You collect a random sample of size n from a population and calculate a 95% confidence

with an increased margin of error?

O Use a 98% confidence level.

O Use a 90% confidence level.

O increase the sample size.

O Use the same confidence level, but compute the interval n times. Approximately 5% 0

© Nothing can guarantee that you will obtain a larger margin of error. You can only say

In a recent telephone survey, 5,000 randomly selected teenagers were asked to cite their primary social network site. Six of 10 teenagers said they use Mybook as

their primary social network site. A 95% confidence interval to estimate the true proportion of teenagers who use Mybook as their primary social network site is found to

be (0.2964, 0.9036). Which of the following is a correct interpretation of the confidence level?

O Ninety-five percent of all samples of this size would yield a confidence interval of (0.2964, 0.9036).

O There is a 95% chance that the true proportion of teenagers who use Mybook as their primary social network site is in the interval (0.2964, 0.9036).

O Ninety-five percent of the time, the procedure used to generate this interval will capture the true proportion of teenagers who use Mybook as their primary

social network site

O Ninety-five percent of all the samples of size 5,000 lie in the confidence interval (0.2964, 0.9036).

O There is a 95% chance that randomly selected teenagers will be part of the 60% who use Mybook as their primary social network site.

(05.04 MC)

Leon wants to estimate the proportion of the seniors at his high school who like to watch football. He interviews a simple random sample of 50 of the 200 seniors on

campus. He finds that 32 like to watch football. Is the random condition for finding confidence intervals met? Explain.

O Yes, the random condition for finding confidence intervals is met because np = 32 and n(1 - P) = 18, both of which are greater than 10

O Yes, the random condition for finding confidence intervals is met because the sample size is greater than 30.

O No, the random condition for finding confidence intervals is not met because np = 32 and n(1 - p) = 18, both of which are less than 100

O No, the random condition for finding confidence intervals is not met because the sample is not at least 50% of the population

O Yes, the random condition for finding confidence intervals is met because the sample is a simple random sample.

Cilantro tastes like soap to some people. This soap taste is inherited through the olfactory receptor gene OR6A2. About 14% of the population has this gene. You wa

to estimate the proportion of Americans who have this gene. How large a sample must you test, with a 3% margin of error and 95% confidence, to estimate the

proportion of people who carry the OR6A2 gene?

O 514 people

O 134 people

O 17.689 people

O 35 people

01.028 people

The faculty members at McAdenville High School want to determine whether there are enough students to have a Winter Formal. Sixty-three of the 150 students

surveyed said they would attend the Winter Formal. Construct and interpret a 99% confidence interval for p.

O The 99% confidence interval is (0.3162, 0.5238). We are 99% confident the true proportion of students attending the Winter Formal is between 31 62% and

52.38%.

© The 99% confidence interval is (0.3162, 0.5238). There is a 99% chance that a randomly selected student who will attend the Winter Formal lies between

31.62% and 52.38%

O The 99% confidence interval is (0.3162, 0.5238). Ninety-nine percent of all samples of this size will yield a confidence interval of (0.3162, 0.5238).

O The 99% confidence interval is (0.4762, 0.6838). We are 99% confident that the true proportion of students attending the Winter Formal is between 47 62%

and 68.38%

O The 99% confidence interval is (0. 4762, 0.6838). Ninety-nine percent of all samples of this size will yield a confidence interval of (0.4762, 0.6838)

PLEASE HELP ASAP, please don’t waste my time if you don’t know what your talking about THANK YOU GUYSSS

with an increased margin of error?

O Use a 98% confidence level.

O Use a 90% confidence level.

O increase the sample size.

O Use the same confidence level, but compute the interval n times. Approximately 5% 0

© Nothing can guarantee that you will obtain a larger margin of error. You can only say

In a recent telephone survey, 5,000 randomly selected teenagers were asked to cite their primary social network site. Six of 10 teenagers said they use Mybook as

their primary social network site. A 95% confidence interval to estimate the true proportion of teenagers who use Mybook as their primary social network site is found to

be (0.2964, 0.9036). Which of the following is a correct interpretation of the confidence level?

O Ninety-five percent of all samples of this size would yield a confidence interval of (0.2964, 0.9036).

O There is a 95% chance that the true proportion of teenagers who use Mybook as their primary social network site is in the interval (0.2964, 0.9036).

O Ninety-five percent of the time, the procedure used to generate this interval will capture the true proportion of teenagers who use Mybook as their primary

social network site

O Ninety-five percent of all the samples of size 5,000 lie in the confidence interval (0.2964, 0.9036).

O There is a 95% chance that randomly selected teenagers will be part of the 60% who use Mybook as their primary social network site.

(05.04 MC)

Leon wants to estimate the proportion of the seniors at his high school who like to watch football. He interviews a simple random sample of 50 of the 200 seniors on

campus. He finds that 32 like to watch football. Is the random condition for finding confidence intervals met? Explain.

O Yes, the random condition for finding confidence intervals is met because np = 32 and n(1 - P) = 18, both of which are greater than 10

O Yes, the random condition for finding confidence intervals is met because the sample size is greater than 30.

O No, the random condition for finding confidence intervals is not met because np = 32 and n(1 - p) = 18, both of which are less than 100

O No, the random condition for finding confidence intervals is not met because the sample is not at least 50% of the population

O Yes, the random condition for finding confidence intervals is met because the sample is a simple random sample.

Cilantro tastes like soap to some people. This soap taste is inherited through the olfactory receptor gene OR6A2. About 14% of the population has this gene. You wa

to estimate the proportion of Americans who have this gene. How large a sample must you test, with a 3% margin of error and 95% confidence, to estimate the

proportion of people who carry the OR6A2 gene?

O 514 people

O 134 people

O 17.689 people

O 35 people

01.028 people

The faculty members at McAdenville High School want to determine whether there are enough students to have a Winter Formal. Sixty-three of the 150 students

surveyed said they would attend the Winter Formal. Construct and interpret a 99% confidence interval for p.

O The 99% confidence interval is (0.3162, 0.5238). We are 99% confident the true proportion of students attending the Winter Formal is between 31 62% and

52.38%.

© The 99% confidence interval is (0.3162, 0.5238). There is a 99% chance that a randomly selected student who will attend the Winter Formal lies between

31.62% and 52.38%

O The 99% confidence interval is (0.3162, 0.5238). Ninety-nine percent of all samples of this size will yield a confidence interval of (0.3162, 0.5238).

O The 99% confidence interval is (0.4762, 0.6838). We are 99% confident that the true proportion of students attending the Winter Formal is between 47 62%

and 68.38%

O The 99% confidence interval is (0. 4762, 0.6838). Ninety-nine percent of all samples of this size will yield a confidence interval of (0.4762, 0.6838)

PLEASE HELP ASAP, please don’t waste my time if you don’t know what your talking about THANK YOU GUYSSS

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A strategy which would produce a new

confidence intervalwith anincreasedmargin of erroris to: A. use a98% confidence level.## What is a confidence interval?

A

confidence intervalsimply refers to a range ofestimated valueswhich defines theprobabilitythat apopulationparameter will lie within it.## How to calculate the margin of error?

Mathematically, a

margin of erroris calculated by using this formula:From the above formula, we can deduce that the

margin of erroris directly proportional to thecritical valueand inversely proportional to the square root ofsample size.This ultimately implies that, increasing the

confidence levelfrom 95% to98%wouldincreasethecritical valueand by extension themargin of errorand vice-versa.Read more on

confidence intervalhere: https://brainacademy.pro/question/25779324#SPJ1