To calculate the random error in a set of measurements this is what I would do.1. Get the standard deviation of the measurements: σ = 1 N − 1 Σ i = 1 N ( x i − x ¯ ) 2 Where N is the number of repeats, x i is the value of each sample, x ¯ is the mean, and σ is the standard deviation.2. The standard error of the mean: s x ¯ = σ N 3. Obtain the random error using a t distribution: ε r n d = ± t N − 1 ∗ s 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is, if I have a measurement with a single reading of a given value, this method for random error calculation is clearly no longer applicable.Therefore, how would you go about calculating the random error in a for a sample set where N=1?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?

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To calculate the random error in a set of measurements this is what I would do.1. Get the standard deviation of the measurements:  σ  =            1              N        −        1                    Σ              i        =        1            N        (          x      i        −                  x        ¯                    )      2      Where   N is the number of repeats,       x    i   is the value of each sample,             x      ¯       is the mean, and   σ is the standard deviation.2. The standard error of the mean:      s                  x        ¯              =      σ          N      3. Obtain the random error using a t distribution:      ε          r      n      d        =  ±      t          N      −      1        ∗      s                  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is, if I have a measurement with a single reading of a given value, this method for random error calculation is clearly no longer applicable.Therefore, how would you go about calculating the random error in a for a sample set where N=1?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?

Anika Stevenson · Accepted Answer

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  (  μ  ,      σ    2    ). If data follows a normal distribution, then the most reasonable estimates are given by            μ      ^        =            x      ¯        =      1    n        ∑          i      =      1        n       and               σ      ^        =            1              n        −        1                    ∑              i        =        1            n        (          x      i        −                  x        ¯                    )      2        ,but in this model it is quite obvious, that we cannot give a reasonable estimate of   σ when we only have a single observation.Another common model is to assume that data follow a poisson distribution   poisson  ⁡  (  λ  ). A remarkable thing about the poisson distribution is, that if   X  ∼  poisson  ⁡  (  λ  ), then  E  ⁡  [  X  ]  =  λ  =  Var  ⁡  (  X  )  .So for poisson data we have   μ  =      σ    2   and we would thus estimate            μ      ^        =            λ      ^        =            x      ¯           and     σ  =                    λ        ^              =                    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             x      1       respectively without any problems. Asymptotic theory even tells us that if   λ is fairly large, then the poisson distribution is close to a normal distribution, so we could even make an approximate   (  1  −  α  )  ⋅  100% confidence interval for   λ as      x    1    ±      z          1      −      α              /            2                  x      1                (                  z                  1          −          α                      /                    2                       is a normal quantile)  using just a single observation.

To calculate the random error in a set of measurements this is what I would do. 1. Get the standar

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