margin of error - a statistic expressing the amount of random sampling error in the results of a survey.

Formula: σ_{p}+ z-score

a confidence interval for a population mean has a margin of error of 0.081

a confidence interval for a population mean has a margin of error of 0.081

a confidence interval for a population mean has a margin of error of 0.081

Margin of error = Interval Size/2
0.081 = Interval Size/2
Cross Multiply:
Interval Size = 0.081 * 2
Interval Size = [B]0.162[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places)
Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get
[B]58.89 < u < 63.11[/B]

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find t

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find t

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

Confidence Interval of a Proportion

Given N, n, and a confidence percentage, this will calculate the estimation of confidence interval for the population proportion π including the margin of error. confidence interval of the population proportion

Find Necessary Sample Size

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

Margin of Error from Confidence Interval

Given a confidence interval, this determines the margin of error and sample mean.

Point Estimate and Margin of Error

Given an upper bound and a lower bound and a sample size, this calculate the point estimate, margin of error.

Solve Problem

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

Solve Problem

A sample of 71 college students yields a mean annual income of $3595. Assuming that ? = $898, find the margin of error in estimating µ at the 99% level of confidence

Solve Problem

[URL]http://www.mathcelebrity.com/marginoferror.php?num=60%2C66&pl=Calculate+Margin+of+Error+and+Sample+Mean[/URL]

Solve the problem

a confidence interval for a population mean has a margin of error of 0.081. Determine the length of the confidence interval

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128

standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and

Standard Error (margin of Error) = Standard Deviation / sqrt(n)
128 = 545/sqrt(n)
Cross multiply:
128sqrt(n) = 545
Divide by 128
sqrt(n) = 4.2578125
Square both sides:
[B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognit

The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the bran. How many adults must he survey in order to be 90% confident that his estimate is within seven percentage points of the true population percentage?
[IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5
1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5
margin of error (E) = 0.07
At 90% confidence level the t is,
alpha = 1 - 90%
alpha = 1 - 0.90
alpha = 0.10
alpha / 2 = 0.10 / 2 = 0.05
Zalpha/2 = Z0.05 = 1.645
sample size = n = (Z[IMG]https://ci4.googleusercontent.com/proxy/mwhpkw3aM19oMNA4tbO_0OdMXEHt9juW214BnNpz4kjXubiVJgwolO7CLbmWXXoSVjDPE_T0CGeUxNungBjN=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Calpha[/IMG] / 2 / E )2 * [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] * (1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] )
= (1.645 / 0.07)^2 *0.5*0.5
23.5^2 * 0.5 * 0.5
552.25 * 0.5 * 0.5
= 138.06
[B]sample size = 138[/B]
[I]He must survey 138 adults in order to be 90% confident that his estimate is within seven percentage points of the true population percentage.[/I]

The monthly earnings of a group of business students are are normally distributed with a standard de

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.

The monthly earnings of a group of business students are are normally distributed with a standard de

The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.