Question

a. Assume that a sample is used to estimate a population mean μ . Find the 80% confidence interval for a sample of size 1101 with a mean of 54.3 and a standard deviation of 20.4. Enter your answer as a tri-linear inequality accurate to 3 decimal places.

b. You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p = 0.39. You would like to be 98% confident that your esimate is within 4% of the true population proportion. How large of a sample size is required?

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=69.7 . You would like to be 99% confident that your estimate is within 10 of the true population mean. How large of a sample size is required?

Answer 1

a)

df = n - 1 = 1101 - 1 = 1100

t critical value at 0.20 significance level with 1100 df = 1.282

80% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

54.3 - 1.282 * 20.4 / sqrt(1101) < < 54.3 + 1.282 * 20.4 / sqrt(1101)

53.51 <   < 55.09

b)

Sample size = Z2/2 * p ( 1 - p) / E2

= 2.32632 * 0.39 ( 1 - 0.39) / 0.042

= 804.6

Sample size = 805 (Rounded up to nearest integer)

c)

Sample size = (Z/2 * / E)2

= ( 2.5758 * 69.7 / 10)2

= 322.32

Sample size = 323 (Rounded up to nearest integer)