put the value of sqrt((card field f1 * card field f2) / (card field f3)) into card field f6
put the value of (card field f1 - card field f4) into card field f7
put the value of (card field f2 - card field f5) into card field f8
put the value of (1.96 * card field f6) into card field f9
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-- part contents for background part 1
----- text -----
STANDARD DEVIATION #1
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Suppose you self two Drosophila flies heterozygous for brown eyes. You would expect a 3:1 phenotypic ratio of wild vs brown in the F1. Suppose however out of 50 flies, you observe 20 that are brown and 30 that are wild. Is this the sort of spread you could reasonably expect if the phenotypic spread anticipated 37.5 wild and 12.5 brown (the expected 3:1 ratio)? Click on the CLEAR button and then assign the expected phenotypic frequencies to the values p and q. If p is the wild frequency then the expected value should be 0.75 of the total and the q frequency 0.25 of the total. Now, enter those figures into the boxes on the right and test the above data against those expectations. The correct answer is given below. Scroll to the point where it appears.
The answers should take these values:
Standard deviation = 0.061
obs p - exp p = 0.15
obs q - exp q = -0.15
1.96 stand. dev. = 0.120
Did your calculation coincide with this. If not, did you enter observed p and q frequencies or the actual observed values? If your data did coincide with the answers given, then your data spread does not fit the spread of a normal distribution. The value is well beyond two standard deviations from the mean. Although this card allows you to compare sampled data against an expected spread, its value for our purposes is limited. Sometimes it is useful to determine what the standard deviation of a normal distribution is in an effort to learn something about the spread of the curve. That computation can be made on the next card. If you wish to analyze data in this fashion, flip to it.
