Statistical Techniques

Let fp( .| `theta` )g be a family of probability densities indexed by a parameter `theta` .Let y = (y1,....., yn) be a sample from p(.|`theta` ) for some unknown `theta` . Let T(y) be a statistic such that the joint distribution factors as follows

`prod` p(yi | `theta` ) = g(T(y); `theta` )h(y):

for some functions g and h. Then T is called a sufficient statistic for `Theta``theta` .

The  statistical techniques involve to study the two important parts namely mean deviation and standard deviation.

Mean Deviation for un grouped data:

Let x1,x2,x3,………xn be the given n observation. Let  `barx`   be the AM and M be the median .then

i.) MD(x¯) = `(sum_(i= 1)^n)/(n)` | xi –x|           where `barx`

Statistical Techniques in Mean, Mean Deviation

1.) Find the mean deviation about the mean for the given data.

15,17,10,13,7,18,9,6,14,11.

Sol: Let the mean of the given data be `barx` Then,

`barx` = `(sum_(i= 1)^n)/(n)` | xi –x|
= `(120)/(10)`  = 12    [n = 10]

so, the values of (xi – x) are:

3, 5, 2, 1,-5,6,-3,-6,2,-1

So, the values of | xi –x| are:

3 , 5 , 2 , 1 , 5 , 6 , 3 , 6 , 2 , 1

MD(`barx` ) =   `(sum_(i= 1)^n)/(n)` | xi –x|


=  `(34)/(10)`  = 3.4

Hence the MD (`barx` ) = 3.4

Statistical Techniques in Variance with Example:

Variance and Standard Deviation for statistical techniques:

Definition of variance :

Mean of the squares of the deviations from the mean is called the variance , to be denoted by σ2. The variance of n observations x1 , x2 ,.....xn is given by σ2 =   `(1)/(n)` `sum_(i=1)^n` (xi - x¯)2 .

Definition  of Standard deviation :

The  positive square root of variance is called standard deviation and it is denoted by σ  .

σ =`sqrt((sum_(i=1)^n( x i-x)^2)/(n))`

Ex: Find the mean, standard deviation and variance of first n natural numbers

Sol:

Mean,x = ( 1 + 2 + 3 +........+ n)    = 1  .1   n( n + 1 ) =  1  ( n + 1 )
n               n   2                      2

Variance, σ2 = `(sum)/(n)` x-2
=  x¯2



Variance, σ2 = `(sum)/(n)`  n2  =  { 1/2 (n + 1) }2



=          `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=     `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=  `((n+1))/(n)` ×   `(n(2n+1))/(6)`  -  `((n+1))/(4)`

= `((n+1))/(1)`   × `((2n+1))/(6)`  -  `((n+1))/(4)`

=`((n+1)(n-1))/(12)`

Variance,σ2  = `((n^2 -1))/(12)`

Standard deviation ,σ2 = `((sqrt(n^2 -1)))/(12)`

= `((1/2) (sqrt(n^2 -1)))/(3)`