Question

An issue of Time Style and Design reported on a poll conducted by Schulman Ronca & Bucuvalas Public Affairs about the shopping habits of wealthy Americans. A total of 603 interviews were conducted among a national sample of adults with household incomes of at least $150,000. Of the adults interviewed, 410 said they had purchased clothing, accessories, or books online in the past year.

a. Find a 95% confidence interval for the proportion of all U.S. adults with household incomes of at least $150,000 who purchased clothing, accessories, or books online in the past year.

b. Interpret the confidence interval.

Answer 1

Answer)

Given sample size = n = 603

Point estimate (p) = 410/603

Now, first we need to check the conditions of normality, that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 410

N*(1-p) = 193

As both are greater than 5, conditions are met, and we xan use standard normal z table, to estimate the interval

From z table, critical value z for 95% confidence interval is 1.96

Margin of error (MOE) = Z*√{P*(1-P)/N}

MOE = 0.0372349423704

Confidence interval is given by

P-MOE < P < P + MOE

0.6426987226378 < P < 0.7171686073787

B)

We are 95% confident that true population proportion lies in between 0.6427 and 0.7172