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Math Help - ranks

  1. #1
    Member princess_anna57's Avatar
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    ranks

    Individuals from a business training programme are monitored in their first year of business, and their achievement in that first year quantified using a range of tests. The training programme was based on a mentoring system, with young trainees paired in every case with older, more experienced people who were training for a career change. The data for the individuals are below.

    Team 1 2 3 4 5 6 7 8
    Young Trainee 30 44 21 17 25 22 16 28
    Mentor 23 37 7 18 10 15 13 29

    a) Draw a scatterplot of the data, using the mentor's scores to predict the young trainee's scores.
    b) Calculate Pearson's correlation for the data, and interpret your figure.
    c) Now calculate Spearman's correlation coefficient for the data.
    d) Which correlation is most appropiate for the situation, and why?

    NOTE: For the data, 30 and 23 under 1, 44 and 37 under 2, etc.

    Please help. Thank you!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Here. The equation gives a regression line which uses the mentor's scores to predict the young trainee's scores. So you would plug in the mentor's scores for  x to obtain the predicted mentors scores. In this case, it would be better to use Pearson correlation, because we cannot assume that there are rankings. Also the linear regression line is linear. Spearmans coefficient is  0.6667 . Here is a helpful page that calculates the spearman coefficient. Also there is always uncertainty in measurements, so we could also add error bars.
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    Last edited by tukeywilliams; July 30th 2007 at 04:21 PM.
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  3. #3
    Member princess_anna57's Avatar
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    Would you be able to show the working out in relation to the answer you got for Pearson? You're being so helpful, thank you.
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  4. #4
    Senior Member tukeywilliams's Avatar
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    Here it is.  r = \frac{\sigma^{2}_{xy}}{\sqrt{\sigma^{2}_{x} \cdot \sigma^{2}_{y}}} . The covariance is  \frac{\sum X \cdot \sum Y}{n^2} .

    So:  r = \frac{\frac{152 \cdot 203}{64}}{\sqrt{\frac{152^2}{64} \cdot \frac{203^2}{64}}} = 0.80781
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  5. #5
    Member princess_anna57's Avatar
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    Ohh I see, thank you!

    When I wrote up the table for Spearman, for the squared difference rank, I got zero. Hmm.. it's not right, is it.
    Last edited by princess_anna57; July 30th 2007 at 05:10 PM.
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  6. #6
    Senior Member tukeywilliams's Avatar
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    no its not right.
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