Question

calculate PART A PART B PART C:

PART A Express the confidence interval 74.3%±8.6%74.3%±8.6% in the form of a trilinear inequality.

% <p<<p< %

PART B: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 44 had kids. Based on this, construct a 90% confidence interval for the proportion p of adult residents who are parents in this county.

Provide the point estimate and margin of error. Give your answers as decimals, to three places.

p =  ±±

PART C: You measure 50 textbooks' weights, and find they have a mean weight of 62 ounces. Assume the population standard deviation is 2.7 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

< μμ <

Answer 1

a) 74.3 - 8.6 < p < 74.3 + 8.6

or, 65.7% < p < 82.9%

b) Point estimate = p = 44 / 400 = 0.11

Z for 90% confidence interval = Z_0.05 = 1.645

Margin of error = Z_0.05 * sqrt(p * (1 - p) / n) = 1.645 * sqrt(0.11 * (1 - 0.11) / 400) = 0.026

c) Z for 99% confidence interval = Z_0.005 = 2.575

confidence interval

                              

                              

                               = (61.02 < µ < 62.98)