To calculate the random error in a set of measurements this is what I would do.
- Get the standard deviation of the measurements:
$$ \sigma=\sqrt{\frac{1}{N-1}\Sigma_{i=1}^N(x_i-\bar{x})^2} $$ Where $N$ is the number of repeats, $x_i$ is the value of each sample, $\bar{x}$ is the mean, and $\sigma$ is the standard deviation.
The standard error of the mean: $$ s_\bar{x}=\frac{\sigma}{\sqrt{N}} $$
Obtain the random error using a t distribution: $$ \varepsilon_{rnd}=±t_{N-1}*s_\bar{x} $$
The problem is, if I have a measurement with a single reading of a given value, this method for random error calculation is clearly no longer applicable.
Therefore, how would you go about calculating the random error in a for a sample set where N=1? Is that even possible?
Thanks for your time!
Usually it depends heavily on the model, that we are considering. A very common model is to assume that data follows a normal distribution $N(\mu,\sigma^2)$. If data follows a normal distribution, then the most reasonable estimates are given by $$\hat{\mu}=\bar{x}= \frac1n \sum_{i=1}^n x_i \quad \text{ and } \quad \hat{\sigma} = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2},$$ but in this model it is quite obvious, that we cannot give a reasonable estimate of $\sigma$ when we only have a single observation.
Another common model is to assume that data follow a poisson distribution $\operatorname{poisson}(\lambda)$. A remarkable thing about the poisson distribution is, that if $X \sim \operatorname{poisson}(\lambda)$, then $$\operatorname{E}[X]=\lambda = \operatorname{Var}(X).$$ So for $\operatorname{poisson}$ data we have $\mu = \sigma^2$ and we would thus estimate $$\hat{\mu} = \hat{\lambda}=\bar{x} \quad \text{ and } \quad \sigma = \sqrt{ \hat{\lambda} } = \sqrt{\bar{x}},$$ and in particular if we only have one observation $x_1$, then we could estimate the mean and standard deviation as $x_1$ and $\sqrt{x_1}$ respectively without any problems. Asymptotic theory even tells us that if $\lambda$ is fairly large, then the poisson distribution is close to a normal distribution, so we could even make an approximate $(1-\alpha)\cdot 100$% confidence interval for $\lambda$ as $$x_1 \pm z_{1-\alpha/2}\sqrt{x_1} \quad \quad \text{($z_{1-\alpha/2}$ is a normal quantile)}$$ using just a single observation.