A Probability Question about Confidence Intervals?

Question

A confidence interval estimate for the means is (3,9). If the level of confidence is .95, the sample size is 36, then determine
a) the maximum margin of error
b) the standard deviation
c) the sample mean

I'm stumped

Answer 1

(a) The margin of error is (9-3)/2 = 3 (the estimate says that the population mean is 6 +- not more than 3 with probability 95%).
(c) The smaple mean must be 6 = (3+9)/2 (a confidence interval is constructed so as to have the smaple mean xbar in the middle).
(b) At .95 as the confidence level, the confidence interval extends from xbar - 1.96*std error to xbar + 1.96*std error (IOW, 95% of the normal prob lies within 1.96*std dev. from the mean. Here I am assuming that the normal distr. was used to construct the confidence interval. The sample is large enough for this. For a small sample, you would need to use the t-distr.) So std error (the std dev. of the sample mean) is 3 (half the interval's length) divided by 1.96 = 1.53. Also, std error = sigma/sqrt(n), where sigma = population std dev. and n = sample size. So in this case 1.53 = sigma/sqrt(36) or sigma = 1.53*6 = 9.18.

I'm assuming the question (c) asks for the population std dev. sigma (the std dev. of the sample mean, which is sigma/sqrt(n) and goes down as sample size n increases, is called "std error" precisely to avoid confusing the two).

Answer 2

If the confidence interval is [3,9], then the mean is 6 (halfway between, 6 ± 3). If the confidence level is .95, then (off the top of my head) that's 2 standard deviations each way from the mean, which makes 1.5 one standard deviation. To answer a) I'd have to find my stat book in one of over 40 boxes stacked in my garage.

Answer 3

enable p be the probability that one attempt score is larger than 500. enable X be the attempt scores. Assuming that X is often allotted you have: X ~ commonplace(? = 496, ? = 108) p = P( X > 500) p = P( Z > (500 - 496) / 108) p = P( Z > 0.03703704) p = 0.4852277 enable Y be the kind of tests that have a score of 500 or greater effective. Y has the binomial distribution with n = 5 trials and fulfillment probability p = 0.4852277 P( Y = 5) = (p) ^ 5 = 0.0268985 notes with reference to the above: For any commonplace random variable X with propose ? and oftentimes going on deviation ?, X ~ commonplace(?, ?) you may translate into oftentimes going on commonplace instruments by: Z = (X - ?) / ? the place Z ~ commonplace(? = 0, ? = a million). you may then use the oftentimes going on commonplace cdf tables to get possibilities. in maximum circumstances, if X has the binomial distribution with n trials and a fulfillment probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (a million-p)^(n-x) for values of x = 0, a million, 2, ..., n P[X = x] = 0 for the different fee of x. that's stumbled on by observing the kind of mix of x products chosen from n products and then an entire of x fulfillment and n - x disasters. Or, to be greater precise, the binomial is the sum of n self sufficient and identically allotted Bernoulli trials. /// === question 2 === till now the rest, for small pattern CI's you will possibly desire to have commonplace data. small pattern self belief periods are used to discover a region wherein we are one hundred (a million-?)% beneficial the genuine fee of the parameter is contained in the era. For small pattern self belief periods with reference to the propose you have: xBar ± t * sx / sqrt(n) the place xBar is the pattern propose t is the t score for having ?% of the data contained in the tails, i.e., P( T > |t|) = ? be conscious that the student t has n - a million stages of freedom sx is the pattern oftentimes going on deviation n is the pattern length subsequently you have: propose height = seventy one.52941 oftentimes going on deviation = 2.139441 there are 17 samples so which you have sixteen stages of freedom contained in the t statistic. the severe fee of the t statistics for a95% CI with sixteen stages of freedom is: 2.119905 the CI is: seventy one.52941 ± 2.119905 * 2.139441 / sqrt(sixteen) = (70.42941, seventy two.62941)