If the deviation of a score from the mean is given as 10 and standard deviation as 5, what will be the T-score for the concerned raw score ?
70
Understanding how to convert raw scores into standardized scores like the T-score is important in statistics and psychometrics. The T-score standardizes a score on a scale with a mean of 50 and a standard deviation of 10. To calculate the T-score, we typically first find the Z-score.
We are given the following information:
We need to find the T-score for this raw score.
The Z-score is calculated by dividing the deviation from the mean by the standard deviation. The formula is:
$$Z = \frac{\text{Deviation from Mean}}{\text{Standard Deviation}}$$
Substitute the given values into the formula:
$$Z = \frac{10}{5}$$
$$Z = 2$$
So, the Z-score for the concerned raw score is 2. This means the score is 2 standard deviations above the mean.
Now that we have the Z-score, we can calculate the T-score using the formula:
$$T = 50 + 10Z$$
Substitute the calculated Z-score ($Z = 2$) into the T-score formula:
$$T = 50 + 10 \times 2$$
$$T = 50 + 20$$
$$T = 70$$
Thus, the T-score for the concerned raw score is 70.
This means a raw score that is 10 points above the mean in a distribution with a standard deviation of 5 corresponds to a T-score of 70.
| Given Information | Value |
|---|---|
| Deviation from Mean ($X - \mu$) | 10 |
| Standard Deviation ($\sigma$) | 5 |
| Calculated Score | Formula | Value |
|---|---|---|
| Z-score (Z) | $Z = \frac{X - \mu}{\sigma}$ | 2 |
| T-score (T) | $T = 50 + 10Z$ | 70 |
| Score Type | Mean | Standard Deviation | Formula |
|---|---|---|---|
| Z-score | 0 | 1 | $Z = \frac{X - \mu}{\sigma}$ |
| T-score | 50 | 10 | $T = 50 + 10Z$ |
| IQ Score (Wechsler Scale) | 100 | 15 | $IQ = 100 + 15Z$ |
Standardizing scores allows us to compare scores from different tests or distributions that may have different means and standard deviations. For example, a score of 70 on a test with a mean of 60 and a standard deviation of 5 is much better, relative to the group, than a score of 70 on a test with a mean of 65 and a standard deviation of 10. Converting these raw scores to Z-scores or T-scores provides a common scale for comparison.
In this specific problem, a T-score of 70, corresponding to a Z-score of 2, indicates a score that is significantly above average (two standard deviations above the mean).
Type II error in hypothesis testing is:
While wandering in jungle, King Dushyant married Shankuntala as narrated in the "Abhigyan Shankuntalam". He gave her royal ring which could serve her identity when she would come to meet him, in future. However, she had lost the ring while going to meet him. When she arrived at Dushyant's palace, he failed to recognise her as she did not had the ring.
Which of the following statistical error King Dushyant had committed in this narrative?
Given below is a summary of ANOVA for four groups of students tested in a research project:
| Source of variance | SS (Sum of squares) | df (Degree of freedom) | MS (Mean sum of squares) |
| Between groups | 76 | 3 | 23.33 |
| Within groups | 122 | 16 | 7.62 |
What will be the value of 'F' for the above data?
An investigator used ANOVA to compare four groups of students on numerical ability on the basis of a test. After analysis of raw scores, the following results were obtained:
| Source of variation | df | Sum of Squares |
| Between Groups | 3 | 625.00 |
| Within Groups | 36 | 2128.00 |
The value of F-ratio would be approximate:
In randomly constituted two groups-experimental and control, a researcher obtains the following results after using a parametric 't' test:
Value of t = 3 for N = 300
On the basis of this evidence which decision in respect of substantive research hypothesis and the null hypothesis will be justified?