Lab Exercise 5

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Ethology <strong>Lab</strong>oratory Manualgrey coated animals (using our initial 3:1hypothesis). Our question is, is this observeddiscrepancy too large to be simply attributed tochance (if it is too large, we could conclude thatour original expectation of a 3:1 ratio is incorrectand that coat color is inherited in a differentmanner)? To determine whether our datacorrespond to this initial hypothesis, we use theChi-Square Test, where we subtract eachobserved value from its expected value, square theresult, and divide this quantity by the expectedvalue. We then add together the results for eachcategory. Chi-Square is summarized in thefollowing formula:In our example we can compute the following...2observed expected (o - e) / e78 75 0.1222 25 0.36table you look across that level of freedom untilyou find a value that is just larger than ourcalculated value (in this case 2.706). Looking atthe top of this value on the chart, we see that thereis at least a 10% chance our difference is simplyby chance alone. Because most hypothesis areconsidered invalidated only if there is a 5% chanceor less the data observed is due to chance, we canconclude that we have no statistically relevantreason to reject our initial hypothesis that the ratioof coat color inheritance is 3:1.The Students t-Test:The t-test is a test of significance of thedifference between two population means. Aprerequisite of this form of statistical analysis isthat the data must exist as at least discretevariables (rank variables are not eligible for thistest). The t-test is conducted by calculating thevalue of t and comparing it with a table similar tothe Chi-Square Test above. To calculate the valueof t for this test, the formula is...2summation of X = 0.48Now we can consult a Chi-Square Critical ValuesTable. In problems using Chi-Square, we need todetermine the degrees of freedom present in ourdata. This degrees of freedom = the number ofgroups minus one (d.f. = n-1). In this case there isone degree of freedom and when you look at thewhere 1 and 2 are the sample means, n 1 and n2are the sample sizes, and s and s2 21 2are thesample variances. The degrees of freedom is givenby n 1 + n 2 - 2. The hypothesis is rejected if thecalculated value of t is equal to or greater than thetabulated value of t.10
<strong>Lab</strong>oratory 5 Analysis of BehaviorThe following example shows the use ofthe t-test statistic: Female voles were collectedfrom two locations and their total length weremeasured. Sample A had a mean length of 21.84mm, a variance of 4.2 mm , and n = 172. SampleB had a mean length of 23.45 mm, a variance of4.5 mm, and n = 135. We want to know if themean length of these two populations isstatistically different. In this case the hypothesiswe are testing is that the length of sample Aindividuals does not differ from sample Bindividuals. Substituting our variables into theequation we obtain...t = 6.7 ; d.f. = 305To find the tabular value of t, look in the tableprovided under alpha = 0.05 (or standardacceptance point of 5%). The table value is 1.960.Since our calculated value is far greater than this,we can reject our initial hypothesis that thepopulations are equal.Questions:1. If money, time, and all the hassles of life were no object, what type of behavior and in what type ofanimal(s) would you most like to study?11
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Lab Exercise 5

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