size and p = proportion of "successes"

for problem 1 you are given n = 237, p = 0.658 and z-critical is z-score corresponding to 0.999 probability (3.090).

for problem 2, p is computed as number of successes/sample size, where number of successes is given as114 and sample size is given as 247. z-critical is z score corresponding to 0.99 probability (2.326).

confidence interval = p +/- margin of error

for problem 3, p is computed as number of successes/sample size, where number of successes is given as 300 and sample size is given as 285. z-critical is z score corresponding to 0.95 probability (1.645)