1. Normal data, variance known: If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and known standard deviation $\sigma,$ then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm 1.95 \sigma/\sqrt{n}.$ This is the only situation in which the z interval is exactly correct.
2. Nonnormal data, variance known: If the population distribution is not normal and the sample is 'large enough', then $\bar X$ is approximately normal and the same formula provides an approximate 95% CI. The rule that $n \ge 30$ is 'large enough' is unreliable here. If the population distribution is heavy-tailed, then $\bar X$ may not have a distribution that is close to normal (even if $n \ge 30).$ The 'Central Limit Theorem', often provides
reasonable approximations for moderate values of $n,$ but it is a limit theorem,
with guaranteed results only as $n \rightarrow \infty.$
3. Normal data, variance unknown. If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and standard deviation $\sigma,$ with $\mu$ estimated by the sample mean $\bar X$ and $\sigma$ estimated by the sample standard deviation $S.$ Then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm t^* S/\sqrt{n},$ where $S$ is the sample standard deviation and
where $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n - 1$ degrees of freedom. This is the only situation in which the t interval is exactly correct.
Examples: If $n=10$, then $t^* = 2.262$
and if $n = 30,$ then $t^* = 2.045.$ (Computations from R below; you could also use a printed 't table'.)
qt(.975, 9); qt(.975, 29)
[1] 2.262157 # for n = 10
[1] 2.04523 # for n = 30
Notice that 2.045 and 1.96 (from Part 1 above) both round to 2.0. If $n \ge 30$ then $t^*$ rounds to 2.0. That is the basis for
the 'rule of 30', often mindlessly parroted in other contexts where it is not relevant.
There is no similar coincidental rounding for CIs with confidence levels other than 95%. For example, in Part 1 above
a 99% CI for $\mu$ is obtained as $\bar X \pm 2.58 \sigma/\sqrt{n}.$ However,
$t^*=2.76$ for $n = 30$ and $t^* = 2.65$ for $n = 70.$
qnorm(.995)
[1] 2.575829
qt(.995, 29)
[1] 2.756386
qt(.995, 69)
[1] 2.648977
4. Nonnormal data, variance unknown: Confidence intervals based on the t distribution (as in Part 3 above) are known to be 'robust' against moderate departures from normality.
(If $n$ is very small, there should be no far outliers or evidence of severe skewness.) Then, to a degree that is difficult to predict, a t CI may provide a useful CI for $\mu.$
By contrast, if the type of distribution is known, it may be possible
to find an exact form of CI.
For example, if $n = 30$ observations from a (distinctly nonnormal)
exponential distribution with unknown mean $\mu$ have $\bar X = 17.24,\,
S = 15.33,$ then the (approximate) 95% t CI is $(11.33, 23.15).$
t.test(x)
One Sample t-test
data: x
t = 5.9654, df = 29, p-value = 1.752e-06
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
11.32947 23.15118
sample estimates:
mean of x
17.24033
However,
$$\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\text{shape}=n,\text{rate}=n),$$
so that $$P(L \le \bar X/\mu < U) = P(\bar X/U < \mu < \bar X/L)=0.95$$
and an exact 95% CI for $\mu$ is $(\bar X/U,\, \bar X/L) = (12.42, 25.16).$
qgamma(c(.025,.975), 30, 30)
[1] 0.6746958 1.3882946
mean(x)/qgamma(c(.975,.025), 30, 30)
[1] 12.41835 25.55274
Addendum on bootstrap CI: If data seem non-normal, but the actual population
distribution is unknown, then a 95% nonparametric bootstrap CI may be the best
choice. Suppose we have $n=20$ observations from an unknown distribution, with $\bar X$ = 13.54$ and values shown in the stripchart below.
The observations seem distinctly right-skewed and fail a Shapio-Wilk normality test with P-value 0.001. If we assume the data are exponential and use the method in Part 4, the 95% CI is $(9.13, 22.17),$ but we have no way to know whether the data are exponential.
Accordingly, we find a 95% nonparametric bootstrap
in order to approximate $L^*$ and $U^*$ such that
$P(L^* < D = \bar X/\mu < U^*) \approx 0.95.$ In the R code below
the suffixes .re indicate random 're-sampled' quantities based on
$B$ samples of size $n$ randomly chosen without replacement from among the
$n = 20$ observations. The resulting 95% CI is $(9.17, 22.71).$ [There are
many styles of bootstrap CIs. This one treats $\mu$ as if it is a scale
parameter. Other choices are possible.]
B = 10^5; a.obs = 13.54
d.re = replicate(B, mean(sample(x, 20, rep=T))/a.obs)
UL.re = quantile(d.re, c(.975,.025))
a.obs/UL.re
97.5% 2.5%
9.172171 22.714980
- The first case was answered in detail in this question.
- One example of the second case is shown here, where the authors apply a normal approximation to the binomial distribution used in calculations of the first case.
The third case is given by Hahn and Meeker in their handbook Statistical Intervals (2nd ed., Wiley 2017):
A two-sided $100(1-\alpha)\%$ confidence interval for $x_q$, the $q$ quantile of the normal distribution, is
$$
\left[\bar{x}-t_{(1-\alpha/2;\,n-1,\,\delta)}\frac{s}{\sqrt{n}},\;\bar{x}-t_{(\alpha/2;\,n-1,\,\delta)}\frac{s}{\sqrt{n}}\right]
$$
where $t_{(\gamma;\,n-1,\,\delta)}$ is the $\gamma$ quantile of a noncentral $t$-distribution with $n-1$ degrees of freedom and noncentrality parameter $\delta = -\sqrt{n}z_{(q)}=\sqrt{n}z_{(1-p)}$.
Here, $z_{(q)}$ denotes $\Phi^{-1}(q)$, the $q$ quantile of the standard normal distribution.
For example, let's assume we drew $n=20$ samples from a normal distribution with unknown mean and standard deviation. The sample mean was $\bar{x}=10.5$ and the sample standard deviation was $s=3.19$. Then, the two-sided $95\%$ confidence interval for the $q=0.25$ quantile $x_{0.25}$ would be given by $(6.42; 9.76)$.
Here is some R code and a small simulation to check the coverage. By changing the parameters, you can run your own simulations:
normquantCI <- function(x, conf_level = 0.95, q = 0.5) {
x <- na.omit(x)
n <- length(x)
xbar <- mean(x)
s <- sd(x)
ncp <- -sqrt(n)*qnorm(q)
tval <- qt(c((1 + conf_level)/2, (1 - conf_level)/2), n - 1, ncp)
se <- s/sqrt(n)
xbar - tval*se
}
# Simulate the coverage
set.seed(142857)
q <- 0.25 # Quantile to calculate the CI for
conf_level <- 0.95 # Confidence level
true_mean <- 100 # The true mean of the normal distribution
true_sd <- 15 # True sd of the normal distribution
sampsi <- 20 # The sample size
trueq <- qnorm(q, true_mean, true_sd) # The true quantile
res <- replicate(1e5, {
citmp <- normquantCI(rnorm(sampsi, true_mean, true_sd), conf_level = conf_level, q = q)
ifelse(citmp[1] < trueq & citmp[2] > trueq, 1, 0)
})
sum(res)/length(res)
[1] 0.95043
Best Answer
Since you only know $n$ and $\bar{X}_n$ why not choose a $Y_n$ that isn't a function of anything else but (at most) those two things?
You chose a $Y_n$ without considering what you have to work with and then struck a problem ... but you didn't reconsider whether you might have chosen a different $Y_n$ that avoided the problem.