Question

There are 16 students taking a test and the scores are:

80, 72, 96, 48, 90, 64, 100, 88, 92, 90, 76, 93, 84, 78, 96, 90

How could data analysis be used to show the test was a good test, or was not a good test?

To answer this, consider the following methods: standard deviation, z-scores, distribution, box and whisker plots, histograms.

There is more than one right answer - just be able to justify your answer. Regardless of your answer, you NEED TO SHOW YOUR WORK! You can do this by hand and submit an image if you'd like. If you use a calculator, you still need to EXPLAIN the process with the calculations.

Answer 1

For the scores

80, 72, 96, 48, 90, 64, 100, 88, 92, 90, 76, 93, 84, 78, 96, 90

use R software

mean(testscores)
[1] 83.5625
> median(testscores)
[1] 89

mean=83.5625

median=89

standard deviation= 13.57924

mean is not equal to median

therefore distribution is not normal.

Distribution is skewed since mean value is less than median value.

Negative skewed distribution.

and in practical terms this means a higher level of occurrences took place at the high end of the distribution. Think of a test administered where there were a great many easy scores and very few lower scores from your class.

the test was a good test,

There is outlier which can be seen form boxpplot and histogram.

test was a good test.

histogram is