Use of standardized scores or z scores
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Z-scores allow comparison of values from different data sets by measuring the number of standard deviations an observation is above or below the mean. A positive Z-score indicates how many standard deviations above the mean, while a negative score indicates how many below. Examples show how to calculate Z-scores from raw data and vice versa, and how Z-scores can be used to compare performances relative to the mean across different samples or events. Care must be taken to determine if a higher or lower score is preferable based on the situation.
Use of standardized scores or z scores
- 2. USE OF STANDARDIZED SCORES OR Z-SCORES Z Scores allow you to compare values from different data sets. A Z scores measures how many standard deviations an observation is above or below the mean. 1. Z Scores is unitless. 2. A positive Z Scores measures the number of standard deviations an observation is above the mean. 3. A negative Z Scores measures the number of standard deviations an observation is below the mean. NOTE: Care needs to be taken to determine whether a higher or lower z- score is preferable. In some situation high z- score is better and In some situation low z- score is better.
- 3. ∑ HOW TO FIND A Z SCORE Example-1 Suppose Jawwad score 75 in a test. The test has an average 60 with a standard deviation of 10. Find Z score Solution: Z = = =1.5 ∑ How TO FIND A RAW SCORE WHEN A Z SCORE IS GIVEN Example-2 Suppose you had an average marks in maths 60 with a standard deviation of 10.How much you score in maths if your z score is 1.5. Solution: Z = Z× = − = + (Z × ) = 60 + (1.5 × 10) = 75
- 4. ∑ TO COMPARE TWO SCORES THAT ARE FROM DIFFERENT SAMPLES Example-3. Jawwad and Humza’s Maths marks are shown in the table below along with the mean and standard deviation for the whole class. Jawwad Marks Humza Marks Mean Marks Standard Deviation Maths 75 65 60 10 Who has the higher relative score Solution: Jawwad Z Score is = =1.5 Humza Z Score is = = 0.5 Comments: Jawwad has the higher relative score. (In this case far from the mean is better)
- 5. Example-4. Jawwad and Humza’s Maths marks are shown in the table below along with the mean and standard deviation for the whole class. Jawwad Marks Humza Marks Mean Marks Standard Deviation Maths 55 50 60 10 Who has the higher relative score Solution: Jawwad Z Score is = = −0.5 Humza Z Score is = = -1 Comments: Jawwad has the higher relative score. (In this case nearest from the mean is better)
- 6. Example-5. Jawwad and Humza’s Maths and English marks are shown in the table below along with the mean and standard deviation for the whole class. Jawwad Marks Humza Marks Mean Marks Standard Deviation Maths 75 50 60 10 English 65 40 50 05 Work out Jawwad and Humza’s standardized scores and comment on their performance. Solution: Jawwad’s standardized score can be shown in a table X Jawwad Z Score Maths 75 Z= = =1.5 English 65 = = =3 Humza’s standardized score can be shown in a table X Humza Z Score Maths 50 = = = - 1 English 40 = = = - 2 Comments: Jawwad did better than Humza overall. The best mark is Jawwad’s English mark of 65 because it has the highest standardized score, while worst mark is Humza’s English mark of 40.
- 7. Example-6. Humza compared his running times in the 100m and 200m events. In 100m he ran in16 seconds, while the age average was 20 seconds with a standard deviation of 4 seconds. In 200 he ran in 39 seconds, while the age average was 44 seconds with a standard deviation of 2.5 seconds. In which event did he do better, relative to other runners his age? Solution: For 100m x = 16 sec. µ = 20 sec. and = 4 . = − = 16 − 20 4 = −1 For 200m x = 39 sec. µ = 44 sec. and = 2.5 . = − = 39 − 44 2.5 = −2 Comments: In 200m he did better. (In this case far from the mean is better)
- 8. Example-7. Humza free style swiming record is 34 seconds with mean 42 seconds and a standard deviation of 3.5 seconds but in butterfly his record is 38 seconds with mean 45 seconds and a standard deviation of 2.5 seconds. Decide in which style his performance is better ? Solution: Free style x = 34 sec. µ = 42 sec. and = 3.5 . = − = 34 − 42 3.5 = −2.29 Butterfly x = 38 sec. µ = 45 sec. and = 2.5 . = − = 38 − 45 2.5 = −2.80 Comments: In butterfly he did better. (In this case far from the mean is better)