# Best Point Estimation Examples

The best point estimator here is based on Laplace, Jeffrey, Wilson, Maximum Likelihood Estimators. The best point estimation is also used to reduce error and increase accuracy. The point estimate can be calculated from sample data. It is used to estimate the true unknown value in the population called parameter. Here are two simple Best point estimation examples to calculate the best point estimate.

## Laplace, Jeffrey, Wilson Maximum Likelihood Estimation

###### Example 1:

Let us consider the problem with Number of Success (S) as 4 , Number of Trials (T) as 7 and Confidence Interval (P) as 2.

###### Solution:

We can calculate the Best Point Estimation using the given formula.

#### Formula:

MLE = S / T Laplace Estimation = (S+1) / (T+2) Jeffrey Estimation = (S+0.5) / (T+1) Wilson Estimation = (S+(z2/2)) / (T+z2) Where, MLE = Maximum Likelihood Estimation S = Number of Success T = Number of Trials z = Z-Critical Value

Substituting the values in the formula,

The value of Best Point Estimation = S/T = 4/7 = 0.5714 The value of Maximum Likelihood Estimation = S/T = 4/7 = 0.5714 The value of Laplace Estimation = (S+1) / (T+2) = (4+1) / (7+2) = 0.5556 The value of Jeffrey's Estimation = (S+0.5) / (T+1) = (4+0.5) / (7+1) = 0.5625 The value of Wilson Estimation = (S+(z2/2)) / (T+z2) = (4+(22/2)) / (7+22) = 0.5714.

###### Example 2:

Let us consider the problem with Number of Success (S) as 8 , Number of Trials (T) as 23 and Confidence Interval (P) as 5.

###### Solution:

We can calculate the Best Point Estimation using the above given formula.

Substituting the values in the formula,

Therefore, The value of Best Point Estimation = S/T = 8/23 = 0.3479 The value of Maximum Likelihood Estimation = S/T = 8/23 = 0.3479 The value of Laplace Estimation = (S+1) / (T+2) = (8+1) / (23+2) = 0.36 The value of Jeffrey's Estimation = (S+0.5) / (T+1) = (8+0.5) / (23+1) = 0.3542 The value of Wilson Estimation = (S+(z2/2)) / (T+z2) = (8+(32 / 2)) / (23+32) = 0.3479.