Question

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 Facebook as a news
source, at the **95%** 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 Facebook as a news
source, at the **99%** level of confidence. *Round
all answers to 2 decimal places.*

Margin of Error (as a percentage):

Confidence Interval: % to %

The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our estimate

- increases
- decrease
- stays roughly the same

2. Assume that a sample is used to estimate a population
proportion *p*. Find the 95% confidence interval for a
sample of size 186 with 56% successes.

- Enter your answer as an
**open-interval**(*i.e.*, parentheses) using decimals (not percents) accurate to three decimal places.Confidence interval =

- Express the same answer as a tri-linear inequality using
decimals (not percents) accurate to three decimal places.
<

*p*< - Express the same answer using the point estimate and margin of
error. Give your answers as decimals, to three places.
*p*= ±±

3. We wish to estimate what percent of adult residents in a
certain county are parents. Out of 600 adult residents sampled, 330
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
places.

p = ±±

Answer #1

A poll of 2752 U.S. adults found that 90% 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 Facebook...

1. If n=290 and X = 232, construct a 99% confidence
interval.
Enter your answer as an open-interval
(i.e., parentheses) using decimals (not percents) accurate
to three decimal places.
Confidence interval =
Express the same answer as a tri-linear inequality using
decimals (not percents) accurate to three decimal places.
< p <
Express the same answer using the point estimate and margin of
error. Give your answers as decimals, to three places.
p = ±±
2. We wish to estimate...

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,...

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...

In a simple random sample of 1180 U.S. adults from a large
population, it is found that 816 “regularly” spend more than 40
minutes per day checking email. Use this sample to estimate the
true percentage, p, of all U.S. adults in this population who
“regularly” spend more than 40 minutes per day checking email.In
each question below, give answers as percents, rounded to at least
two decimal places.
(a) Find a 95% confidence interval for p
% < p...

A) Express the confidence interval
66.4%<p<81.2%66.4%<p<81.2% in the form of
ˆp±Ep^±E.
_____% ± ± %_____
B) Express the confidence interval 0.69±0.0510.69±0.051 in open
interval form (i.e., (0.155,0.855)).
C) In a survery of 181 households, a Food Marketing Institute found
that 141 households spend more than $125 a week on groceries.
Please find the 98% confidence interval for the true proportion of
the households that spend more than $125 a week on groceries.
Enter your answer as an open-interval
(i.e., parentheses)...

A
poll conducted in 2013 found that the proportion of of U.S. adult
Twitter users who get at least some news on Twitter was 0.52. The
standard error for this estimate was 0.024, and a normal
distribution may be used to model the sample proportion.
1. Construct a 99% confidence interval for the proportion of
U.S. adult Twitter users who got some news on Twitter. Round to
three decimal places.
to
2. Identify each of the following statements as true...

in our Condor Cafe, the Oxnard College Culinary Program offers
the most delicious and affortable breakfast and lunch choices among
the 152 Community Colleges in California. This week, a random
sample of 110 orders were recorded, and 52 orders were for the
$5-meals. Please find the 99.9% confidence interval for the true
proportion of the orders for the $5-meals.
Enter your answer as an open-interval
(i.e., parentheses) using decimals (not percents) accurate
to three decimal places.
Confidence interval =
Express...

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