Statistics (Confidence Interval)?
Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval. For large sample confidence intervals about the mean you have: xBar ± z * sx / sqrt(n) where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size The sample mean xbar = 58.3 The sample standard deviation sx = 3 The sample size n = 25 The z score for a 0.95 confidence interval is the z score such that 0.025 is in each tail. z = 1.959964 The confidence interval is: ( xbar - z * sx / sqrt( n ) , xbar + z * sx / sqrt( n ) ) ( 57.12402 , 59.47598 ) == -- == --- == -- == -- == -- == B) The sample mean xbar = 58.3 The sample standard deviation sx = 3 The sample size n = 100 The z score for a 0.99 confidence interval is the z score such that 0.005 is in each tail. z = 2.575829 The confidence interval is: ( xbar - z * sx / sqrt( n ) , xbar + z * sx / sqrt( n ) ) ( 57.52725 , 59.07275 ) == -- == -- == -- == C) To find the sample size needed for a confidence interval of a given size we need only to concern ourselves with the error term of the CI. We know that the interval is centered at xbar so we need to find the value of n such that z * sx / sqrt(n) = width. The z-score for a 0.99 confidence interval is the value of z such that 0.005 is in each tail of the distribution. z= 2.575829 The equation we need to solve is: z * sx / sqrt(n) = width n = (z * sx / width) ^ 2. n = ( 2.575829 * 3 / 1 ) ^ 2 n = 59.71407 Since n must be integer valued we need to take the ceiling of this solution. Always take the ceiling so that the size of the CI will be correct. n = 60
College stats? confidence interval?
Wife doesn't want kids: The 1996 GSS asked, "If the husband in a family want children, but the wife decides that she does not want any children, is it all right for the wife to refuse to have children? Of the 708 respondents, 576 said yes. A) Find a 99% confidence interval for the population proportion who would say yes. Can you conclude that the population proportion exceeds 75%? Why? B) Without doing any calculation, explain whether the interval in (a) would be wider or narrower than a 95% confidence interval for the population proportion who would say yes. A) First, calculate the sample proportion: p = x / n Where: p = population proportion x = number of favorable outcomes n = number in the sample p = 576/708 = 48/59 = .813559 CI = p ± z * sqrt(p * (1- p)/n) where: CI = Confidence interval p = sample proportion z = z value of .95 n= number in the sample CI = p ± z * sqrt(p * (1- p)/n) CI = .813559 ± 1.96 * sqrt(.813559 * (.186441)/708) CI = .813559 ± 1.96 * sqrt(.000214238) CI = .813559 ± 1.96 * .014637 CI = .813559 ± .028688 CI = .784871 to .842245 We can conclude that the population proportion exceeds 75% because the lowest number in our confidence interval is greater than .75 .784871 > .75 Good luck in your studies, ~ Mitch ~ P.S. - I may be reaching a bit because it's been a few years since I took Stats I. - Hopefully, I've given you the correct answer.
Statistics question with confidence intervals?
We are given the interval [21%, 24.9%]. Since the confidence interval is defined as CI = p ± z*SE we can conclude that p itself must be halfway between the two ends of the interval as z*SE fixed. Therefore: p = (21% + 24.9%) / 2 = 22.95% = 0.2295 q = 1 - p = 1 - 0.2295 = 0.7705 Furthermore, by comparing p to either of the two limits of the interval we find the value of z*SE: z*SE = 0.249 - 0.2295 = 0.0195 Now, using your calculator or a table of values you find that z* ≈ 1.96 for 95%, hence we have got the following: 0.0195 = 1.96 * sqrt(pq / n) = 1.96 * sqrt(0.2295 * 0.7705 / n) Rearranging this to solve for n: 0.0195 / 1.96 = sqrt(0.2295 * 0.7705 / n) (0.0195 / 1.96)^2 = 0.2295 * 0.7705 / n n = (0.2295 * 0.7705) / (0.0195 / 1.96)^2 ≈ 1786
What are confidence intervals used for in statistics?
There are three common definitions of confidence intervals, and people often confuse them.A Bayesian confidence interval is a betting range. For example, in a study of a new weight-loss program, participants lost an average of five pounds more than a matched control sample of people. If a Bayesian says a 95% confidence interval for the long-run average difference caused by the program is 2 pounds to 6 pounds, that means she’s willing to bet $95 if the true long-term average is outside that range in order to earn $5 if the true long-term average is inside that range.For the same confidence interval, some frequentists would say it mean if you did a very large number of similar experiments, 95% of them would result in a confidence interval that contained the true mean, and 5% of them would result in a confidence interval that did not include the true mean.Another type of frequentist would say, for any true mean less than the minimum of the interval (that is, less than 2 pounds) the probability of observing a sample mean of 5 pounds or greater is less than 5%; and for any true mean above the maximum of the interval (that is, greater than 6 pounds), the probability of observing a sample mean of 5 points or less is less than 5%; and for any true mean inside the interval (that is, from 2 to 6 pounds) neither of those statements is true (that is the probability of observing a sample mean less than 5 pounds is more than 5% and less than 95%).There is a fourth meaning which few people defend in theory, but is easier to compute than the three above and is often used. It says that if the true mean is the observed sample mean (5 pounds) then there is a 95% chance of observing a sample mean inside the interval (from 2 to 6 pounds).In simple cases, these four definitions often result in similar intervals, but there are cases in which the intervals can be significantly different. So you should always ask what type of interval you have been given.
If the bottom of one 95% confidence interval for Population 1 just barely overlaps with the top of another 95% confidence interval for Population 2, then does it necessarily imply that the two samples have means that are statistically indistinguishable from each other?
You're making a binary decision, "Is B better than A?", so what you want is a hypothesis test. If you wanted to figure out the quality of your values of 14% and 17%, you would use confidence intervals. A hypothesis test is also better since you'll be calculating a p-value, which answers your follow up question. A p-value is the probability of getting the data you actually got, if B is really no better than A. So it's the probability that you have seen an improvement due to random chance, rather than because B is actually better than A. If the p-value is very small, than you can safely conclude that B is better than A. However, keep in mind that a hypothesis answers a question of statistical significance, not practical significance. You could be almost certain that B is better than A, but if the improvement is small, it might not be worth additional system resources or whatever.How do you perform the hypothesis test? 1) State your hypotheses:Null hypothesis (what we want to disprove): B is no better than A. Alternate hypothesis (what we want to prove): B is better than A.2) Now we do a statistical version of proof by contradiction:Assume null hypothesis, then find the p-value, the probability of current result given null-hypothesis, called a "two-proportion z-test". The mechanics of the test are given here: Difference in Proportions. (Use the two-tailed test, since we are testing for an improvement, not simply any difference.)Of course, hypothesis tests and confidence intervals are connected. So you could calculate a 90% confidence interval for the proportion difference. If 17-14 = 3 lies outside to the right of the interval, then you would conclude at 95% that B is better than A. However, I do not think you can directly use the two different confidence intervals you calculated-- you need a confidence interval for the difference.
95 Percent Confidence Interval Statistics?
1. An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 216 of these passed the probe. Assuming a stable process, calculate a 95 percent (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.) 2. A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.49. Calculate a 95 percent large-sample CI for the true average percentage elongation μ. (Round your answers to three decimal places.)
Are 95% confidence intervals more meaningful than statistical significance?
These two concepts are closely related but tell different things.See my answer of What's the difference between significance level and confidence level?
What are some every day uses for confidence intervals in statistics?
For most things in daily life, you'll never need a formal CI. However, we humans use informal CIs all the time. A formal CI is a mathematical function; an informal CI is a pretty good guess of a range of values. For instance, you may not know how much money is in your pocket down to the penny, but you do know if it's between $100 and $1. That's an informal one, and you'd get pushback from many statisticians about whether it's a CI.But let's say that you want to maintain a certain range of cash, so at the end of every day you count all the cash you're carrying, and after several months you calculate the mean and standard deviation. Then you calculate out a CI at whatever confidence level you like. That's a pretty good indicator for how much money you're likely to carry on average from here forward, all other things being equal. The months of data collection can be seen as a sample of all the months to come, and even arguably months from the past too. The same process can be undertaken with any kind of data in your life: tire pressure, spending on video games (and video scores), rainfall, time to drive to work or school, prices of grocery items, volume of music heard from next door (in db's), electric bills, prices of gasoline, time spent studying, amounts spent for lunches, times running the 100-meters, time spent with a BFF, and on and on.
How to interpret 95% Confidence Interval in a statistics?
The first answer above is a common misconception. Firstly, it is NOT the random variable itself that the C.I. is intended to include - rather, the C.I is intended to include a population parameter, usually the mean, which is assumed to be fixed. Secondly, because a population parameter doesn't bounce around, it cannot fall into the C.I. some percentage of the time. If a number of experiments were conducted, the C.I. would do the bouncing, bracketing the true value of the parameter some percentage of the time. From Wikipedia: "A confidence interval does not predict that the true value of the parameter has a particular probability of being in the confidence interval given the data actually obtained. (An interval intended to have such a property, called a credible interval, can be estimated using Bayesian methods; but such methods bring with them their own distinct strengths and weaknesses)." "A confidence interval with a particular confidence level is intended to give the assurance that, if the statistical model is correct, then taken over all the data that might have been obtained, the procedure for constructing the interval would deliver a confidence interval that included the true value of the parameter the proportion of the time set by the confidence level." "More specifically, the meaning of the term "confidence level" is that, if confidence intervals are constructed across many separate data analyses of repeated (and possibly different) experiments, the proportion of such intervals that contain the true value of the parameter will approximately match the confidence level; this is guaranteed by the reasoning underlying the construction of confidence intervals." So, the short answer to your question, "I have two numbers (52.2, 90.7) - what do they mean?" is: The probability that the interval, (52.2, 90.7) contains the true value of the population mean is 95%.
How do you interpret hazard rates in statistics? Confidence intervals too?
Definition of the hazard ratio: The hazard ratio compares two treatments. If the hazard ratio is 2.0, then the rate of deaths in one treatment group is twice the rate in the other group. It looks like your hazard ratio is .94? That would mean for every .94 deaths in one group there is 1 death in the other or for every 94 there are 100 in the other But that is just a (point) estimate. Then you have a margin of error for that .94 point estimate. It looks like the margin of error was .02 subtract .02 from .94 and add .02 to .94 to get the confidence interval .92-.96 You are always looking for the parameter of the population. But you cant survey the population (too large) so you take a sample. You take a sample find the hazard ratio and say that the hazard ratio of the sample was .94 and so the hazard ratio of the population is .94 plus/minus. The plus/minus is the margin of error and when you do the plus/minus you get the confidence interval. There should have been an associated % (95%?) with the confidence interval. If you increase the confidence level (say to 99%) then your margin of error will increase, and so will your confidence interval. If it's decreased to 90%, your CI will decrease.