What is the effect of a larger sample size onthe interval estimate of a population mean?
373106
The sampling error in the mean estimate is reduced by increasing the the size of the sample.\n\nThe standard deviation of the sample does not change as one poster suggested.  The standard deviation and variance should stay roughly the same, independent of sample size because they are measures of the population.\n\nHowever, the standard deviation of the sampling distribution -- the distribution of sample means from repeated samples -- is called the standard error.\n\nA property of the standard error is that it may be approximated by the stdev / sqrt(N) where N is the sample size.  This says that the standard error is proportional to 1/sqrt(N).  An implication of this is that you need to have a sample size that is 4X larger than your current sample size to get a standard error that is 1/2 the current standard error.
479884
Well you don't need a sample if you already have access to an entire population of records! You would use a sample because you can't get a lot of records, and use that sample to give you a rough idea. If you wanted to test a hypothesis though, you should test it at a 5% significance level, or 1% to be really careful. However you need to know if there is an underlying distribution of the data, i.e. normal, etc. If what you have isn't the population of the records then use what you have, the larger the sample size the better! With more values your 'confidence interval' will be narrower, and you can be more certain of the results of your statisical analysis.
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
728472
samples or sample size; usually designated "n"
446411
Usually when  you pull a sample, you know the size of the sample you pulled.  Otherwise, how would you calculate any of the test statistics much less the margin of error?
915189
To my knowledge, there is no min sample size.  In fact, that is one reason K-S might be used, in that Chi Square g-o-f tests can require a relatively larger sample size.
871703
Sorry, if the result can be greater than 1 then you ain't estimating a proportion.  For proportions each unit in the sample gives you either 0 or 1 as a result.\n\nYou need to swith to the general case where you calculate the standard deviation from the sum of squared differences from the mean, etc. PQ/n or whatever cannot be applied.
570786
First of all, if the population is normal, the distribution of sample means is normal regardless of sample size.  So I. is true.\n\nSecond, if the population is not normal, the distribution of sample means is approximately normal if the sample size is large.  So III. is also true.\n\nII. cannot be determined to be true unless we know what the population looks like.\n\nSo I would go with C.\n\nUPDATE: Apparently ghakh hasn't heard of the Central Limit Theorem.  You know, the "Fundamental Theorem of Statistics"?  In layman's terms, it states that if the sample size is large enough, then the sample means will be approximately normal if the sample size is large enough REGARDLESS OF WHAT THE POPULATION LOOKS LIKE!!!!!!  All you need is a finite variance, which for elementary statistics students is not something they need to be concerned with.\n\nMaybe you should take my statistics class, ghakh.  You might learn something!
979650
For any random variable X, if a is a positive number, it is clear that the probability that X will be larger than a goes to 0 as a grows without limit. This is because a probability distribution has integral 1, and therefore the area under the probability density curve (continuous distribution) to the right of a, or the sum of the probabilities for values of X larger than a (discrete distribution) have to go to 0 as a goes to infinity.\nHope this helps -
699932
It is how the values (data samples) are laid out over the numbers in your population.  I'm assuming you are referring to the empirical distribution function, which is a step function showing, at each X, for example, how many data points are less than X.\n\nTry this website for more definitions:\nhttp://www.stat.berkeley.edu/~stark/SticiGui/Text/gloss.htm
303069
Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
929935
variations matter greatly in interpreting data.  Something might seem like it is returning significantly different results than the control but stat evaluation may show that the variations are minimal at best
181967
As with any sample, you generally discard the highest and lowest extremes to reduce the rate of error. Millionaires can be found in any field and are atypical of the population anyways, unless you are talking about Wall Street :)
969083
That statement is false.\n\nIf those statements above are true and assuming n is large, then the standard error would be 5, because by the empirical rule, it would take two standard errors above and below expected to get 95% in the middle.\n\nThe standard error for the sum is equal to sqrt(n)*SD, so in order to double the standard error, you would have to actually multiply the sample size by 4.  That is because\n\nsqrt(4n)*SD = sqrt(4)sqrt(n)*SD\n= 2*sqrt(n)*SD\n\nSo doubling the sample size would not be enough to double the standard error for the sum.  Since the standard error for the second part of the problem would be 10 (using the same reasoning as before, and using the fact that doubling the sample size would double the expected sum), then doubling the sample size would not be enough.\n\nIn other words, in order for that statement to be true, you would have to quadruple the sample size, not double.\n\nedit: The population should be large as compared to the sample size also.  That probably should have been stated in the problem, but, given the wording of the problem, it is probably safe to assume that.\n\nUm, rev, 400 is the expected sum, not the number of draws.  There's a difference.  Way to pretend you know what is going on here.
399013
As it happens, I just got back from a test on statistics. We covered that very question. \n\nHere goes:\nthe sample size is dependent on how confident you want to be about your results. In the following formula, then Z is the confidence level. Use 1.65 for 90% confidence level, 1.96 for 95%, 2,33 for 98% and 2.58 for 99%. \nYou also need to know the standard deviation, which should be given. and the error allowable, for instance 3%, or 5% or such. That number goes into the formula as E. \n\nso number n equals = (Z * standard deviation/E)^2
95550
r - standard error = the significance of the estimate
826746
It's not completely to do with the weighting efficiency itself.  It depends how large your actual sample size is. If you only had 3 people in the sample then it wouldn't be very reliable regardless of the weighting; if you have 3 million people in the sample then it'd still probably be as reliable as you could hope for even with the heaviest weighting.\n\nI think (I'm not sure of this, but fairly sure) that your effective sample size is your achieved sample size multiplied by the weighting efficiency, so multiply your sample size by 0.58\n\nEg. if you had 1,000 in your sample, then your results have an effective sample size of 580, so your answers would be as reliable as if you'd asked 580 people, so use this base size in significance tests rather than the full 1,000.  ("reliability" in this case would rely on the % confidence you want to rely on, the actual percentages that you're comparing, and the effective sample sizes of the groups you're comparing).