Skewness help badly needed

postalservice

Hi there,
I have a project due tomorrow and i am stuck on one part.

We were given data based on test results of a class of 100 students
I calclulated the skewness value to be 72.4, mean of 38.6, median of 35.4 and mode of 32.5...and they are correct
The final part of the section however asks......
"A second class of 100 students took the same exam and a skewness value of minus 10 was calculated from their marks. Can you determine which class did better overall in the exam? Explain."

Any help would be greatly appreciated. Thanks:o

hivizman

postalservice wrote: »

Hi there,
I have a project due tomorrow and i am stuck on one part.

We were given data based on test results of a class of 100 students
I calclulated the skewness value to be 72.4, mean of 38.6, median of 35.4 and mode of 32.5...and they are correct
The final part of the section however asks......
"A second class of 100 students took the same exam and a skewness value of minus 10 was calculated from their marks. Can you determine which class did better overall in the exam? Explain."

Any help would be greatly appreciated. Thanks:o

In a positively skewed distribution, the mean is larger than the median. This means that more than half of the test scores fall below the mean score. The mean is "pulled up" because of the effect of a few very high scores, which counteracts the effect of a large number of mediocre scores. A similar distribution arises for personal income - mean income is higher than median income because of the effect of a small number of individuals earning several multiples of average income (by the way, this is why it's important, when using the word "average", to say whether you intend "mean", "median" or "mode").

In a negatively skewed distribution, the mean is smaller than the median. This comes about because a few very poor scores "pull down" the mean, even though the majority of students are achieving marks greater than the mean. This situation often happens with examination marks where students who don't attend are given zero marks, and these zero marks are included in the statistics.

From the skewness alone, you can't comment on whether the first class has done better or worse than the second class - you also need the mean and standard deviation. For example, if the mean mark in the second class is 75, then you'd conclude that the class performed better irrespective of the negative skewness. However, if the second class has the same mean and standard deviation as the first, then the negatively skewed class has more "good" performances than "poor" performances, while the positively skewed class is the other way round.

MathsManiac

postalservice wrote: »

We were given data based on test results of a class of 100 students
I calclulated the skewness value to be 72.4, mean of 38.6, median of 35.4 and mode of 32.5...and they are correct.

Are you sure?

I find it hard to believe that you could get a skewness coefficient that large given the mean, mode, and median you have quoted.

hivizman

MathsManiac wrote: »

Are you sure?

I find it hard to believe that you could get a skewness coefficient that large given the mean, mode, and median you have quoted.

Yes, I agree. I've done a couple of quick simulations and given your mean, median and mode I'd be surprised if your skewness should be more than about 1.5 to 2. That also puts into question the skewness of -10 for the other class - a set of marks is presumably constrained to fall between 0 and 100, and if you have 99 marks at 100 and one at zero, the skewness is -9.95, so that class would undoubtedly be better than the first class. This sounds like the sort of answer that would fit the question actually set!

So check that you've calculated skewness correctly for the first set of marks. There's a function in Excel that calculates skewness for you, which you could use as a comparison with your own calculation.