It may or it may not. The question does not give enough information, or context, for a sensible answer.
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the purpose and function of standard error of mean
When the sample size is doubled from 100 to 200, the width of the confidence interval generally decreases. This occurs because a larger sample size reduces the standard error, which is the variability of the sample mean. As the standard error decreases, the margin of error for the confidence interval also decreases, resulting in a narrower interval. Thus, a larger sample size leads to more precise estimates of the population parameter.
yes
Yes
The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.The standard error should decrease as the sample size increases. For larger samples, the standard error is inversely proportional to the square root of the sample size.
The margin of error increases as the level of confidence increases because the larger the expected proportion of intervals that will contain the parameter, the larger the margin of error.
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0.0016
Standard error is random error, represented by a standard deviation. Sampling error is systematic error, represented by a bias in the mean.
It is 6.1, approx.
It would help to know the standard error of the difference between what elements.
standard error for proportion is calculated as: SE = sqrt [(p)(1-p) / n ] so let us say that "p" is going to represent the decimal proportion of respondents who said YES.... so... p = 20/25 = 4/5 = 0.8 And... we then are going to say that the complement of "p" which is "1-p" is going to represent the decimal proportion of respondents who said NO ... so... 1-p = 1 - 0.8 = 0.2 Lastly, the "n" in the formula for standard error is equal to 25 because "n" represents the sample size.... So now all you have to do is plug the values you found for "p" and for "1-p"... (remember "p = 0.8" and "1-p = 0.2")... and "n=25".... Standard Error (SE) = sqrt [(p)(1-p) / n ] ............................ = sqrt [(0.8)(1-0.8) / 25 ] ............................ = sqrt [(0.8)(0.2) / 25 ] ............................ = sqrt [0.16 / 25] ............................ = sqrt (0.0064) ............................ = +/- 0.08
Standard error is a measure of precision.
The standard error is the standard deviation divided by the square root of the sample size.
Standard error (which is the standard deviation of the distribution of sample means), defined as σ/√n, n being the sample size, decreases as the sample size n increases. And vice-versa, as the sample size gets smaller, standard error goes up. The law of large numbers applies here, the larger the sample is, the better it will reflect that particular population.
the purpose and function of standard error of mean