explain confidence interval in biostatistics.?
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.
189782
It's a calculation called the r value.  The link below shows how to calculate it.  If you take a college level statistics class you'll learn how to do it with a good calculator.
43798
I'm assuming this is related to Prob/Stat.\n\nLevel of significance, is simply the accuracy at which you want to perform your tests at. \n\nType 1 & 2 are different kinds of errors associated with Probability. Type 1, if I remember correctly, is if you say the null hypothesis is true, when it is in fact not. A Type 2 error, is saying the null hypothesis is incorrect, when it is in fact correct.
108115
What are the main topics of Biostatistics? ANOVA?
479884
Well you don't need a sample if you already have access to an entire population of records! You would use a sample because you can't get a lot of records, and use that sample to give you a rough idea. If you wanted to test a hypothesis though, you should test it at a 5% significance level, or 1% to be really careful. However you need to know if there is an underlying distribution of the data, i.e. normal, etc. If what you have isn't the population of the records then use what you have, the larger the sample size the better! With more values your 'confidence interval' will be narrower, and you can be more certain of the results of your statisical analysis.
861521
you would need to look at table for t values.\nyou are looking for a 95% confidence level, so you would divide .5 (which is 1-.95) by 2 and get .025 that is the column you find the value for t.\nyour n number, the row for the t number, is 15-1 or 14 and 12-1 or 11.\nLook at the table IV in colum t=0,025 and row 14 write down the number. that number is the t value for the first equation.\nLook at table IV in column t=0.025 and row 11, that is your t value for the second equation.\nNow that you have the t values you can plug them into the formula for confidence interval.\n\nGood luck
1109248
First make sure you want a prediction interval for the default rate and not a confidence interval.  The confidence interval will be for the unobservable rate (average rate) while the prediction interval will be for the individual occurances of either default or not default per random 1000 observations.\n\nSecond, make sure you want the interval for the rate per 1000 or the interval for the rate with a sample of 1000: these are actually different!\n\n\nIf prediction interval for the rate per 1000 observation is your poison then that can be approximated by:\n\np +- A*S*sqrt(1+1/n)\nwhere n is the number of samples of 1000\np is the average default rate of the samples\nA is the 95%ile of the student -t distribution (yes! 95th for a 90% PI: 5% on either end of the +-)\nS is the standard deviation of the rate (think binomial).\n\nYou can look up many ways of creating the interval for a binomial proportion but this is the basic theme.
830991
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.
106983
Here is how to do a confidence interval for a proportion.  It is based on the binomial distribution.\n\n100 (1-alpha)% confidence interval for p is:\n[Y/n +or- z(alpha/2)*\nsqrt([Y/n*(1-Y/n)]/n)]\n(wouldn't fit on one line)\nY is the number of "successes," n is the number of "trials" and z(alpha) can be found on a table (Normal Distribution).\np is the proportion you want the confidence interval for.\n\nHere are some tips on how to set up your problem:\nY, the number of successes, will be 300.  n=450.\nalpha is equal to 0.05.  If memory serves me correctly, the value of z(0.025) is 1.96.  \n\nHope this helps.
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
266644
I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
303094
OK.  From any given population, you can take many, many different samples of size n.  For each of those samples, you can find the mean.  So if you take 1000 samples of size 50,  Then you would have 1000 sample means.  Now each of those sample means are probably going to be different from the overall population mean, since samples are going to look different from the population.  Also, each of the 1000 sample means are going to look different from each other because the mean is coming from different samples.\n\nSo if I were to draw a histogram (hopefully you know what that is; it is a bar graph that you can use to look at the distribution of numerical data).  Anyway, if I were to draw a histogram of my 1000 sample means,  You should notice the following three things...\n\n1.  The center of your histogram should be the same as you population mean (so if the mean of the population is 20, then the histogram of my 1000 sample means should center around 20).\n\n2.  As the sample size gets larger, the spread of the histogram should get smaller (so the spread of the means of samples of size 50 would be smaller than for the means of samples of size 25).  This is because with larger samples, you would have more information about the population, and thus you should be able to get closer to the population mean.\n\n3.  As the sample size gets larger, the shape of the histogram will look more and more like a normal curve.\n\nThis is because of the Central Limit Theorem, that states that if you take a sample of size n from any population with a finite variance, the distribution of the mean of the sample would get closer and closer to a normal curve as n gets larger.  The mean of the normal would be the same as the population mean, and the standard deviation of the normal (usually called the standard error) would  be the standard deviation of the population divided by the square root of n.  Notice that as n gets larger, the standard error gets smaller, just as described before.  So typically you can approximate these histrgrams with a normal curve if n > 25.\n\nThis phenomenon is hard to describe in words alone.  Looking at pictures of these histograms that I described would help you understand it better.
578047
This depends on the focus of your study.  What or who are you trying to assess the risk of?  Biostatistics has several methods for assessing risk when it comes to diseases.  Quality Assurance methods assess the risk of malfunction and ways of reducing the likelihood of it happening.  I'm sure there are many other forms of assessing risk as well, so I'm afraid your question needs to be a little more specific.
30830
You can't have an interval, like Male +/- (What?).  You can have a confidence, like 75% probability that respondant is male.
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)
280200
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.
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.