Question

Just need to solve Problem 4

4.2.14. Let X denote the mean of a random sample of size 25 from a gamma-type distribution with a = 4 and 3 > 0. Use the Central Limit Theorem to find an approximate 0.954 confidence interval for pl, the mean of the gamma distribution. Hint: Use the random variable (X - 43)/(432/25)/2 = 5X/23 - 10. 21 TL11C1L

Answer 1

Mean and variance of Gamma, where $Z \sim Gamma(\alpha,\beta)$

E(2)- fr£)d* Z -1 Z E(22 Z - Z 2 (tds f(arzp)) VE) E(2+)_Ee) : «* ..

Defene, P(z7 To fiud za keo Z~N(,1 aue to are(,lowo tad =F) R Stadio Cede Xiamma(ote 2,B) E(X) 2P v(xi) 2B .: Вусьт |2 2 5 No, loo1-) 0 [oso: codae kendl .: 645 [ RCode: uor(o.05, lower tal- F) ] = 644854

P(-145 2) -72 P(-645x -2< 1-645 x 7 (-5.81595+2 < X< 5-8\595 2 21- <в < 3.31595 7-21595 25 3-81595 T-815A5
R Code:

The Normal Distribution R Script Console Terminal Description Density, distn bution function quantile fucion and rardom generaton for the normal distritnion with mean equal to mean and standard devation equal to sd qnorm(0.05,lower. tail = F) [1] 1.644854 Usage dnorm(x, mean = 0, sd = 1, log = FALSE) pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE) qnorm (p, mean = 0. sd 1, lower. tail - TRUE, log. p = FALSE) 0, sd 1) rnorm (n, mean Arguments vector of quantles vector of probabiltbes length (n) > 1 te ength is taken to be te number reaured number of observations vector of means mean