# problem-8.16

problem-8.16 We do this all at once using a list for storage

    > lst = list()
    > m = 250; n = 10
    > for(i in 1:m) {
    + lst$exp[i]  = t.test(rexp(n),     mu=1,  df=n-1)$p.value
    + lst$unif[i] = t.test(runif(n),    mu=.5, df=n-1)$p.value
    + lst$t4[i]   = t.test(rt(n, df=4), mu=0,  df=n-1)$p.value
    + }
    > sapply(lst, function(x) sum(x<0.05)/length(x))
      exp  unif    t4
    0.088 0.052 0.044

When the population distribution is skewed, the sampling distribution of T may differ a lot from the t-distribution.