Theory of realibility.

Reliability from error component.

Any obtained score consists of two components a true value and an error,. Therefore, o=bserved value for an individual is decomposed as.

$$O_i=T+e.......(1)$$

where,

$$O_i=Total\,score\,observede\,for\,an\,individual$$

$$T=True\,score\,(which\,is\,never\,known)$$

$$e=Error$$

Since, T is never known, it is considered as the mean of observations of a large number of admistrations of a test to the same person.

$$i,e.,T= \frac{O_1+O_2+......O_n}{n}$$

$$If \,T \,and \,e \,are \,uncorrelated \,then \,Var (O_i)=Var(T)+Var(e).....2$$

$$Now\,reliability\,is \,defined\,as\,r_w= \frac{Var\,(T)}{Var\,O_i}.....3$$

From above, we have

$$r_w=1-\frac{Var\,(e)}{Var\,O_i}.....4$$

Since T is never known formula 4 is used to estimate reliability coefficient.

Example.

Suppose four items (statement are tested to five individual on a six point scale. The score observed for each of the individual found as shown below.

Two way alalysis of varience is carried out to find the varience or erroe component.

$$Correction \,factor\,(c.f)=\frac{T^2}{N}$$

$$where\,T=\sum{O_i}$$

$$N=r×c$$

$$r=Number\,of\,rows\,$$

$$c=Number\,of\,column$$

$$In\,above\,problem\,T=68\,,N=20$$

$$C.F=\frac{68^2}{20}=231.20$$

$$Now\,Total\,sum\,of\,square\,TSS=\sum{O_i^2}-{c.f}$$$$

$$=(6^2+6^2+.....1^2+2^2)-231.20=288.00-231.20=56.80$$

$$Sum\,of\,square\,due\,to\,(Colums)=\frac{\sum t_c^2 }{r}-cf$$

$$=\frac{18^2+19^2+19^2+12^2}{5}-231.20=6.80$$

$$Sum\,of\,square\,due\,to\,individual\,(row)=\frac{\sum t_r^2 }{c}-cf$$

$$=\frac{21^2+18^2+14^2+10^2+5^2}{4}-231.20=40.30$$

Example.

Find the reliability coefficient of the previous example by using Split-half method.

Solution.

From above table.

$$\sum{X}=37$$

$$\sum{X^2}=319$$

$$\sum{Y}=31$$

$$\sum{Y^2}=235$$

$$\sum{XY}=266$$

$$n=5$$

Therefore, the correlation coeficient bwtween X an Y is defined as.

$$r(X,Y)=\sum{XY}-{\fra{\sum{X}{n}$$

$$Mean\,(or\,\overline{X})=\frac{\sum X\sumY }{n}$$