how do I calculate the 95% confidence interval and p value for two rate ratios? rate per 1,000 persons?
438145
Chisq test for proportional differences or a One-way Chi-sq same thing though.  The difference in your proportions +/- 1.96 x standard error.
761495
The first sample proportion is p1 = 30/100. Its variance is (30/100)*(1-30/100)/100.\n\nThe second sample proportion is p2=60/80. Its variance is (60/80)*(1-60/80)/80.\n\nWhen you look at the difference, p1-p2, and assume that the two proportions are independent then the standard deviation is\nsqrt(var1+var2)=\nsqrt(  (30/100)(1-30/100)/100 + \n          (60/80)(1-60/80)/80)\nwhich is approximately 0.06666146.\n\nHere you can use the normal approximation for both proportions, and the difference is \n30/100-60/80 = -.45.\nIf you use a two sided test then the z-score is -.45/0.06666146 = -6.75, and the p-value is 1.473087e-11. It is highly likely that the difference is zero! Please note that this last part is not quite right. If you want to do a formal hypothesis test then you need to do a pooled estimate.
1097980
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
865854
I assume that you mean that 1.06*se was added to xbar.  I think that's what you mean.  Otherwise your interval was not well defined.  The minimum value is larger than the maximum.\n\nSince you have a normally distributed population with a known sd sigma, then xbar would have a normal distribution.  You subtracted 0.96 standard errors and added 1.06 standard errors, so you would need to look up the area between -0.96 and 1.06 on the standard normal table.\n\nP(xbar - 0.96(sigma/sqrt(n)) < mu < xbar + 1.06(sigma/sqrt(n)))\n= P(- 0.96(sigma/sqrt(n)) < mu - xbar < 1.06(sigma/sqrt(n)))\n= P(-0.96 < (mu-xbar)/(sigma/sqrt(n)) < 1.06)\n= P(-0.96 < z < 1.06)\n\nThat would turn out to be 0.8554-0.1685 = 0.6869.  That would be the probablity that mu would be in one of these confidence intervals.\n\nTo find the probability that mu will be in at least 690 of the 1000 intervals, you would need to recognize that this is a binomial.  There are n = 1000 trials, each trial has probability of success p = 0.6869, and the trials are independent.  np and n(1-p) are both greater than or equal to 5, so the sample size is large enough.  Using the normal approximation to the binomial, the mean would be 686.9 and the standard deviation would be sqrt(1000*.6869*(1-0.6869)) = 14.67.  Using the continuity correction, I would need to start at a value of 689.5.\n\nz = (689.5-686.9)/14.67 = 0.18\n\nLooking that up on your table, the area to the right is 1-0.5714 = 0.4286.  There is a 42.86% chance of getting at least 690 intervals that contain mu.
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.
789510
Well a prediction interval follows a Normal Distribution with mean\nB0+B1*x' and variance 1+1/n+[(x'-xbar)^2]/sum(xi-xbar)^2\n\nB0 = beta sub 0 ... 1... etc\nx' is the value you pick to predict at\nand sum(xi-xbar)^2 = SSxx = sum(x^2)-nxbar^2\n\nso if you wanted to create a 95% prediction interval with this information you would have to use the formula\n\nB0+B1*x'(+/-)Z(alpha/2)sqrt{1+1/n+[(x'-xbar)^2]/SSxx}
458114
The confidence interval is the gap you set to determine a level of certainty in your answer. For example, in public health, we usually set alpha=0.05, meaning we have a confidence interval arbitrarily set to 95%.\n\nIf the p-value we get for whatever statistic we just crunched is less than 0.05, then we say that we are at least 95% certain of the validity of our answer. The lower out p-value is, the more certain we are. If we get p=0.00002, then that's 99.99998% confidence, which is higher than the 95% threshold we set.\n\nIn short, it's a measure of validity for our results.
828638
Assuming the population is normal, to calculate the percentage of subjects obtaining 65 or below, we find the Z score that corresponds to 65:\n\nZ score \n= ( score - mean ) / standard deviation\n= (65 - 80) / 6\n= -2.5\n\nLooking up a Z-score of -2.5 in a 'table of standard normal probabilities' (or using statistical software), gives 0.0062 which is 0.62%. \n\nAs for the confidence interval.. it is 'estimate' +/- 'margin of error'\n80 +/- 1.95 * 6 / sqrt(100)\n80 +/- 1.176\nThe 95% confidence interval (78.824,81.176)\n\n(I.e we can be 95% confident that the true population mean lies between 78.824 and 81.176, and this is based our the sample of 100 subjects which had a mean of 80)
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
545082
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.
713968
The p value is an arbitrary number usually set at .05. It could be anything else, that's just tradition. What you are trying to do is ensure that if a difference in means is noticed, that it is because there is a true difference, not just a difference that can occur by chance. The null hypothesis is saying that there is no difference in means, the alternate hypothesis says there is a difference in means that is greater than chance. That chance is the .05 p value set by you. Therefore, if you're p value is greater than .05, you must accept the null hypothesis, saying that there is greater than a 5% chance that a difference in means is still related to chance alone. Your p value must be less than .05 to reject the null hypothesis.
303069
Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
747582
required rate of return = risk free rate + beta (market rate - risk free rate)\n                                    = 0.075 + 2(0.15 - 0.075)\n                                    = 0.225 or 22.5%
380813
1-i) P(h>200)=P(S>(200-178)/10)=P(S>2.2)=0.5 - 0.4861 = 0.0139 = 1.39 %   (Where S denotes Standard normal distribution value, h the non standard value).\n1-ii) P(h<158)=P(S<(158-178)/10)=p(S<-2)=0.5 - 0.4772 = 0.0228 = 2.28 %\n2)\n  i)P(39.5 < h <40.5)= P((39.5-37)/3.5 < S < (40.5-37)/3.5)\n                   = P( 0.71 < S < 1)\n                   = P( 0 < S < 1 ) - P( 0 < S < 0.71)\n                   = 0.3413 - 0.2611 = 0.0802 = 8.02 %\n  ii)The limits are 38.5 : 39.5\n iii)P(38.5 < h <39.5)= P((38.5-37)/3.5 < S < (39.5-37)/3.5)\n                   = P( 0.43 < S < 0.71)\n                   = P( 0 < S < 0.71 ) - P( 0 < S < 0.43)\n                   = 0.2611 - 0.1664 = 0.0947 = 9.47 %\nNote : S = (h - mean)/sd.
407360
yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
799156
Use Binomial distribution:\n\n(n nCR x) p^x (1-p)^(n-x)\n\nn nCR x = n!/(x!(n-x)!)\np=probabiliy (.10)\nn=patients (10)\nx= desired\n\na) P(X<=2) = P(X=0) + P(X=1) + P (X=2)\nb) 1- P (X<=2)... 1-prevoius answer
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.