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.