How to calculate standard statistic for confidence interval?

Confidence interval for Population Mean is

Sample Mean +/- Confidence coefficient * SD/sqrt n

Confidence coefficient is the critical value of t for 98% confidence for 24 degrees of freedom = 2.492 (because n < 30, degrees of freedom = n-1 = 25-1 = 24)

CHOICE (C) is the answer...

Is it asking for the t-score? If so, then you will need to use a t-score distribution chart.

Since your sample size is 25, there are 24 degrees of freedom.

The corresponding t-score for a 98% confidence interval at 24 degrees of freedom is C. 2.492

http://www.medcalc.org/manual/t-distribution.php

permit p be the threat that one attempt score is larger than 500. permit X be the attempt rankings. Assuming that X is often disbursed you have: X ~ established(? = 496, ? = 108) p = P( X > 500) p = P( Z > (500 - 496) / 108) p = P( Z > 0.03703704) p = 0.4852277 permit Y be the type of exams that have a score of 500 or greater suitable. Y has the binomial distribution with n = 5 trials and success threat p = 0.4852277 P( Y = 5) = (p) ^ 5 = 0.0268985 notes with regard to the above: For any established random variable X with advise ? and classic deviation ?, X ~ established(?, ?) you are able to translate into familiar established gadgets by ability of: Z = (X - ?) / ? the place Z ~ established(? = 0, ? = one million). you are able to then use the familiar established cdf tables to get possibilities. usually, if X has the binomial distribution with n trials and a success threat of p then P[X = x] = n!/(x!(n-x)!) * p^x * (one million-p)^(n-x) for values of x = 0, one million, 2, ..., n P[X = x] = 0 for the different fee of x. it is stumbled on by ability of watching the type of combination of x gadgets chosen from n gadgets and then a entire of x success and n - x disasters. Or, to be greater precise, the binomial is the sum of n self reliant and identically disbursed Bernoulli trials. /// === question 2 === earlier the rest, for small pattern CI's you will desire to have established records. small pattern self belief periods are used to locate a area wherein we are one hundred (one million-?)% confident the actual fee of the parameter is interior the era. For small pattern self belief periods with regard to the advise you have: xBar ± t * sx / sqrt(n) the place xBar is the pattern advise t is the t score for having ?% of the records interior the tails, i.e., P( T > |t|) = ? word that the student t has n - one million stages of freedom sx is the pattern familiar deviation n is the pattern length in this occasion you have: advise top = seventy one.52941 familiar deviation = 2.139441 there are 17 samples so which you have sixteen stages of freedom interior the t statistic. the intense fee of the t information 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)