# do you know how to do this statistics problem? Need help please!?

this is the problem:

Compute the confidence interval for a population mean given the following data: Confidence level = 96% x-bar = 13.5, standard deviation = 3.04, and n = 45.

### 2 Answers

- Dr JLv 51 decade agoBest Answer
the CI is centered at xbar and has a margin of error

E= alpha*s/sqrt(n)

where alpha is the critical value for a two-tailed test

Since n=45 is > 30, we might just assume a normal distribution for xbar, so alpha=2.0537.

E=2.0537*3.04/sqrt(45)=0.9307

CI is (xbar-E, xbar+E)

If you do this on a TI-83, use Tinterval or Zinterval

stats

(12.54, 14.46) Tinterval

(12.57, 14.43) Zinterval

Technically, Tinterval is better, but the difference is slight.

If you are drawing from a normal distribution and the variance is unknown, then T is the way to go.

- wilmaLv 43 years ago
(r,b) = A(3,4) = B(4,4) P(X = n) is the risk of n pink ball in container A after the technique. P(Ypq) is the risk whilst container Y is chosen as 1st container, then ball of the colour p transfered from Y into the different, and ball of the colour q is decrease back into container Y. P(X = 2) = P(Arb) + P(Bbr) = a million*3*4/2*7*9 + a million*4*3/2*8*8 = 6/sixty 3 + 6/sixty 4 = 127 / 672 P(X = 3) = P(Arr) + P(Abb) + P(Brr) + P(Bbb) = a million*3*5/2*7*9 + a million*4*5/2*7*9 + a million*4*4/2*8*8 + a million*4*5/2*8*8 = 5/18 + 9/32 = 161 / 288 P(X = 4) = P(Abr) + P(Brb) = a million*4*4/2*7*9 + a million*4*4/2*8*8 = 8/sixty 3 + a million/8 = 127 / 504 answer, E(pink marbles in A) = ? nP(X = n) = 2 * (127 / 672) + 3 * (161 / 288) + 4 * (127 / 504) = 127 * [a million/336 + a million/126] + 161/ninety six = (127/40 two) * [a million/8 + a million/3] + 161/ninety six = (127/40 two) * (11/24) + 161/ninety six = (127/21) * (22/ninety six) + (161/ninety six) * (21/21) = [127 * 22 + 161 * 21] / ninety six * 21 = 6175 / 2016 = 3.062996 = 3.063