Statistics - Best Point Estimation

Point estimation involves the use of sample data to calculate a single value (known as a statistic) which is to serve as a "best guess" or "best estimate" of an unknown (fixed or random) population parameter. More formally, it is the apppcation of a point estimator to the data.

Formula

${MLE = frac{S}{T}}$

${Laplace = frac{S+1}{T+2}}$

${Jeffrey = frac{S+0.5}{T+1}}$

${Wilson = frac{S+ frac{z^2}{2}}{T+z^2}}$

Where −

${MLE}$ = Maximum Likephood Estimation.

${S}$ = Number of Success .

${T}$ = Number of trials.

${z}$ = Z-Critical Value.

Example

Problem Statement −

If a coin is tossed 4 times out of nine trials in 99% confidence interval level, then what is the best point of success of that coin?

Solution −

Success(S) = 4 Trials (T) = 9 Confidence Interval Level (P) = 99% = 0.99. In order to compute best point estimation, let compute all the values −

Step 1 −

$ {MLE = frac{S}{T} \[7pt] , = frac{4}{9} , \[7pt] , = 0.4444}$

Step 2 −

$ {Laplace = frac{S+1}{T+2} \[7pt] , = frac{4+1}{9+2} , \[7pt] , = frac{5}{11}, \[7pt] , = 0.4545}$

Step 3 −

$ {Jeffrey = frac{S+0.5}{T+1} \[7pt] , = frac{4+0.5}{9+1} , \[7pt] , = frac{4.5}{10}, \[7pt] , = 0.45}$

Step 4 −

Discover Z-Critical Value from Z table. Z-Critical Value (z) = for 99% level = 2.5758

Step 5 −

$ {Wilson = frac{S+ frac{z^2}{2}}{T+z^2} \[7pt] , = frac{4+frac{2.57582^2}{2}}{9+2.57582^2} , \[7pt] , = 0.468 }$

Result

Accordingly the Best Point Estimation is 0.468 as MLE ≤ 0.5

Statistics - Best Point Estimation

Formula

Example

Result

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