I have taken $N$ measurements $x_i$ of the time taken by a computer program to run a particular calculation, and would like to calculate a 95% confidence interval for the mean runtime. We can calculate the sample mean and standard deviation in the usual way:
$$
\bar{x} = \frac{1}{N} \left( \sum_{i=1}^N {x_i} \right); s = \sqrt{\frac{\sum_{i=1}^N (x_i – \overline{x})^2}{N-1} }
$$
Now, as I understand it, if $N \ge 30$, we can appeal to the Central Limit Theorem and say that our sample mean $\bar{x}$ has come from a normal distribution with mean $\mu$ and standard error $\sigma / \sqrt{N}$, where $\mu$ and $\sigma$ are the population mean and standard deviation respectively. A 95% confidence interval for the mean runtime is therefore bounded by, for some $k_1, k_2$:
$$
\bar{x} \pm k_1 \frac{\sigma}{\sqrt{N}}\ or\ \bar{x} \pm k_2 \frac{s}{\sqrt{N}}
$$
We can take $k_1$ from the normal distribution: $k_1 = Z_{0.975} = 1.96$. However, if we need to calculate a confidence interval without knowing $\sigma$, what value should we use for $k_2$?
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According to the paper Statistically Rigorous Java Performance Evaluation, we can also take $k_2$ from the normal distribution: $k_2 = Z_{0.975} = 1.96$. However, my understanding is that we should instead use Student's t-distribution with $N – 1$ degrees of freedom: for $N = 30$ this gives $k_2 = t_{(0.975, 29)} = 2.045$. Which of these is correct?
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The same paper claims that the above method (using the t-distribution) is valid even for $N < 30$. Is this true, given that we cannot apply the Central Limit Theorem in this case? If not, is there anything we can say about a confidence interval for the population mean in the case $N < 30$?
Best Answer
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'.)
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 $n = 70.$
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 $(11.33, 23.15).$
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 $(\bar X/U,\, \bar X/L) = (12.42, 25.16).$
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
.reindicate 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.]