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Why Use Z scores?

Why Use Z scores?. Percentages can be used to compare different scores, but don’t convey as much information

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Why Use Z scores?

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  1. Why Use Z scores? • Percentages can be used to compare different scores, but don’t convey as much information • Z scores also called standardized scores, making scores from different distributions comparable; Ex: You get two different scores in two different subjects(e.g Statistics 28 and English 76). They are not yet comparable, so lets turn them into percentages( e.g 28/35=80% and 76/100, 76%). Relatively you did better in statistics.

  2. Percentages Verse Z scores • How do you compare to others? From percentages alone, you have no way of knowing. Say µon English exam was =70 with ó of 8 pts, your 76 gives you a z-score of .75, three-fourths of one stand deviation above the mean; Mean on statistics test is 21, with ó of 5 pts; your score of 28 gives a z score of 1.40 standard deviations above mean; Although English and statistics scores were similar, comparing z scores shows you did much better in statistics

  3. Using z scores to find percentiles • Prof Oh So Wise, scores 142 on an evaluation. What is Wise’s percentile ranking? Assume profs’ scores are normally distributed with µof 100 and ó of 25. X-µ 142-100 z= 1.68 ó 25 Area under curve ‘Small Part’ = .0465, equals those who scored above the prof; 1–.0465= 95.35th percentile. Oh so wise is in top 5% of all professors. Not bad at all. Never use from mean to z only to find percentile!! We’re only concerned with scores below a certain rank

  4. Starting with An Area Under Curve and Finding Z and then X… • Using the previous parameters of µ of 100 and ó of 25, what score would place a professor in top 10% of this distribution? After some algebra, we have X=µ+z (ó) • 100(µ) + 1.28(z)(25)(ó)=132 (X). A score of 132 would place a professor in top 10 %; • What scores place a professor in most extreme 5% of all instructors?

  5. What does ‘most extreme’ mean? • It is not just one end of the distribution, but both ends, or 2.5% at either end; • X= µ + z(ó)= 100+ 1.96(25)= 149 • 100 +-1.96(25)=51; 51 and 149 place a professor at the most extreme 5 % of the distribution;

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