sample size


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sample size - measures the number of individual samples measured or observations used in a survey or experiment.

30 people are selected randomly from a certain town. If their mean age is 60.5 and ? = 4.6, find a 9
Assume a small sample t-test since sample size is 30 or less: [URL]http://www.mathcelebrity.com/normconf.php?n=30&xbar=60.5&stdev=4.6&conf=95&rdig=4&pl=Small+Sample[/URL]

As the sample size increases, we assume:
As the sample size increases, we assume: a. ? increases b. ? increases c. The probability of rejecting a hypothesis increases d. Power increases [B]d. Power increases[/B] [LIST] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size also increases power [/LIST]

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

Confidence Interval for the Mean
Free Confidence Interval for the Mean Calculator - Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean

Determine ux and sigma(x) from the given parameters of the population and sample size u = 76, sigma
Determine ux and sigma(x) from the given parameters of the population and sample size u = 76, sigma = 28, n = 49 ux = ? sigma(x) = ? [B]u = ux = 76[/B] sigma(x) = sigma/sqrt(n) so we have 28/sqrt(49) = 28/7 = [B]4[/B]

Find Mean 106 and standard deviation 10 of the sample mean which is 25
mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x

Find Mean 106 and standard deviation 10 of the sample mean which is 25
Do you mean x bar? mean of 106 inches and a standard deviation of 10 inches and for sample of size is 25. Determine the mean and the standard deviation of /x If so, x bar equals the population mean. So it's [B]106[/B]. Sample standard deviation = Population standard deviation / square root of n 10/Sqrt(25) 10/5 [B]2[/B]

Find Mean and standard deviation
one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

Find Mean and standard deviation
one trunk can carry 5068.8 lb. weight of boxes that it carries have a mean of 75lb and a standard deviation of 16 Ib. For Sample size of 64 ,find the mean and standard deviation of /x

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.

HELP SOLVE
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 20.5, n = 11, ? = 7, H0: = 18.7 , Ha: ? 18.7 , ? = 0.01

HELP SOLVE
sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. x = 3.7, n = 32, ? = 1.8, H0: = 4.2 , Ha: ? 4.2 , ? = 0.05

HELP SOLVE
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test about the mean, , of the population from which the sample was drawn x = 3.26 , S = 0.55, ?N= 9, H0: = 2.85, Ha: > 2.85 , ? = 0.01

Point Estimate and Margin of Error
Free Point Estimate and Margin of Error Calculator - Given an upper bound and a lower bound and a sample size, this calculate the point estimate, margin of error.

Proportion Sample Size
Free Proportion Sample Size Calculator - This calculator determines a sample size to select to meet certain criteria related to a confidence percentage, reliability percentage, and a p value proportion. Simply enter your values not using percentage signs. This works whether p^ is known or not known.

Sample Size Reliability for μ
Free Sample Size Reliability for μ Calculator - Given a population standard deviation σ, a reliability (confidence) value or percentage, and a variation, this will calculate the sample size necessary to make that test valid.

Sample Size Requirement for the Difference of Means
Free Sample Size Requirement for the Difference of Means Calculator - Given a population standard deviation 1 of σ1, a population standard deviation 2 of σ2 a reliability (confidence) value or percentage, and a variation, this will calculate the sample size necessary to make that test valid.

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

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

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 IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What i
The IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. a) What is the probability that a randomly person has an IQ between 85 and 115? b) Find the 90th percentile of the IQ distribution c) If a random sample of 100 people is selected, what is the standard deviation of the sample mean? a) [B]68%[/B] from the [URL='http://www.mathcelebrity.com/probnormdist.php?xone=50&mean=100&stdev=15&n=1&pl=Empirical+Rule']empirical rule calculator[/URL] b) P(z) = 0.90. so z = 1.28152 using Excel NORMSINV(0.9)
(X - 100)/10 = 1.21852 X = [B]113[/B] rounded up c) Sample standard deviation is the population standard deviation divided by the square root of the sample size 15/sqrt(100) = 15/10 =[B] 1.5[/B]

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.

The monthly earnings of a group of business students are are normally distributed with a standard de
[URL]http://mathcelebrity.com/community/threads/standard-deviation-of-545-dollars-find-the-sample-size-needed-to-have-a-confidence-level-of-95-and.450/[/URL]

Which of the following can increase power?
Which of the following can increase power? a. Increasing standard deviation b. Decreasing standard deviation c. Increasing both means but keeping the difference between the means constant d. Increasing both means but making the difference between the means smaller [B]b. Decreasing standard deviation[/B] [LIST=1] [*]Power increases if the standard deviation is smaller. [*]If the difference between the means is bigger, the power is bigger. [*]Sample size increase also increases power [/LIST]

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of
You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]