A. Parametrie bootstrap simulation for the distribution of the t-statistic in no
ID: 3062032 • Letter: A
Question
A. Parametrie bootstrap simulation for the distribution of the t-statistic in non-normal samples. Consider the nonnormal (and rather skewed) setup: Let X1....Xn i.i.d. from a Gamma distribution with parameters: shape a and scale-s (both positive and unknown). The mean and variance are denoted E(X)-Imu and Var(X) ,Sigma^2. Let n=9. Assume that you observed a sample mean bar: 10 and a sample variance (using n-1 in the denominator) hatSigma"2-25 i) Construct a 95% confidence interval for the meanmu based on a t statistic and the t tables or R command t Note that the t-tables are NOT appropriate here since the density of the X's is not Normal. However, we can still use the t-statistic,ie., T-sqrtin)^barX-mu)/ hatSigma , but we need to approximate its distribution (and quantiles) by bootstrap. To do this, we proceed as follows (ii) Estimate the parameters a and s using Method of Moments (MOM). Recall that for a Gamma distribution: E(X) = a*s and Var(X) = a*s^2. ii) Use the R command rgamma) to generate B-500 pseudo-samples (each of size n) using the estimated values for the shape - a and scale s. From each pseudo-sample, compute the value of the bootstrapped t-statistic T* sqrtfn^(^barX-ImuhatSigma* where arX* and hatSigma*2 are the sample mean and sample variance of the pscudo-sample, and Imu* is the mean of the bootstrap data generating mechanism (which in this case equals 10). Collect the 500 T* values in a vector called TV. Plot a histogram of TV; does this appear skewed? (iv) Use the R command quantile(TV, c(0.025,0.975)) to estimate T (0.025) and T (0.975), i.e., the 0.025 and 0.975 quantiles of the distribution of T* from the 500 bootstrap replicates Construct a 95% bootstrap confidence interval for l u by solving for lnu in the inequality: *(0.029Explanation / Answer
Given that,
sample size (n) = 9
sample mean (Xbar) = 10
sample variance (sigma2 )= 25
sample standard deviation (s) = sqrt(25) = 5
Confidence level (C) = 95% = 0.95
95% confidence interval for mu is,
Xbar - E < mu < Xbar + E
where E is margin of error.
E = tc * (s / sqrt(n))
where tc is the critical value for normal distribution.
tc we can find in excel.
syntax :
=TINV(probability, deg_freedom)
where probability = 1 - C
deg_freedom = n-1 = 9-1 = 8
tc = 2.31
E = (2.31 * 5) / sqrt(9) = 3.84
95% confidence interval for mu is,
10 - 3.84 < mu < 10 + 3.84
6.16 < mu < 13.84
95% confidence interval for mu is (6.16, 13.84)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.