Question

Problem(5) (a) (1 pt) Let Z~ Normal(0, 1). Recall the definition of z-value, i.e., P(Z > zr) = r. Find the probability of P(-70/2 < 3 < 2a/2). (b) (4 points) Let X1, X2, ... , Xn be a random sample from some population with (un- known) mean u and (known) variance o?. Based on the Central Limit Theorem and part (a) above, show that the confidence intervals for the population mean y can be estimated by <ll< X + = 1-a. Vn 70/2 n o o PX - 7/27 p(* (c) (3 points) Find za/2 for each of the following: (i) a = 0.10 (ii) a = 0.05 (iii) a = 0.01 (d) (3 points) What is the confidence level of each of the following confidence interval for u? (i) * £1.960 (m) (ii) + +1.645 () (ii) 7 +2.575 (2)

(b) (c) and (d) please

Answer 1

The solution.is below solved step by step

16) Given XI, X₂ . Xn with ñ a random sample N = mean Nos variance size of sample ME X = sample Mean for A the sample The Central Limit Theorem states that for Large samples, the distribution approximately distributed with a distributed with a mean:- Nx = p = X and standard deviation means is al TE I a Now, Z= X-N N(0,1) (This means X approximately follows the Normal (0,1) distribution) To obtain 100(1-6)9%. Confidence Confidence Interval

As we know, nu part a Pl-<ZcKay) - 1-4 pl-2012 <X-N olun 7 - H2 ) 1-2 pl - 1-2 In -24/2 7 7 EX-ps Zala Adding it on both sidu (-7-26 ES-5-7-24) P ( 5 + 2 a 3 p 2 X 2 ) = 1-0 1- or = for P ( X-7, MA IN S X+ Z2 ) Therefore the 100(1-«)"}. Confidence Interval Nis X-24/2 X + Zapz toho And the confidence Interval for the population ľ can be estimated by P(x-72 or <N<X + L Fors) 7 h 7 mean 1-2

(c) To 22/2 find (0) 2 = 0.10 for 0.05 Using 7-table, the critical value or the area under standard Normal curve. Loos 1.64 lii) X= 0.05 - 0.025 20.025 1.96 & Critical value at 0.05 (iii) <= 0.01 a = 0.005 Looos : 2.58 – Critical value at 0.005

lo) To find the confidence level for N 0) MI 1.960 1-9601 Here Lala = 1.960 with the (componing formulae durived And 100/1%)% confidence given by in part ) hend is At kahit an Using the y-table i we observe that the area tt under the curve Corruponding to the row and column value is 0.9150 =1-a Thus confidence devel 10.9950).100 % 21.5"). similarly, () a I 1.645 SLE) (1-x) = 0.9495 –959%

mu d n 9919%. (ii) 차 2575 285 이트 We have (x = 19941 Thus