Question

We wish to estimate what percent of adult residents in a certain county are parents. Out...

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 156 had kids. Based on this, construct a 95% confidence interval for the proportion ππ of adult residents who are parents in this county.

Give your answers as decimals, to three places

Homework Answers

Answer #1

Solution :

Point estimate () = x / n = 156 / 400 = 0.39

1 - = 0.61

Z/2 = 1.96

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

= 1.96 * ((0.39 * 0.61) / 400)

= 0.048

A 95% confidence interval for population proportion is ,

- E < < + E

0.39 - 0.048 < < 0.39 + 0.048

0.342 < < 0.438

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 128 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. ____ < p < ____ Express the same answer using the point estimate and margin of error. Give your answers as...
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 40 had kids. Based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. < p <  Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three...
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 360 had kids. Based on this, construct a 90% confidence interval for the proportion p of adult residents who are parents in this county. Give your answers as decimals, to 4 places. Answer = ± (Note this is NOT in interval form, so check your answer carefull
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 400 had kids. Based on this , construct a 90% confidence interval for the proportion of sdult residents who are parents in this country.
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 500 adult residents sampled, 235 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. < p < Express the same answer using the point estimate and margin of error. Give your answers as decimals, to...
We wish to estimate what percent of adult residents in a certain county are parents. Out...
We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 270 had kids. Based on this, construct a 90% confidence interval for the proportion p p of adult residents who are parents in this county.
1. We wish to estimate what percent of adult residents in a certain county are parents....
1. We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 296 had kids. Based on this, construct a 95% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. __< p <__ Express the same answer using the point estimate and margin of error. Give your answers as decimals,...
QUESTION PART A: Assume that a sample is used to estimate a population proportion p. Find...
QUESTION PART A: 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 300 with 44% successes at a confidence level of 99.9%. M.E. = % Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places. Round final answer to one decimal place QUESTION PART B: We wish to estimate what percent of adult residents in a...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news...
1. A poll of 2234 U.S. adults found that 31% regularly used Facebook as a news source. Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places. Margin of Error (as a percentage): Confidence Interval: % to % Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use...
Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her...
Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 65 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 16.3 and a standard deviation of 3.8. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to 4 decimal places. [ , ] (Report and...