When the sample proportion is 0.52, Confidence Interval 95% and Populaton Size 900\nLower Limit0.487359477\nUpper Limit0.552640523\nUsing these figures can we accept the poll result? Kindly help me to interpret the result?
203120
your confidence interval of .4873 to .5526 indicates the range of the true mean with 95% confidence.  Since values above and below .50 are in this range, the poll is a tossup.
436889
when the probability is 0.2:\n\nP((β-12%)*Sqrt(n)/σ>=1.301)=20%\n1.301 is the T value under degree of freedom of 44 and α =10%\n\nI did not pay attention in leaasons of statistics, I dont know if it's correct.
479987
The bigger the sample size, the more accurate the estimate.\n\nSay I was estimating the percentage of red m&m's in a pack.\n\nIf I sampled 10 m&m's, I might I might find that the sample contains 1 red one. This could lead me to think that a pack contained 10% red m&m's.\n\nHowever, if I sampled 100 m&m's, I might find that the sample contains 15 red m&m's, leading me to think that a pack contains 15% red m&m's. This answer is impossible using the sample size of 10, as you can't get 1.5 m&m's of a particular colour.\n\nIf I sampled 1000 m&m's I might find that I have 143 red ones, leading me to think that the pack contains 14.3% red m&m's.\n\nSo, as you can see, the bigger the sample size, the closer the estimation comes to the actual answer.
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)
247098
33% is equal to approximately 3 out of 10(3.3)\nTherefore, if six out of ten were to be selected, that would equal 66% of ethnics - more than twice the probability.\nIt follows then that in iself, the odds would be just over two-to-one against this happening.\nSo the probability of more than of no more than six would be two tone ON.\nDoes that make sense?
258538
would calculate your chance of me calculating a statistics question for you at 1234.367:1
254561
It is actually binomial with parameter p=0.5 and n=400. Since 400 is quite a large number, u can approximate this by the Normal distribution with the same mean and variance, which are, np=400 and np(1-p)=100.\n\nX ~ N(200,100)\n\nu wanna test that if 216 is within 95% interval.\n\nVariance=100\nSDev=10\n\n216-200=16,\n16/stdev=1.6\n\nso, 216 is 1.6 stdev from the mean 200. which is within +/-1.96(95% confidence interval. hence, the null hy is accepted, with 95% significance level.\n\n\n\nbtw, drbinsomething's solution is equivalent to mine. he directly test P as a random variable while I tested mean nP as the random variable. He didn't mention it, but he took P to be asymptotically Normal as I did. '1.96' only applies when the distribution is Gaussian/Normal.
418914
A representative sample should have the same characteristics of the population from which it is drawn.  For example, if you wanted to find out the average height of all Americans, a sample that included mostly basketball players would NOT be representative, nor would a sample that included only women. Instead, a representative sample would include people of various races, both sexes, and all ages--so it would have the same variety of heights that can be found in the whole population.\n\nStatistical theory says that the type of sample most likely to be representative is a random sample, a sample where random numbers or games of chance assure that every member in the population has an equal chance of being selected.  Since this is often impractical to do, other types of "good" (usually representative) samples include systematic samples (where you number off and, for instance, call every 100th name in the phone book) and stratified samples (where you carefully include various sub-groups, like people of all different races in your sample).\n\nAmong the least representative samples are internet polls, where anyone who wants to can answer.  Usually only those with strong opinions on a subject bother responding to such polls.
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!
528866
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.
652503
Use a t-test since the sample size is small.\nThe discrepancy measure is given by new mean - old mean = 8 - 6 = 2\nThe test statistic is (discrepancy - hypothesize value)/standard error\nhere, the hypothesized value is 0, ie: no difference between the 2 techniques\nthe standard error is 4.2/sqrt(12) = 1.212\n\nso, T = 2/1.212 = 1.649\n\nusing the t-test, the p-value is P(t_11 > 1.649) = 0.06369\n\nso, the conclusion is to not reject the hull hypothesis since the p-value is greater than alpha.
439527
I'm wondering why you created the data 1-3500?  I would think you would use the set of data available and figure the standard deviation and average.  With that information, you would be able to predict tomorrows value at a certain confidence level.  I forget is that the p test or the t test.  its been a while since I took statistics.
116117
If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
233058
> How many surveys should be sent out to get a given response rate?\n\nObviously you can't control the response rate by sending out more or fewer surveys.  The response rate is determined by how many of the recipients choose to respond.\n\nIf you believe each response is independent of any other and that the responders are a representative sample, then variance = N p q = 390 * (1/2) * (1/2) = 97.5 (for a standard deviation of 9.87 which is about 2.53% of the 390).\n\n95% is about +/- 1.96 standard deviations, right?  That exceeds +/- 3% ... it is more like +/- 4.96%, you need more responses.
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
186366
you're kidding right? I hope you don't use any of the data you collect on here for a school project. This is about the worst sample you can have for a survey. Your results are going to be pretty much worthless.
616006
the others are correct, it is 1/120.\n\nIf you are not sure how to do the calculation, it is 3,000,000 divided by 25,000.\n\nOr you can cut it down to:\n\n  25  \n------\n3000\n\n  5\n-----\n600\n\n\nThis cancels down to 1/120.\n\nThat means that the probability is 1 in 120, or 1:120  (one person out of every 120 people has VD) or 8.33% of the population has VD.   (The 3 is recurring - you show that by putting a dot above the last 3)\n\nWhy did you need to ask us what the answer was when you could work it out yourself?