Question

In
a recent random sample, where 600 adults were surveyed, 16.4%
indicate that they have fallen asleep watching television programs
at night. Find the 98% confidence interval for the population
proportion of people who sleep watching television at night.

Answer #1

Given in the question

No. of adults surveyed = 600

P(they have fallen asleep watching television programs at night) =
0.164

P(they did not fall asleep watching television programs at night) =
1 - 0.164 = 0.836

Here we will use one proportion Z test because np= 98.4 and np(1-p)
=82.3 is greater than 10

We need to calculate 98% confidence interval so alpha= 0.02

alpha/2 = 0.01

Zalpha/2 from Z table is 2.3264

98% confidence interval can be calculated as

p +/- Zalpha/2*Sqrt(p*(1-p)/n)

0.164 +/- 2.3264*sqrt(0.164*(1-0.164)/600)

0.164 +/- 2.3264*0.015

0.164 +/- 0.035

So 98% confidence interval is 0.129 to 0.199

Therefore based on data provided 98% confidence interval for the
population proportion of people who sleep watching television at
night is 0.129 to 0.199.

A random sample of n = 1000 voters is taken and 650 of those
surveyed indicate that they like pizza. What is the value of the
sample proportion of those who like pizza? Provide a 95% confidence
interval estimate of the proportion of individuals in the
population who like pizza. Interpret your results.

A random sample of 815 adults surveyed resulted in 20% preferred
honey to jam/jelly on their peanut butter sandwich. Find the 99%
confidence interval for the proportion of adults that prefer
honey.
What is the UL, Upper limit of the confidence interval?

Based on a random sample of 1120 adults, the mean amount of
sleep per night is 8.91 hours. Assuming the population standard
deviation for amount of sleep per night is 3.2 hours, construct
and interpret a 95% confidence interval for the mean amount of
sleep per night.
A 95% confidence interval is (___,___)

In a recent survey of 100 adults, 56 stated that they were
getting the recommended 8 hours of sleep each night. Assuming the
distribution is approximately normal, find the point estimate and
standard error for the proportion of adults who are getting the
recommended amount of sleep each night.
Round your answers to three decimal places, as needed.

3- A recent Gallup poll of a random sample of 1,015 US adults
reported that a 95% confidence interval for the population
proportion of adults who frequently worry about being the victim of
identity theft is (0.32, 0.40).
(a) We are 95% confident that what parameter is contained in
this interval? (Circle your answer.) Can we say that there is a
probability of 95% that this parameter is actually contained in
this interval?
A) sample proportion B) population proportion C)...

In a recent poll, 600 people were asked if they liked soccer,
and 10% said they did. Based on this, construct a 90% confidence
interval for the true population proportion of people who like
soccer.
As in the reading, in your calculations:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.
Give your answers as decimals, to 4 decimal places.
[________, ________]

A random sample of 1005 adults in a certain large country was
asked "Do you pretty much think televisions are a necessity or a
luxury you could do without?" Of the 1005 adults surveyed, 524
indicated that televisions are a luxury they could do without.
Complete parts (a) through (e) below. Click here to view the
standard normal distribution table (page 1). LOADING... Click here
to view the standard normal distribution table (page 2). LOADING...
(a) Obtain a point estimate...

A random sample of 1028 adults in a certain large country was
asked "Do you pretty much think televisions are a necessity or a
luxury you could do without?" Of the 1028 adults surveyed, 528
indicated that televisions are a luxury they could do without.
Complete parts (a) through (e) below. Click here to view the
standard normal distribution table (page 1).LOADING... Click here
to view the standard normal distribution table (page 2).LOADING...
(a) Obtain a point estimate for the...

In a random sample of 200 people, 154 said that they watched
educational television. Find the 90% confidence interval of the
true proportion of people who watched educational television. If
the television company wanted to publicize the proportion of
viewers, do you think it should use the 90% confidence
interval?

In a recent poll of 50 randomly selected adults in the U.S., 14
were able to name ten or more presidents.
a) What is the sample proportion?
b) What is the margin of error for a 95% confidence interval?
(round to three places)
c) What is the 95% confidence interval for this scenario? (round to
three places)
d) Would you support 35% as possible population proportion of U.S.
adults who can name ten or more presidents? Why or why not?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 24 minutes ago

asked 24 minutes ago

asked 38 minutes ago

asked 47 minutes ago

asked 50 minutes ago

asked 50 minutes ago

asked 50 minutes ago

asked 51 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago