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Numerical Related to Estimation of True Score

For the estimation of true score
$$T=r×O+(1-r)\overline{X}$$
$$where\,r=reliability\,coefficient$$
$$O=Observed\,frequency$$
$$\overline{X}=Mean\,score$$

$$T=True\,score$$. For confidence interval (C.I)

$$C.I=\overline{X}±t_a/_2S.E$$
$$Where\,C.I=Confidence\,interval$$
$$\overline{X}=mean\,score$$
$$t_a/_2=Critical\,value\,for\,two\,tail\,test \,at\,α\,level\,of\,significance$$
$$S.E=standard\,error$$
Here, S.E is calculated by using the relation.
$$S.E=σ\sqrt{(1-r)}r$$
$$r=reliability\,error$$
$$σ=standard\,deviation$$

Numerical
The reliability coefficient of a test is found to be 0.75 and its found to be 0.75 and its mean score is 60. Also, the standard deviation of the test is found 10. A obtained score of 50 on the test. What is his true score? Also, compute 95% confidence interval for the true mean.

$$Solution$$.
$$Given\,that\,,r_w=0.75$$
$$\overline{X}=60\,,s.d=10$$
$$Observed\,score(o)=50$$
$$True\,score(T)=?$$
$$T=r_w×O+(1-r_w)\overline{X}$$
$$T=r_w×O+(1-0.75)\overline{60}$$
$$=52.3≈53$$
$$S.E=σ\sqrt{r_w(1-r_w)}$$
$$S.E=10\sqrt{0.75(1-0.75)}$$
$$=4.3≈4$$
$$95\%\,confidence\,interval\,of\,score$$
$$=T±1.96×4$$
$$=53±1.96×4$$
$$=53±7.48$$
$$=(60.84, 45.16)$$
$$=(61,45)$$

The reliability coefficient of a test is found as 0.65 and the mean score as 5o with standard deviation 5. Mr x obtains a score on the test. What is his true score? Also, compute 95% confidence interval for the true score.

$$Solution$$
$$Given$$.
$$Given\,that\,,r_w=0.65$$
$$\overline{X}=50\,,s.d=5$$
$$Observed\,score(o)=40$$
$$True\,score(T)=?$$
$$Letα\,=5\%, \,then\,t_a/_2=1.96$$
$$T=r×O+(1-r)\overline{X}$$
$$T=0.65×40+(1-0.65)×50$$
$$=43.5$$
$$S.E=σ\sqrt{r(1-r)}$$
$$S.E=5\sqrt{0.65(1-0.65)}$$
$$=2.38$$
$$95\%\,confidence\,interval\,of\,score$$
$$=\overline{X}±t_a/_2×S.E$$
$$=50±1.96×2$$
$$=50±3.92$$
$$=(53.92,46.08)$$
$$=54,46$$
$$95\%\,confidence\,interval\,of\,score$$
$$=54,46$$

The reliability coefficient of a test is found as 0.55 with the mean score of 45 with standard deviation3. Mr X obtains a score of 35 on this text. What is his true score? Also, compute 95% confidence interval for the true score.

$$Solution$$
$$reliability\,coefficient(r)=0.55$$
$$Mean\,score\overline{X}=45$$
$$Observed\,score(o)=35$$
$$standard\,deviation(σ)=3$$
$$True\,score(T)=?$$
$$Letα\,=5\%, \,then\,t_a/_2=1.96$$
$$T=r×O+(1-r)\overline{X}$$
$$T=0.55×35+(1-0.55)×45$$
$$=39.5$$
$$S.E=σ\sqrt{r(1-r)}$$
$$S.E=3\sqrt{0.55(1-0.55)}$$
$$=1.5$$
$$95\%\,confidence\,interval\,of\,score$$
$$=\overline{X}±t_a/_2×S.E$$
$$=45±1.96×2$$
$$=45±3.92$$
$$=(48.92.41.08)$$
$$=49,51$$
$$95\%\,confidence\,interval\,of\,score$$
$$=49,51$$

The reliability coefficient of a test is found as 0.75 with a mean score as 55 and s.d of 5. Mr X obtains a score of 45 on the test. What is his true score? Also, compute 95% confidence interval for the true value.

$$Solution$$
$$reliability\,coefficient(r)=0755$$
$$Mean\,score\overline{X}=55$$
$$Observed\,score(o)=45$$
$$standard\,deviation(σ)=5$$
$$True\,score(T)=?$$
$$Letα\,=5\%, \,then\,t_a/_2=1.96$$
$$T=r×O+(1-r)\overline{X}$$
$$T=0.75×45+(1-0.75)×55$$
$$=47.5$$
$$S.E=σ\sqrt{r(1-r)}$$
$$S.E=5\sqrt{0.75(1-0.75)}$$
$$=2.17$$
$$95\%\,confidence\,interval\,of\,score$$
$$=\overline{X}±t_a/_2×S.E$$
$$=55±1.96×2$$
$$=55±3.92$$
$$=(58.92,51.08)$$
$$=,59,51$$
$$95\%\,confidence\,interval\,of\,score$$
$$=59,51,$$




Reference

Kerlinger, F.N. Foundation of Behavioural Research. New Delhi: Surjeet Publication, 2000.
Kothari, C.R. Research Methodology. India: Vishwa Prakashan, 1990.
Singh, M.L. and J.M Singh. Understanding Research Methodology. 1998.
Singh, Mrigendra Lal. Understanding Research Methodology. Nepal: National Book centre, 2013.

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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