Statistics Elementary and Probability Analysis
PROBABILITY: the number of times a particular event occurs, divided by the total number of opportunities for the event to occur. Example: If a coin is flipped an infinite amount of time the probability that you would get heads is 0.5 and the probability that you would be tails is 0.5 . SUM RULE: the combined probability of two (or more) mutually exclusive events occurring is the sum of their individual properties. Example: heads and tails are examples of mutually exclusive events. In other words, if a tail is flipped, that excludes the possibility of a head being flipped on the same coin at the same time (also the same with gametes). Prob. (head or tail) = prob, (head) + Prob. (tail) 1.0 = 0.5 + 0.5 PRODUCT RULE: the joint probability of both of two independent events occurring simultaneously is the product of their individual probabilities. Example: What is the probability of obtaining a head on both a nickel and a penny? Prob. of two heads = Prob. (head on penny) x Prob. (head on nickel) (0.5) x (0.5) = 0.25 OR: What is the probability of obtaining rr from the cross Rr x Rr? Based on the principle of segregation, the probability of an "r" from the female is 0.5 and the probability of an "r" from the male is 0.5. Therefore, the probability of obtaining a "r" simultaneously from the female and male gametes to yield an rr offspring is (0.5) x (0.5) = 0.25 BINOMIAL EXPANSION: combinations of events can be calculated by expanding the binomial (p +q)n in which n is the group size, p is the prob. of event 1 and q is the prob. of event 2. Any term in this binomial expansion and therefore any combination of events can be determined using the following formula: n! pxqn-x x! (n-x)! (4!=1x2x3x4) Example: What is the prob. of getting 6 head & 2 tails if you flip 8 coins? n=8 x=6 p = 0.5 q = 0.5 8! 6! (2!) 0.56 (0.52) 28 (0.016)(.25)= 28/256
Goodness Fit:
Mendel's of data closely resembled a 3:1 ratio in the F2 generation of the monohybrid cross. But since it was not a perfect 3:1 ratio, would this be due to chance or because the underlying hypothesis, the principle of segregation, was incorrect? For this reason, it is necessary to have a statistical test to measure the "goodness of fit" of the observed and expected or predicted data - whether the numbers observed are adequately close to the expected numbers according to a particular hypothesis. CHI-SQUARE TEST - commonly used test to measure the goodness of fit; the observed numbers are compared to the expected numbers. (O-E) 2 /E
Example:
Parental cross
GENERATION
round, yellow vs. wrinkled, green round, yellow
RATIO
F1 generation round, yellow F2 wrinkled, yellow round, green wrinkled, green SEED TYPE round, yellow wrinkled, yellow round, green wrinkled, green OBSERVED 315 101 108 32 556
SEED TYPE________OBSERVED
315 101 108 32 EXPECTED 313 104 104 35 556
9.84 3.16 3.38 1.00 X2 0.01 0.08 0.15 0.25 X = 0.49
2
The significance of the chi-square value can be determined by comparing it to a theoretical value from a chi-square table that has the same degrees of freedom. The term degree of freedom refers to an integer that is often equal to the number of categories in the data, minus one. If the chi-square value exceeds the value listed in the table,, then the underlying hypothesis may not be correct or significant experimental error occurred, as the deviation from the expected results are not due to chance atone! In this example, there are three degrees of freedom, and at the 0.05 level the maximum acceptable chi-square value is 7,815. From this we can assume that our hypothesis is correct, and that no significant experimental error occurred. df = n-1 where n is the number of categories or classes in the data probability range - __> p>__ For the problem above 0.95>p>0.90