T-Scale and constriction of T-Scale.

T-Scale.

T-Scale simply the five times the σ Scale with zero point at-56.

T-Score at normalised standard scores converted into a distribution with a near 50 and ( σ) of 10. In 6-scaling, the mean is zero and ( σ) is 1.00 I,e The point of reference is zero and unit of measurement is one. If the point of reference is ( σ) scale is moved from the mean of the normal curve to a point 5 below the mean, the new reference point becomes the zero in scale and the mean becomes 5.

The ( σ) divisions below the mean become 4, 3, 2, 1 and 0 and the ( σ) division above the means becomes 6, 7,8,9 and 10 with s,d of the distribution remaining equal to 1.

The T-scale beings at -56 and ends at +56. But 6’s are multiply by 10. In T-Scale, the mean becomes 50 and divides 0,10,20,30,40,50,60,70,80,90 and 100. This figure shows the relationship between ( σ) score and T-Scale.

Construction of T-Scale.

The main construction for T Scale is as below.

1. A large and representative group of test items are compiled, which vary difficulty from easy to hard. These items are administrated to a sample of subjects for whom the final test is intended.
2. The percentage of those passing each item is computed. The items are arranged in an order of difficulty in terms of those percentage.
3. The test is administrated to a representative sample and the distribution of total scores is tabulated.
4. The cumulative frequency (c.f) is found as in column (3) of the illustration given below, then the average of two consecutive c.f’s is computed starting below from the highest score.
5. The percentage of total subjects are expressed and placed in column(5) .
6. The percentage are expressed in decimal points.
7. The 0.5 is subtracted from each of such figures from the area under a normal curve, this gives the ( σ) Scale value.
8. In order to convert ( σ) scale value to T-Scale, the ( σ) Scale value is multiplied by 10 and 50 is added. The final figures are T-Scale values.

Illustration.

Total=62.

T-Series values in above table are obtained by

For instance 99.2%=0.992.

0.5 is subtracted from 0.992, then we get 0.492 from normal curve table, z value corresponding to 0.492 is found to be 2.41.

This values is multiplied by 10 and added to 50 to get 24.1+50=74.

That is,74 is the T-score value for the percentage.

Corresponding to 99.2%.

Similarly,we calculate the T-Score Values.

How Percentile rank is calculated? Illustrate with the example.

Solution.

In many scaling, percentile method is often used. The scale value is the percentile Rank or PR which is calculated by reversing the method of com[putting the percentage in group data and dividing the results by N(sample size).

We have,

The distribution of 60 subjects according to the score obtained is found as.

Find the percentile rank for the score 157 and 186

Now, by using formula

$$\frac{iN}{100}=\frac{(Pi-L)×f}{h}+c.f$$

$$\frac{iN}{100}=Percentile\,rank\,of\,given\,solve$$

$$Pi=Score\,given\,whose\,percentile\,rank\,is\,to\,be\,found$$

$$L=lower\,limit\,of\,the\,class\,where\,Pi\,lies$$

$$f=frequency\,of\,the\,corresponding\,class\,where\,pi\,lies$$

$$h=height\,or\,class\,interval\,of\,the\,class$$

C.f upto the lower limit of the class interval.

For percentile rank of 157

157 lies in the (154.5-159.%) class interval.

Then.

$$\frac{iN}{100}=\frac{(Pi-L)×f}{h}+c.f$$

$$\frac{iN}{100}=\frac{157-154.5×6}{5}+9$$

$$=12$$

$$\frac{iN}{100}=12$$

Then percentile rank for 157 is

$$\frac{iN}{100}=\frac{12}{N}=\frac{12}{60}=0.2=20\%$$

Similarly,

186 lies class (184.5-189.5).

$$\frac{iN}{100}=\frac{(Pi-L)×f}{h}+c.f$$

$$=\frac{(186-184.5×3}{5}+54$$

$$=54.9$$

Percentile rank for 186

$$\frac{iN}{100}=\frac{54.9}{60}$$

$$=0.915$$

$$=91.5\%$$

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.