Question

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.

Answer #1

**Answer:-**

**Given that:-**

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 adult residents who are parents in this country.

Sample Proportion = 400/500=0.8

Margin of error

90% of interval

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.

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

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 of 200 adult residents sampled, 150 had
kids. Based on this, construct a 99% confidence interval for the
proportion ππ of adult residents who are parents in this
county.
Give your answers as decimals, to three places.
< π <

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 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 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...

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 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...

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...

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