Summary from last time:
Refer to document on dispersions for more detail.
The calculation of dispersion in a distribution is very important because it represents a uniform way to determine probabilities and therefore to determine if some event in the data is expected (i.e. probable) or is significantly different than expected (i.e. improbable).
For the last 25 years the mean annual rain in Eugene is 51.5 inches with a dispersion of 8 inches.
During this same period, the mean annual rain in Seattle was 39.5 inches with a dispersion of 7 inches.
On average, does it rain significantly more in Eugene than Seattle?
Here is the wrong way to do this problem:
But this is not the correct procedure to use when comparing two separate distributions. It is only the correct procedure to use when comparing one data point to the rest of the same distribution.
In this example, the number of data points is 25 and the square root of 25 is 5. Hence, for Eugene, the error in the mean value of 51.5 inches is 8/5 = 1.6 inches.
The difference in mean rainfall between Seattle and Eugene is 12 inches which is 12/1.6 = 7.5 dispersion units.
Thus there is a highly significant difference in the mean annual rainfall between Eugene and Seattle.
Note this method is only an approximation. A more exact and proper way to compare two sample means will be given later.
Another way to look at this rainfall comparison is as follows:
We have already determined that 65 inches is not a significant amount of rainfall in Eugene compared to the normal value of 51.5 inches. Would 65 inches be a significant amount of rain in Seattle?
For the case of Seattle, 65 inches is 65-39.5 = 26.5 inches above normal. The dispersion in the Seattle data is 7 inches and so 26.5 inches is 26.5/7 = 3.8 dispersion units above the mean. This is highly significant which again reinforces the notion that there is a significant difference in mean rainfall between
Eugene and Seattle (note also this difference in community web pages).
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