Could somebody explain how to work this problem? (Confidence Interval - Statistics)?
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.
17841
You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
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/
860672
Use Statistics\n\ntake your numbers and find the standard deviation this is your uncertainty (±.23). Average all the numbers this is your beaker mass (106.49). Beaker mass plus uncertainty is best estimate with uncertainty (106.49±0.23)
729266
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)
455303
To obtain 95% confidence intervals for the population variance we use the result:\n\n[(n-1) S²] ÷ (sigma²) ~ chi-square(n-1)\n\nWhere sigma² is the population variance\nand S² is the sample variance\n\nFor the males we have n = 40, so using the result we have a chi-square(39).\n\nFor a 95% confidence interval we want the values that contain 95% of the distribution between them (the lower limit will have 2½% below it and the upper limit will have 2½% above it).\n\nUnfortunately my tables don't have a chi-square 39 listed - so you will have to look these up in your tables - I will use the chi-square 40 results just to show you the method.  The chi-square 40 results are 24.43 and 59.34.\n\nNow we have these values (called "critical values") - we stick them into our thrilling result:\n\n[39 × 11.3²]/(sigma²) = 24.43\n\nand:\n\n[39 × 11.3²]/(sigma²) = 59.34\n\nRearranging each of these gives us the sigma²:\n\n(sigma²) = [39 × 11.3²]/24.43 = 203.844\n\n(sigma²) = [39 × 11.3²]/59.34 = 83.922\n\nNow you want the confidence interval for the standard deviation (not the variance), so we need to root these:\n\nsigma = 14.3\n\nsigma = 9.16\n\nSo our 95% confidence interval is (9.16, 14.3).\n\nYou can do a very similar thing for the females and then can compare these two limits to see if there is any overlap between them.\n\nHope this helps.
806818
1. If you know the variances you'd know the mean, because you need to calculate the mean to find the variance. Newer stat texts have finally started to get smart and quit putting stupid topics like this in there!\n\n2. If you suspect that the population variances are equal, you can combine the samples to get one estimate of that variance. That's called 'pooling.' When variances are equal use the formula that combines those variances for the calculation.\n\n3. The p-value comes from this calculation so you don't use it to determine which formula to use. Your background knowledge is what you would use to determine whether variances are likely to be equal.\n\n4. It's never wrong not to pool. Pooling when appropriate does give a test more power, but not pooling is always OK.
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
986574
if you are using a TI-83, use the Z-Interval test funcition.\n\nManual working:\n\nSampling error = 2.06(SD/sqrt n)\n\n\nlower boundary =  20 - 2.06(3/sqrt 25) = 18.764\nupper boundary =  20 + 2.06(3/sqrt 25) = 21.236\n\n96% confidence limits is\n18.764 < mean < 21.236
262206
range = mean +- (t/sqrt(N))*st.dev.\n\nWhere t is found in a chart according to the confidence level in question, N is the number of data points, and st.dev. is the standard deviation of the data set.
660150
That would not be possible without the sample values or some other information.  You can make some statements about possible sample means.  For example, since the standard error of the sample mean would be 0.29/sqrt(25) = 0.058, then\n\nThe probability that the sample mean will be between 3.02-0.058 and 3.02+0.058 is 68%.\nThe probability that the sample mean will be between 3.02-0.116 and 3.02-0.116 is 95%.\n\nand so on.
650317
If you have a computer program that can compute binomial probabilities, you can use that.\n\nIf you are doing this by hand, however, I would suggest using the normal approximation to the binomial here.\n\nn = number of trials = 200\np = prob. of success = 1/6 if we consider getting a five to be a success.\n\nThen np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal.  This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.\n\nNow for the continuity correction.  The rectangle over 40 goes from 39.5 to 40.5.  Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.\n\nz = (39.5-33.33)/5.27 = 1.17\nArea to the right of 1.17 on the standard normal is 0.121.
758530
I think the answer would be 1), though it is a stupid question. The right way to frame the question would be around a confidence interval.
488365
Try again Paul C.  The population doesn't need to look normal.\n\nThis all comes down to the Central Limit Theorem.  Let's assume that I take many samples of size 1004 from the population.  Well, each sample will have a percentage that is different from the population percent, because samples are going to look different from the population.  Also, each sample is going to look different from each other, because they are different samples.  If we were to draw a graph that looked at the "behavior" of these many samples, we would notice the following:\n\n1.  If I were to average out all of these sample percents, I would get the population percent.  Hopefully you can see why this would happen.  Sample percents will be around the population percent, but just be off by a little bit.\n\n2.  If I were to look at the spread of the percents, the spread would be smaller as bigger samples were taken.  In other words, if I had taken samples of size 500, the sample percents would be more spread out than for samples of size 1004.  That is because if were take larger samples, then we have more information about the population, so we should expect to get closer to the population percent.\n\n3.  As I take larger and larger samples, the shape of the graph of sample percents would look more and more like a normal curve.  A normal curve is shaped like a bell, where the middle of the bell is were the population percent is.\n\nSo Paul C, it doesn't matter that the population doen't look normal; all we need is a sample that is large enough.\n\nNow the margin of error on polls is typically going to be\n\n2*sqrt(p(1-p)/n))*100%\n\nwhere p is the sample proportion (the sample percent divided by 100%) and n is the sample size.  It can be shown that this margin of error would be at its largest if p = 0.5, so the largest the margin of error would be in this case would be\n\n2*sqrt(0.5*0.5/1004)*100% = 3.2%\n\nThis is typically associated with 95% confidence.  This basically means that if I were again to take many, many samples of size 1004,  95% of the sample percents would be within a margin of error away from the true population percent.\n\nThis of course depends on the sample that is taken.  The online polls that you see would not be good estimates for the population percent.  Below I have a link to a story that MSNBC has on each of its online polls.  The title is "How 1,000 people can be more representative than 200,000."  I think that it is a very good explaination of this topic.
458174
Actually, statistics books are notorious for being written in such a way that no normal human can understand them. The trick, generally, is trying to find a way to relate the problems to something you're interested in. For example, if you like basketball: one possible sampling distribution would be the number of free throws individual players make over a season. You could take a sample of, say, 10 of the players and find the mean and standard deviation. You could compute a 95% confidence interval with that information, which would allow you to say that you're 95% confident that the average number of free throws made by NBA players in a season will be between this number and that number. You could find the probability that a given player would make at least a certain number of free throws in a season by finding the z-score and looking it up on the normal probability distribution table, etc. If you're really struggling with the terminology, I recommend visiting your math help center on campus (most of them have one), or contacting a tutor.
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.
803485
I think it's p=problem, i=infere' g=gather info? t=test e=expirement\n           good luck