7.4: Analysis with finite statistics

7. Finite Statistics with Student \(t\) distribution
   1. Normal distribution assumes an infinite population, but only limited samples are available for calculating the mean and variance of a sample set
      1. Sample mean \[ \bar{x} = \frac{1}{N} \sum_{i=1}^{N} x_{i} \] \[ \bar{x} \neq x' \ \mathrm{explicitly} \]
      2. Sample variance \[ S_{x}^{2} = \frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^{2} \]
      3. Since another statistical term was used in \(S_{x}\), the degree of freedom is considered \[\nu = N-c\] where \(N\) is number of samples and \(c\) number of statistical constraints used in calculations
   2. If accidentally substituting sample statistics into Normal distribution \[ p(x) = \frac{1}{S_{x}\sqrt{2\pi}}\mathrm{e}^{\frac{-(x-\bar{x})^{2}}{2S_{x}^{2}}} \] then
      1. Comparison of the distribution of normal (blue) and finite statistics (red)
      2. An overestimate of probability near sample mean
      3. Underestimates the probability far from the tail
      4. More complex function so two-sided Table format is provided
   3. Normalizing the \(t\) variable (similar to \(z_{1}\))
      1. Non-dimensional term \[ t = \frac{x-\bar{x}}{S_{x}} \] depends on two parameters
         1. \(\nu\) degree of freedom (i.e. # of samples)
         2. \(P\) confidence interval of probability
      2. Method of prediction of next selection of random variable \[ x_{N+1} = \bar{x} \pm t_{\nu,P}S_{x}\  (\%\ P)\]
         1. Same confidence interval \(\%\ P\) but fewer samples increases range of uncertainty
         2. Same sample size but smaller \(\%\ P\) reduces range of uncertainty
         3. As \(N\rightarrow \infty\), then \(t_{\nu, P}\rightarrow z_{p}\) since the sample statistics approach infinite terms
   4. Example

      Exercise \(\PageIndex{1}\)

      Example of finite version: Consider the same capacitors but mean (\(\bar{x} = 145\ \mu\)F) and standard deviation (\(S_{x} = 33\ \mu\)F) statistics are determined from 31 samples.

      • What is expected value of the next capacitor with 95% confidence?
      • How would fewer data points \(N<31\) affect that range for the prediction?

      Answer

      From the set of \(\nu = N - 1 = 30\) and confidence interval, the \(t_{30,95\%} = 2.042\). Therefore \[x_{32} = 145 \pm 2.042(33) = 77.6 \longleftrightarrow 212.4\ \mu\mathrm{F}\] 

      If the same mean and standard deviation statistics were collected from \(N=24\) data points, \(t_{23,95\%} = 2.069\) then the results would indicate \[ 76.7 < x_{25} < 213.3\ \mu\mathrm{F}\]

   5. Using the Standard Deviation of the Means
      1. The sample statistics \(\bar{x}\) and \(S_{x}\) provide insight on the true properties \(x'\) and \(\sigma\), respectively
      2. Using the \(N\) samples, the central limit theorem offers a range on the prediction of the true mean: \[x' = \bar{x} \pm t_{\nu,P}S_{\bar{x}} \] where \[S_{\bar{x}} = \frac{S_{x}}{\sqrt{N}}\]
      3. Note the difference between next measurement versus and indication of the true mean value with confidence interval
      4. Offers an uncertainty of \(x'\) from the statistics instead of the sensor

      5. Exercise \(\PageIndex{2}\)

         From the prior example, how large is uncertainty of the true mean value from 24 samples at 99.5\% confidence?

         Answer

         The \(t\) multiplier is found from \(t_{23,99.5\%} = 3.104 \). Hence, the range of the true mean would be \[x' = 145 \pm 3.104\left(\frac{33}{\sqrt{24}}\right) = 145 \pm 20.91\ \mu\mathrm{F}\]

         The width of the uncertainty decreases as the confidence level \(\% P\) similarly decreases. 

8. Design of Experiments
   1. Numerous methods define the mechanism to choose range of multi-variable experiments
   2. Approaches available from NIST
   3. Other other author (following LibreText sections)