What is S when calculating z-scores for a large sample w/out information on sample scores or variance?
601821
Now this is just a guess, but it is possible that you are talking about confidence interval for a sample percent/proportion.  My students occasionally mix these up with confidence intervals for means.  It those cases it will say something like "Out of a sample of size 1000, 465 said that they support the governor" or something like that.  The confidence interval would look like this.\n\nphat +- z*sqrt(phat(1-phat)/n)\n\n"phat" is the sample proportion (you know, p hat?).\n\nIf that is the case, the you would do the problem you stated this way.\n\np-hat = 15/100 = 0.15\n\n95% CI:\n0.15 +- 1.96*sqrt(0.15*0.85/100)\n0.15 +- 0.07\n(0.08, 0.22)
461783
This is a Two Tailed test ... in other words, you are trying to determine if you have statistical evidence to suggest that the average change was something other than zero disruptions different between weeks 1 and 2 ... just because you saw a change of 5, doesn't mean anything ... it could just be a random fluke !\n\nHo (Null Hypothesis):  Sample Aveage change in distruptions was zero -- no change, or X-Bar = 0\nHa (alternative):  Sample Average change was not 0, or X-Bar <> 0\n\nAt the 0.05 (5%) level of significance (in total), you would have 0.025 probability in the upper end of the distribution and another 0.025 probability in the other end ... that's a total of 0.05 or 5%\n\nUsing your Z-Score table, you find the Z-Score (the number of Standard Deviations in a Standardized Normal Distribution with mean of 0 and Standard Deviation of 1) associated with 0.025 or 2.5% ... this is:\n\n-1.959961082 Standard Deviations from the mean for the low tail and 1.959961082 standard deviations from the mean for the high tail.\n\nSo, the number of standard deviations from a presumed mean of zero (under the null hypothesis) for the sample at hand is:\n\n( 5 - 0 ) / 2  = 2.5\n\nThat is to say, that if the Null Hypothesis of NO CHANGE in behavior were true, a legitimate value of 5 drawn purely randomly could still happen, but that would be 2.5 Standard Deviations away.\n\nBut, your upper and lower limits were only +/- 1.959961082 !!!\n\nSo 95% of the time, you would expect to see something between +/- 1.959961082 standard deviations from zero (within this range) ... but this time, you saw a value associated with a whopping 2.5 standard deviations from the presumed mean of zero ... (we'll call 1.959961082 = 1.96 here)\n\n<< I ACTUALLY MADE A LITTLE GRAPHIC HERE, BUT IT WON'T DISPLAY PROPERLY >>\n\nSo you would reject Ho.\n\nConclusion:  Based upon the sample taken, beleieved to be truely random and under the assumptions of normality, we have statistical evidience which would tend to suggest that the observed sample difference of 5 is statistically significant at the 95% confidence level, thereby, leading us to conclude, based upon the evidence observed, that the incentive system may have played a roll in the positive inprovement of the respondents.\n\nDoes this help ?   My contact information is in my profile.\n\nRegards,\n\nAlan
892149
This depends on how exact your answer needs to be.  You can use the 68-95-99.7 rule to determine that in the 95% of the time you will fall within two standard deviations (i.e., bet. 6 and 8).  Half of the remaining 5% is usage below 6 liters, so only 2.5% of the time do you run out of gas.\n\nTo be more exact, you can look up the z-score given by:\n(X - mu) / (sigma) = (8 - 7) / .5 = 2 and find that the result is that you run out of gas 2.2275% of the time.\n\nIf you need more explanation, just add that to your additional details.  Good luck!
490371
The t-test requires that the population of interest is normally distributed.  Now this requirement is not totally necessarily needed if the sample size is large, since then the distribution of sample means is approximately normally distrubuted anyway.\n\nNow if the sample size is small, then the normal assumption is more important.  If your population is skewed, then the t-test statistic will not have a t distribution under the null hypothesis.\n\nThat is where the signed-rank test comes in.  The signed-rank test does a test using the ranks as opposed to the original data.  The ranks do not have extreme values, and so the z test statistic will have an approximate normal even if the sample is small (Usually you can use a sample greater than or equal to 10 for the z test; otherwise there are exact tests that you can use to compute the p-value).\n\nNow the signed-rank test does have its flaws.  It is not as powerful as the t-test.  That means that the signed-rank test is less likely to reject the null if the null is false.  Also, if there are ties in the original data, then the distribution of the test statistic is harder to define.\n\nSo, to summarize, use the signed-rank test if the sample is small and the data has extreme values.
928594
If you assume it's normal, then yeah, although I assume there is more data than you have given as you do not have the variance there, although you could treat this as a binomial distribution giving it mean 0.3, variance n0.3*0.7, although when calculating the confience interval you'll want to divide by n^2 for that variance.
906033
first fin the mean with the sample of 280 \n\nmultiply 280 to .184 to get the mean \n\nmultiply 280 to .184 and to .816 (from 1-.184) then take the \n\nsquare root that will be your standard deviation \n\n\ntake the z score you will get   ( 56 - 51.52 )/6.483851\n\n=.6909   find teh valu in the standard normal distribution table \n\nyou will get 0.2549    subtract it from .5  because you need only \n\nthe probability higher than of 56  you will get \n\n24.51%
896594
The width of the confidence interval is 40, so the error about the mean is 20. You have a sample of 25 batteries, so the number of degrees of freedom is 24. The critical t-score for 24 degrees of freedom and a confidence level of .95 is approximately 2.063899.\n\nWe can use the t-score to approximate the sample standard deviation:\n\nt = error / (s/sqrt(n))\nSolving for s we get \ns = error*sqrt(n)/t, \ns = 20*sqrt(25)/2.063899\ns = 48.45199
122629
I believe the difference is that z-statistics are from a normal distribution with a given standard deviation.  Whereas t-statistics is a normal distribution where the standard deviation has been estimated from the selected sample (i.e. you don't know what sigma is, you have to measure it in the same data sample).
662741
In basic statistical hypothesis testing, the P-value represents the probability of encountering a result more extreme than the value that we've observed in our testing.\n\nAs an example - suppose we had conducted a properly designed experiment, and discovered that the mean score of our sample was 103.5. We would evaluate this finding of 103.5 by comparing it to a known "sampling distribution" - a distribution of sample means of size n that was randomly drawn from a population with (for the sake of this example) a mean of 100 and a standard deviation of 1.75.\n\nWhat we want to do is evaluate the likelihood that a sample of size n - randomly drawn from the population - would have a mean as extreme as 103.5.\n\nWe begin with the assumption that the null hypothesis is true - that our sample belongs to the population - and that the deviation of 3.5 from the population mean of 100 is simply due to random error.\n\nIn this situation, the appropriate statistical tool for evaluating the probability of observing a sample mean as extreme as 103.5 is the t-test. We begin by "standardizing" the raw score of 103.5 by converting it to a deviation score and then dividing that deviation score by the standard deviation of the sampling distribution.\n\nThe formula is simply  (103.5 - 100.0) / 1.75  =  2.0\n\nThis means that our sample mean of 103.5 is 2.0 standard deviation units above the mean of the sampling distribution. Now we need to find out the probability of encountering a sample that is 2 standard deviation units from the mean. \n\nBecause the sampling distribution is assumed to be approximately normally distributed, we can refer to the z-table (reproduced in any statistic text), and discover that only 2.28 percent of all the scores in a normal distribution fall above 2 standard deviation units above the mean. The proper interpretation of this result is that the probability of observing a sample this extreme from a population with a mean of 100 and a standard deviation of 1.75 is 2.28% or less.\n\nP-values are used in conjunction with critical values - pre-determined probabilities that we feel are so unlikely to have occurred by chance, that we are comfortable concluding that results this extreme could not have been selected from the population (thereby rejecting the null hypothesis that the sample was selected from the population).\n\nThe common critical values are .05 and .01, meaning, in each case, that we'll reject the null hypothesis if our finding is so extreme that it would only be expected to happen by chance less than 5 percent of the time - or in the case of .01, less than 1 percent of the time.\n\nReferring back to our finding - we concluded that a result of 103.5 would only happen 2.28 percent of the time or less - if our pre-selected critical value had been .05 (5 percent), we would reject the null hypothesis, and conclude that our sample was not selected from the population. If the critical value had been 1 percent, because our finding happens 2.28 percent of the time, we would fail to reject the null hypothesis.\n\nMan, I hope this made some sense!
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
310004
find a better Statstic book, or use CHIDIST / CHIINV functions in MS-Excel. Or use this thing:\nhttp://www.georgetown.edu/faculty/ballc/webtools/web_chi.html\njust make sure it gives same numbers for lower df's as your table
658731
Slovin's formula is used to determine the ideal sample size out of a population (in other words, how large of a percentage of the whole must I sample so that it adequately represents the whole).  The formula is:\n\nn = N/(1 + N * e^2)\n\nwhere n is the sample size, N is the population, and e is the margin of error involved.
1018969
Just google it dear..very simple..calculate mean and average the sum of square of variation of each score from the mean.\nhttp://davidmlane.com/hyperstat/A16252.html
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?
108492
Assuming you are talking about standard deviation...\n\nSigma is the square root of variance.  If you need to add sigma's together, they are added by taking the square room of the sum of squares.
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.
674991
For one-sample t-test, \nt = (sample mean - population mean) / estimated standard error. \n\nIn math notation, t = (X bar - mu not) / (s/sqrt(n))\n\nAssumptions: (normal population or sample size, n, is at least 30) and population standard deviation is unknown.\n\nThere are other variations of t-tests like two-sample(pooled and unpooled) and paired ...