Confidence interval variance among a set of groups,

Question

A psychiatrist is interested in finding a 90% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 13 randomly selected children with Tourette syndrome, with each value of tics-per-hour separated by a comma.

• Child 1: 8 tics per hour
• Child 2: 9 tics per hour
• Child 3: 0 tics per hour
• Child 4: 5 tics per hour
• Child 5: 9 tics per hour
• Child 6: 1 tics per hour
• Child 7: 1 tics per hour
• Child 8: 2 tics per hour
• Child 9: 7 tics per hour
• Child 10: 3 tics per hour
• Child 11: 9 tics per hour
• Child 12: 3 tics per hour
• Child 13: 4 tics per hour

Parts I already solved:

1. The confidence interval was computed with t distribution.
2. With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between 3.035733655917 and 6.348881728698 days.

The part I need help with: If many groups of 13 randomly selected children with Tourette syndrome are observed, then a different confidence interval would be produced from each group. About x percent of these confidence intervals will contain the true population mean number of tics per hour and about y percent will not contain the true population mean number of tics per hour. With this information, help me find the percentages.

Answer 1

If you believe that a 90% confidence interval (CI) is obtained by a reasonable method, so that it truly has 90% 'coverage probability' of the population mean $\mu,$ then you can say that 90% of samples of size $n=13$ from this population will yield CIs of this type that include $\mu.$

My question in your case is whether a t confidence interval is the appropriate type of CI.

If you take the 13 measurements to be normally distributed, then a 90% t confidence interval for the mean number of tics per hour is $(3.04, 6.34).$ [It is customary to round boundaries of a CI to one or two more significant digits than for the data; I see no point in keeping a dozen or so decimal places. Also, 'days' must be a typo.]

t.test(x, conf.lev=.9)$conf.int
[1] 3.035734 6.348882
attr(,"conf.level")
[1] 0.9

However, the data do not seem normally distributed, even though they are roughly symmetrical. A normal probability plot of normal data should be roughly linear, but such a plot for your 13 observations seems distinctly curved:

x = c(8,9,0,5,9,1,1,2,7,3,9,3,4)
qqnorm(x);  qqline(x, col="green", lwd=2)

So I tried a nonparametric 90% confidence interval based on a one-sample Wilcoxon test. Because of ties in the data the CI $(2.5, 6.5)$ is only approximate.

Also, a simple nonparametric bootstrap quantile 90% CI is $(3.23, 6.15)$ as shown below. The sample size $n = 13$ may be too small for a "95%" CI to cover the true population 90% of the time.

set.seed(411)
a.re = replicate(5000, mean(sample(x,13,rep=T)))
quantile(a.re, c(.05,.95))
      5%      95% 
3.230769 6.153846 

Answer 2

Since this is a 90% confidence interval, in the long run, if you make a lot of these intervals, about 90% of them will bracket the correct mean and about 10% will fail.