As I understand 3 factors decide sample size: population variance, confidence interval and level of confidence. There are several formula suggested by textbooks. Is there a universal formula? Thanks.
119853
what the hell does that mean?
635347
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
630388
It is based on mean and standard deviations (statistical stuff). With just the information you gave, it is impossible to tell.\nWe would need to know the mean, standard devaition, and method of calculating scale...but then again if we had all this stuuf you would be able to asnwer your own question, right?
609751
confidence intervals are about what % confident you are that a given person's score falls between x score and x score on any given day.  formulas: \n\n                                   68% CI=X+/-1(Standard error of measurement)\n                                   95%CI=X+/-1.96(SEM)\n                                   98%CI=X+/-2.58(SEM)\n\nto get the SEM:  SD(standard deviation)times the square root of 1 minus the reliability coefficient(rtt)\n\nI might be wrong about the margin of error, but I think it's the SEM(the index of the amount of error that can be expected in obtained test scores due to the unreliability of the test) \n\nhope this helps--statistics suck!
441869
sample space=8\nnumber of outcomes with at\nleast 2 heads=4\n\nP(at least 2 heads)=4/8=1/2\n\nodds:even money\n\ni hope that this helps
110776
To answer that question you would have to know if the differences between "having a variation" and "not having the variation" are normally distributed.  If so, you could make an assumption on the number to look at based on normal distribution and 3 standard deviations.\n\nIf there is no normal distribution to the data, then you would have to determine your own "statistics" for the odds that the deviation would occur, and calculate the number based on the possible number of "variations" that could occur, versus the odds that any one could occur.
437588
Since 50 is not in the confidence interval, then you would reject the null hypothesis.  Remember that a confidence interval is a range of guesses for the population mean.  Since 50 is not one of the guesses, then we would reject the null.  That means that the p-value is less than 0.05.\n\nThis is assuming that the sample is a random sample.  That is rarely the case since students would choose which class they would take.  I would take this confidence interval with a grain of salt.
277087
Since we are not given any raw data, we must assume normality and use the population standard deviation.\n\nThe 90% CI for the population mean is given by:\n(xbar-z(.05)*sigma/sqrt(n), xbar+z(.05)*sigma/sqrt(n))\n\nHere we are told xbar = 12.4, sigma = 6, and n = 54.  Using a table of normal probabilities, we find z(.05) = 1.645.  Simple calculation shows the CI is:\n(11.0569, 13.7431)\n\nFor part (2), we want our margin or error to be 1 with a 90% confidence level.  Our population standard deviation is still 6 and z(.05) is still 1.645.  Thus, we want:\n1.645*6/sqrt(n) = 1\n9.87/sqrt(n) = 1\nsqrt(n) = 9.87\nn = 97.4169\nRound up to get n = 98.\n\nFor your additional comments:\nIncreasing the confidence level also increases the margin of error--not (a).\nThe sample standard deviation could be higher or lower than the population standard deviation, so it could increase or decrease the margin of error, but there is no guarantee--not (b).\nIncreasing the sample size to 100 makes for a better estimate of the mean and would definitely decrease the margin of error.\n\nThus, (c) is the correct choice.
541224
When the population standard deviation is unknown, thats when you use the t-distribution, in calculating the confidence interval. \n\nThe t-distribution (also known as Student's t-distribution) measures the significance of a difference of means of small samples when two distributions are though to have the same variance, but possibly different means.\n\nBut as the sample size increases, the t-distribution gets close to the normal distribution.\n\nHope this helps!
204907
population sample size (n) = 12\nsquare root of sample size = 3.464\nmean of the data = 11.692\nstandard deviation = .777\n\nIt's too hard to type in math notation so, in words, use the formula: \nmean plus or minus critical k-value multiplied by (standard deviation divided by the square-root of n)\n\nSo for your example, it would be:\n11.692 +/- k(.777/3.464) --> take .777 divided by 3.464\n= 11.692 +/- k(.224) so solving for k -->\n11.692 + k(.224) = 13 \nk = 5.84 \n11.692 - k(.224) = 10\nk = 7.6\n\nThis formula is for the confidence interval for the mean of a normal population. It assumes that this was a simple random sample and is just one of the formulas that you could possibly use. You didn't specify if the data was Normally distributed (I'm not going to check and do everything for you). Just to put this answer in perspective, a confidence level of 99.9 % corresponds to a "k-value" of 3.291
498841
a.  The standardized version of the mean is computed\n\n(sample mean - population mean)/(population sd/sqrt(sample size))\n\nso\n\n(108 - 100)/(16/sqrt(4))\n= 8/8\n= 1\n\nb.  The studentized version of the mean is computed\n\n(sample mean - population mean)/(sample sd/sqrt(sample size))\n\nThe only difference is that you use sample sd as opposed to population sd.\n\n(108-100)/(12/sqrt(4)) = 8/6\n= 1.33
350048
See:\n\nThe section, Sample Size For Proportions ("Large" Populations), in the following:\nhttp://www.wku.edu/~david.neal/statistics/misc/sampsize.html\n\n...or:\n\nhttp://www.stat.auckland.ac.nz/~wild/ChanceEnc/Ch08.psampsize.pdf
208779
Wikipedia reveals all.\n\nThe third link lists online statistics textbooks.  The Statistics Homepage is especially good.
363746
Yes.  This would require the use of a t confidence interval.  That is because the sample is small, the population is normal, and the population standard deviation is unknown.  You would have to compute the mean and the standard deviation of the six values to determine the confidence limits.\n\nedit: BTW, the mean is 81.98 and the standard deviation is 4.18.  The confidence interval I get is (77.18, 86.87)
654955
In my opinion, it doesn't take that large of a sample in order for the CLT to take effect.  Although in many classes, it is said that a sample larger than or equal to 25 is needed, that's not always necessary.\n\nIt does really depend on the shape of the population, though.  If the population is roughly symmetric, then a sample of 10 or possibly less might be sufficient.  If not, if the population is skewed and/or has outliers, then a larger sample would be necessary.\n\nAn elemetary stats book by Weiss has some good guidelines.  I don't really remember what exactly they are, but, basically, you should look at the histogram and the normal qq plot of the of the data.  Look at the histogram is roughly symmetric, and the normal qq plot is roughly linear.\n\nOne of my instructors once usually looked at the estimates for the skewness and  the kurtosis to make the determination.  I think that if the skewness is less than 2 in absolute value and kurtosis is less than 2.5, then the t test would be OK.\n\nedit: (This is the email I sent you earlier)\nI said the skewness and the kurtosis is less than those numbers, not greater than.\n\nYou can use Jarque-Bera as well as looking at the skewness and kurtosis.\n\nAlso, when I say kurtosis, I mean that it is the regular kurtosis minus 3, since 3 is the kurtosis of the normal.\n\nIn terms of what is considered minimum, well, the absolute minimum number of values you need is 2 of course.  But again, that's only if the population is considered normal.\n\nThe CLT converges much faster in some populations versus others.  So sometimes 10 is enough, sometimes not.  That's why I didn't set a particular number.\n\nedit: Yes, nonparametric tests can be used for pretty much any sample size.  But again, the sample size in which you transfer from parametric to nonparametric should depnd on the shape of the population.\n\nAlso, alyxsylvr, you would know that standard errors typically do not depend on the sample size as compared to the population size; it depends on the absolute size of the sample.
139872
first you have to understand the diff between consumer n customer.... customer is the one who purchases certian item and may or may not use.... but the consumer is the one who may or may not purchase but always use.\n\nso consumer confidence level is calculated by the seller who sells particular item to the buyer who may be the customer or customer who  will directly consume. repeatition of purchasing certain item by the use can be measured by the seller and he can assume that because of his repeatition his confidence level on ceratain label or brand is high or vise versa... i hope u understand :)
409682
b is the most convincing.\n\n90 is the value farthest from the mean, and a small sample variance indicates that it's unlikely that you just somehow ended up with an unusually high (or low) population.