What is the "rule of thumb" for determining the sample size?
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
124804
For an uncertainty x of 0.01 (which your 99% accuracy would be) you need 1/x^2 samples which would be 10000.  This is true for a large population (>>10000).  I believe you still need that many for 20000 population but am only 99% sure.\nSource: Best recollection of a stat course 50 years ago.
842271
My rule of thumb is that the skewness should be less than 2 in absolute value.  Other people might have different rules of thumb.\n\nYou may find the Jarque-Bera test of interest.  It is a test for normally that uses both the sample skewness and sample kurtosis.  I have posted a link about it below.
714755
A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.
1147020
to determine the ideal sample size for a population, slovin's formula is used which is\n\nn = N / ( 1 + (N*e^2))\n\nWhere:\nn = number of samples\nN = total population\ne = margin of error\n\nSample Calculation:\nA group of students want to know the age of students in a high school but do not have the resources to survey an entire population of 2,500. If they want to use a sample with a 5% margin of error, what should their sample size be?\n\nGiven:\nN = 2,500\ne = 5% = 0.05\n\nRequired: n = ?\n\nSolution:\n\nn = N / ( 1 + (N*e^2))\nn = 2500 / ( 1 + (2500*.05^2))\nn = 2500 / 7.25\nn = 344.83   ~ 355
1186248
In this case 'N', which is the sample size, is 120.  The probability of an event occuring is equal to the number of times it occurred in the sample divided by N, so:\n\nprobability of X=0:  20/120 = 1/6 ~ 0.1667\nprobability of X=1:  65/120 = 13/24 ~ 0.5417\nprobability of X=2:  25/120 = 5/24 ~ 0.2083\nprobability of X=3:  10/120 = 1/12 ~ 0.0833\n\nAlso the average value of X would be 1.2082 using this probability distribution through an infinite number of line checks.
299127
You need to be more specific.\n\nHowever:\nThere are secondary variables - unintentional variations other than those variable you are testing.  The way to avoid this, such as noise, is to either find a way to eliminate it or make it uniform.\n\nA large sample size will make an experiment more accurate.\n\nDouble-blind studies (one where the research assistant administering the test does not know what is being tested for and therefor cannot influence the results) are good.\n\nIn any case, I cannot really answer further unless I know more about the study.
233497
Usually this term is used in parameter estimation. There exists a number (parameter) relative to a population, but you only use a sample of that population to estimate it. You can´t say for sure what the number is, but you can give a confidence interval.\n \nFor example, lets say you want to know the average height of NBA players but only know the height of 50 players. Supposing those 50 players were a random sample, the average of that sample is the best estimate of the average height of all NBA players. Nevertheless, the actual NBA average will likely not be the same. Using the center limit theorem allows you to develop a theory that leads to the confidence interval.\n \nSuppose it turned out that your estimated average was 2.00m and the 95% C.I. was 1.90 - 2.10. What that really means, is that, *if* the actual average is 2.00m, 95% of the time you sample 50 players randomly, the sample average will be in that range. You can turn that statement around (it boggles the mind, I know) and say that, if the actual population is outside the 1.90 - 2.10 interval, the chance of a random sample having a 2.00 average is less than 5%.\n \nIf you really want to understand the why´s and the caveats, you have to get a basic understanding of central limit theorem and draw the normal distribution of the sampled parameter.
921908
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
795974
the sample would be ....for example: if you used a group of 30 women between that ranged from18-22yrs...the sample mean would be women of the age 20.
481083
It depends what you want to predict.  For example, you could want to know how many people at a school go out for lunch instead of eat at the cafeteria.  Instead of asking everyone, you could choose to ask 30 people at random in the school and use that sample to predict the total school population who go out for lunch.\n\nFor example, if 12 out of the 30 people go out to eat lunch and there are 800 people in the school, then you could predict the total number of persons going out to lunch by taking 12/30 and multiplying that to the total population.\n\n(12/30)*800=320\n\nYou use the sample to predict the behavior of the larger group.  The bigger the sample, usually the more accurate the prediction is.
276294
use the formula for confidence interval.. its found on most statistics book.. i would type the formula, but that wouldn't be helping you to be a better student..
951022
Increase the sample size (c); this reduces the standard error.\nRecomendation (b) is counterproductive (nothing is gained and the interval gets even larger), (a) also makes the interval larger (though it does increase the confidence level), and (d) cannot be done.
874169
There appear to be two different definitions of the standard error.\n\nThe standard error of a sample of sample size n is the sample's standard deviation divided by sqrt(n). It therefore estimates the standard deviation of the sample mean based on the population mean (Press et al. 1992, p. 465). Note that while this definition makes no reference to a normal distribution, many uses of this quantity implicitly assume such a distribution.\n\nThe standard error of an estimate may also be defined as the square root of the estimated error variance sigma^^^2 of the quantity,\ns_e=sqrt(sigma^^^2)\n\n\n\n\n    The standard error is the estimated standard deviation of a statistic. The formula depends on what statistic you are talking about. For example, the standard error of a sample mean is just the sample standard deviation divided by the square root of the sample size. For more complicated statistics, the standard error is also more complicated.\n\n    You can use the standard error in two ways. First, the statistic divided by the standard error gives you a way of testing whether the parameter that your statistic is estimating equals zero. You compare this ratio to a t-distribution. In most contexts, the ratio of the statistic divided by its standard error is called a Wald test. For many simple applications, you can also call this a t-test.\n\n    A second application of the standard error is the production of confidence intervals. You can get a crude confidence interval by taking your statistic plus or minus two standard errors. A more precise confidence interval would use percentiles from the t-distribution.\n\n    There are a lot of subtleties in the use of the standard error, especially in more complex problems. Sometimes, for example, the standard error applies not to the statistic itself, but to the logarithm of that statistic. For example, a logistic regression model will compute an odds ratio for your data, but the standard error refers not to the odds ratio, but to the log odds ratio. In this situation, you need to compute the confidence interval on the log scale and then transform the results back to the original scale of measurement.\n&#92;\n\ns
408294
in statistics what you do is take a small sample, do the tests on the small sample and then generalize the results to the whole population.\n\nHere you took a small sample, found the mean to be 0.68\n\nBut the mean in the whole population will not be always equal to 0.68 . It can be 0.68, but it can lie anywhere around it to. \n\nHence, we have our confidence intervals around 0.68\n\nSo, when we say that the confidence intervals are (.63692, .72308), it means that we are 95% sure that the whole population mean will lie between 0.63692 seconds and .72308 seconds. \n\nSince 0.75 is not in that range, the chance of the population mean being 0.75 is less than 5% or negligible. \n\nI hope that helps.
885851
Beta is the probability of a type II error, accepting the null when the alternative is true.\n\n1-beta is called the power or ability recognize that Ho is false.  \n\nA quote from my Marx and Larsen: " ...sample size is the parameter that researchers almost invariably turn to as the mechanism for ensuring that a hypothesis test will have sufficiently high power against a given alternative"...
898945
Considering the data you had shown,\nyou must use a one-tailed test.\nyour null hypothesis would be:\nHo: There is no significant difference between the mean amount of time devoted to patient care per week by doctors using conventional methods and alternative methods.\nHo: mean(conventional) = mean(alternative)\n\nyour alternative hypothesis must be:\nHa: The mean amount of time devoted to patient care per week by doctors using conventional methods exceeds that of doctors using alternative methods\nHa: mean(conventional) > mean(alternative)\n\nNow that you have the hypotheses, next is to set the level of significance but it's all up to you depending on how confident are that you will not commit any kind of error.\n\nAnd your asking about the appropriate test?\nFor me, the choice depends on the given.\nBut you can only choose from the second and fourth test since you have to use a directional test (one-tailed test).\n\nUse Z-test if the standard deviation or variance of the population is known.\nUse t-test if the standard deviation or variance of the population is unknown.\nThat's the general rule of choosing the appropriate test.\nBut there's another consideration in the selection of test.\nyou must also take into consideration the number of your sample, whether it exceeds 30 or not.\n\nThat's all. I hope I've been a great help to you. Please inform me if there is something wrong with my lecture here.\nthank you.