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# Animal Behavior BIOL 4230

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This 12 page Class Notes was uploaded by Derick Crona on Wednesday October 21, 2015. The Class Notes belongs to BIOL 4230 at Tennessee Tech University taught by Christopher Brown in Fall. Since its upload, it has received 17 views. For similar materials see /class/225711/biol-4230-tennessee-tech-university in Biology at Tennessee Tech University.

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Date Created: 10/21/15

Ammal Behavror BIOLW39FS 42305230 Spnng 2008 STATISTICS AND HYPOTHESIS TESTING data O mumes heyw r r rm y u rem u Tu cunductthe rst expenmenL hmd w39mxmw aquot kF th beluw Desaipziwzsmm statistic and me whreh hupefully yuu are all farmhzr mm rs the mm ur the average as us mure pupulzrly called The mean Dr E represean Lhe mddlequot er r mamemauea1 arms Lhecenlml tendency quhe data and rs grvm by me funnula urnmww n M hmar mean medAan and mude aruund the mean Fur example luuk at the data curves beluw Eaeh quhese cmle haveme same mean butthar spreadnessquot vanes qulte 2be Themustcummun Thevananee err2 rs grver by me nrmula ZltX eff n71 standard aevranure err rs than amply are square mut quhe varranee 2 5 Animal Behavior BIOLWFS 42305230 Spring 2008 rt Note that most descriptive statistics can be calculated using a spreadsheet program such as Excel or from higherend calculators using the TC and owl buttons Parameters versus Statistics Before we leave the idea of descriptive statistics you should know a little about an important distinction in mathematics You ll notice that in our running speed example above we did not record speeds for each and every anole of each sex This would have been much too timeconsuming and dif cult Instead what we did was take a sample of running speeds for each sex We then calculated our mean variance etc from this sample These values the mean variance etc are called statistics because they represent one estimate of the true values of the mean and variance So that s what statistics really re estimates of an unmeasured true value for a population You can probably guess that if we went and raced a second sample of anoles the values of the mean and variance would likely differ from what we got for the rst sample The true values for a population are called parameters or parametric values and are usually represented by Greek letters the mean by p mu and the variance by 62 sigma squared We almost never know the parametric values but we assume that our statistics come reasonably close to these parameters In general as long as we randomly sample our populations and have a large enough sample size this assumption will hold However it s always possible that we just happen to sample anoles that are uncommonly fast or uncommonly slow if so then our statistics will likely differ from the parametric values 2 SE L Inferential Statistics and the Logic of Experimental Design At the heart of all good science lies a question in our example we ask the question does running speed of anoles differ between males and females Before we even collect data we can come up with one or more potential answers to this question We call these potential answers hypotheses The most basic hypothesis in most studies suggests that there are no differences between our populations that is male and female anoles run at equivalent speeds and we call this the null hypothesis As researchers we typically hope to prove this hypothesis false and so we can have the following two types of alternative hypotheses 1 the data sets are different from each other but we don t know the direction of the difference larger or smaller or 2 the difference between the data sets is in a predicted direction for example females gt males usually we need previous experiments to make this hypothesis If we somehow knew the parametric values then it would be easy to determine if our populations differed for example if average female speed was 25 cmsec and average male speed was 35 cmsec then we know these differ because we ve raced everV stupid anole and 25 is not equal to 35 However we haven t raced every anole so we have uncertainty in how close our statistics are to the parametric values This is where inferential statistics come in they help us as researchers make objective decisions about differences which might occur between two or more sets of data or to put it another way they help us answer the question quotare the differences we see real or have they occurred simply by chancequot Inferential statistical tests come in a wide variety of forms but nearly all have the same conceptual framework We begin with our null hypothesis male speed is equivalent to female speed in our example and then collect data on running speed from both sexes to determine whether to accept or reject this hypothesis We then choose an appropriate statistical test this may or may not be a simple matter but we ll ignore those types of details for now Many basic types of tests can be done using Excel or other spreadsheets Whatever we choose the test will calculate a test statistic based on your data for example a tvalue for the ttest or an F value for analysis of variance or an rvalue for a correlation You or more likely the computer program will then compare your calculated test statistic to an expected or cutoff value and this will generate something called the probability or P Value Now for the big question what does all of the above mean To answer that let s rst do a thought experiment Suppose we take two different samples from just the female anoles We would expect Animal Behavior BIOLWFS 42305230 Spring 2008 these samples to have similar although not necessarily identical means and variances However what if we happened to collect extremely fast anoles in one sample and extremely slow anoles in the other These means and variances would differ greatly from each even though they both come from the same population with a single true mean and variance simply because we chose very different anoles by chance Inferential statistics and their associated Pvalues attempt to quantify the likelihood that this is going on In essence they answer the following if two populations really have the same mean what is the probability that samples taken from each population will differ by as much as we calculate they do by random chance alone A slightly different way to ask this is what is the probability that our two samples come from populations with the exact same parametric mean value For example suppose you perform a ttest on running speed and get a Pvalue of 025 This indicates that there is a 25 chance that these samples come from populations with the same parametric mean that is a 25 chance that running speed does not really differ between sexes or a 75 chance that running speed really m differ between sexes The question then becomes at what probability level do we say that two samples are different enough to be considered well di erent The answer to this is actually very subjective see below but in general biologists have agreed that a test resulting in a Pvalue of 005 or less will be regarded as demonstrating real differences between groups Thus a test with a Pvalue of 005 or less is said to be signi cant and rejects the null hypothesis while a Pvalue of greater than 005 is said to be nonsigni cant and fails to reject the null hypothesis you never support the null hypothesis Another way of saying this is that with P S 005 we have at least a 95 probability that our two populations are really different Why 005 Well once a test has been done there are four possible outcomes 1 you may fail to reject a true null hypothesis 2 you may reject a false null hypothesis 3 you may reject a true null hypothesis a Type I error or 4 you may fail to reject a false null hypothesis a Type II error The last two conclusions are incorrect We can minimize the chance of making a Type I error by choosing a low cutoff probability level symbolized u for rejection of the null hypothesis Thus we can choose our on 0001 instead of 005 if we want this will ensure that only if there is a 999 chance that differences did not occur by chance will we reject the null hypothesis However by doing so we greatly increase the probability of incorrectly accepting the null hypothesis when it is in fact wrong and thus making a Type II error Biologists have therefore chosen on 005 as a kind of compromise between these two types of error Animal Behavior BIOLWFS 42305230 Spring 2008 Some Statistical Calculations For the females only you can do the males ifyou d like 1 Mode Look for the most common speed in this case it is 38 which appears three times 2 Median First arrange the values from lowest to highest which gives 28 29 313134 38 38 38 45 45 47 49 5152 59 Now look for the number with an equal number of data points below it and above it in line For this dataset that number is 38 there are 7 speeds below and 7 above If the sample size is an even number you have to take the average of the two numbers with equal data points above and below 3Mean 7 3847382945593845284951343l523l X 4lcmsec 15 4Va1iance S 2 38 412 47 412 38 412 29 412 45 412 31 412 9043 15 1 5 Standard Deviation s x9043 951 SE 4 246 15 6 Standard Error of the Mean Animal Behavior BIOLWFS 42305230 Spring 2008 NOTE Most of the rest of this lab handout is borrowed from Elizabeth A Jakob and Marta J Hersek in Learning the skills ofResearch Animal Behavior Exercises in the Laboratory and Field All material copyrighted 2004 by the authors Choosing a Statistical Test To determine which type of statistical test to use on a given set of data we must rst determine whether or not the data t a normal bellshaped distribution If so we can use parametric statistical tests If the data are not normal we must use non parametric tests Since many of the data collected in animal behavior studies are not normally distributed we will focus on nonparametric tests in this lab A ow chart to help you decide which tests to use is given in Figure 1 at the end of the handout Below are some useful de nitions and then a series of worked examples for a number of nonparametric tests Here are some helpful terms ordinal data numerical data such as number of seconds distance or frequency of a behavior categorical data data that can be put into categories such as alivedead or activeinactive unpaired data data points that are independent from each other such as data generated by testing two separate groups of animals paired data data points that are naturally paired in some way most commonly because the same animal was tested more than once These data points should not be treated as independent from one another number of groups the number of different test groups being compared Test 1 39 Wilcoxon Ranked Sums test This test is used to determine the signi cance of differences between two sets of unpaired data A ranking system is used Example You are interested in whether the movement rate of the protozoan Paramecium caudatum is in uenced by whether they are tested under dim or bright light The null hypothesis is P caudatum has the same rate of movement under both conditions You measure movement rate by counting the number of squares in a counting chamber a Paramecium crosses every 10 seconds P candutum movement data squares crossed per 10 sec Dim Light Rank Bright Light Rank 10 7 5 I 11 95 6 2 12 115 7 3 12 115 8 4 15 I3 9 5 16 I4 10 7 17 I5 10 7 11 95 1 Order each group from smallest to largest Next rank the data of the two groups combined The lowest score of both groups gets a value of 1 the next highest of both groups a value of 2 etc In the case of ties for example two values of 12 each value is ranked the ranks are averaged and the average rank is assigned to each ofthe tied scores 11122 115 Ifyou ve done this properly your last rank will equal N the total number of samples 2 Designate the sample size of the larger group as N L and that of the smaller as N S In our example N L 8 and N 7 S 3 Sum the ranks T of each group Animal Behavior BIOLWFS 42305230 Spring 2008 TS795ll5ll5l3l415815 TL123457795385 4 Calculate the test statistics U S and U L USNSNL N N 1 S23 TS25 UL NSNL US 535 5 Choose the greater of the two values of U This is the test statistic Compare it to the critical value in Table 1 The test statistic must be higher than the critical value to be signi cant In this example the higher Uis 535 Look in Table 1 under n 8 and m 7 at the 95 level P 005 NOTE m is the sample size of the group with the smallest ranks sum in this example it is N L n is the sample size of the group with the highest ranks sum in this example it is NS The critical value is 46 since 535 gt 46 we can reject the null hypothesis with a 95 probability that rejection is correct We conclude that Paramecium swim more slowly under bright light Test 2 Kruskal Wallis test The Kruskal Wallis Test is similar to the Wilcoxon rankedsums test but here we have more than two groups Work through the Wilcoxon rankedsums test example before attempting this one Example You are interested in the antipredator behavior of garter snakes You wonder how close you as a simulated predator can get before the snake crawls away Because snakes are poikilotherms and can move more quickly when it is warmer you suspect that this behavior is in uenced by temperature You compare three groups snakes at 23 C 25 C and 27 C The data are closest approach distance in meters The null hypothesis is that snakes tested under these three temperatures do not differ in how close an experimenter approaches before they ee Flight distance of snakes in meters 23 C Rank 25 C Rank 27 C Rank 05 I 075 2 35 7 1 3 325 6 55 12 125 4 4 8 6 I3 3 5 475 10 8 14 425 9 525 11 R1 22 R2 37 R i 46 1 First order and rank the data as described for the Wilcoxon rankedsums test When there are tied scores each score is given the mean of the ranks for which it is tied Compute the sum of the ranks for each group symbolized by R R 1 is the sum of the ranks of group 1 etc 2 Now compute the test statistic H using the following formula 12 R 2 H 7 Z 3N1 N N 1 711 In this formula the 2 Es a summation sign and indicates that you should sum up each R2 value fromR 1 to R 3 Plugging in the appropriate numbers forR N the total number of observations and n the number of observations in each group Animal Behavior BIOLWFS 42305230 Spring 2008 3141 64 5 5 H 2 12 my my 462 14141 4 If you have a large number of ties use the correction for ties Compute H as above then divide by Z r3 t N 3 N where t the number of observations in a tied group of scores and N the total number of all observations 3 Compare your test statistic with Table 2 The dfcolumn corresponds to the degrees offreedom for this test Degrees of freedom is a number that results from the way the data are organized and refers to whether the observations are free to vary For example if all of 50 observations must fall into 2 categories as soon as we know that one category holds 41 data points then the other category holds 9 For every statistical test there are established methods for determining degrees of freedom For the KruskalWallis test we have df ofgroups 7 1 In our example df 31 2 The test statistic must be higher than the critical value to be signi cant H at 64 is greater than 599 so you may reject the null hypothesis at P lt 05 The three groups do differ Test 3 Sign test The sign test is used for twogroups when the data are paired In this test only the signs of the differences are used Another nonparametric test the Wilcoxon matchedpairs signed rank test is more powerful because it uses both the signs and the magnitude of the differences Example You imagine that male mice might bene t from avoiding inbreeding or mating with close relatives Because mice depend on odor for a great deal of their information about the world you decide to present males with soiled litter from the cages of females You test each male twice once with litter from his sister and once with litter from a stranger The females are sexually receptive so the soiled litter should be rich in chemical cues You present the litter in random order so that half the males get their siblings litter rst and half get the stranger s litter rst Since the same males are tested twice a MannWhitney U test is inappropriate Null hypothesis The number of sniffs per minute will be the same when males are exposed to the litter of their sisters vs that of strangers Male ID Number of SniffsMin Number of SniffsMin with Sign ofthe Difference Number with Sister s Litter Stranger s Litter 1 9 2 8 3 3 3 5 4 20 11 5 15 9 6 35 21 7 4 6 8 11 10 9 41 20 10 22 21 11 16 16 0 12 18 17 13 7 0 14 11 5 1 Subtract one data column from the other to determine the sign of the difference It doesn t matter which you subtract from which just be consistent Animal Behavior BIOLWFS 42305230 Spring 2008 2 Note the most frequent sign In this case the most frequent sign is positive and there are 11 The test statistic therefore equals 11 3 Determine N the number of pairs which showed a difference Here we disregard male 11 so N 13 4 Look at Table 3 for N 13 Note the critical value at the 95 level In this test the test statistic must be equal to or greater than the critical value In this example we can reject the null hypothesis at P 005 Test 4 Wilcoxon s Signed Ranks Test As with the sign test this is used for paired data Example Juvenile scorpions may climb vegetation when in the presence of an adult We placed juvenile scorpions in the presence or absence of an adult and measured how high into vegetation they climbed Each juvenile was tested under both conditions with the order of testing randomized among juveniles some were tested with an adult rst others without an adult rst Height Climbed by Juvenile Scorpions in m Adult Present Adult Absent D Rank R 032 9 260 241 019 8 243 239 004 2 290 285 005 3 294 282 012 7 270 273 003 1 268 258 010 6 298 289 009 5 285 278 007 4 1 Compute the difference D between the paired observations 2 Rank the D s from smallest to largest ignoring the sign After you have ranked the data go back and assign each rank the original sign from the D column 3 Sum the positive and negative ranks independently this gives T 44 and T 1 Take the smaller of these two values and compare it to the value in Table 4 for n 9 where n is the sample size If our smaller value is less than or equal to the table value at P 005 our results our signi cant We conclude that juvenile scorpions climb higher in the presence of an adult Test 5 Chi Square test of39 J r J and 7 WM 5 d f t test Tests using the chisquare statistic are useful when you have nominal data categories rather than numbers For example a category might be large vs small laid eggs vs did not lay eggs etc First we will look at the chisquare test of independence This test helps us determine whether two variables are associated If two variables are not associated that is they are independent knowing the value of one variable will not help us determine the value of the other variable Example When snails sense the presence of a nearby star sh a predator via chemicals in the water they will climb We can look at three groups of snails the rst group is the control group with the snails exposed to plain sea water the second group is exposed to water scented by a sea urchin an herbivore and the third group to sea water scented by a predatory star sh The data collected for each snail is whether it climbed or not These are categorical data the snail could do one thing or the other The categories are mutually exclusive the snail could not climb m not climb If the variables are independent there will be no Animal Behavior BIOLWFS 42305230 Spring 2008 relationship between the type of water the snail is exposed to the rst variable and how it responds the second variable Note if instead of making categories of climb and not climb the experimenter had measured the distance each snail moved the chisquare test would be inappropriate 1 Make a table of observed frequencies the data actually collected in the experiment Observed frequencies Source ofTest Water Behavior Control Sea urchin Predator Row totals Climb 12 14 24 50 Not climb 28 23 15 66 Column totals 40 37 39 Grand total 116 Note The grand total of the rows should equal the grand total of the columns 2 Calculate and tabulate the expected frequency for each category that is the frequency expected in each category if there is no relationship between the variables column total x row total grand total Expected frequencies Source ofTest Water Behavior Control Sea urchin Predator Row totals Climb 172 16 168 50 Not climb 228 21 222 66 Column totals 40 37 39 Grand total 116 3 Calculate the value of chisquare X2 2 0 E2 I Z E where O the observed frequency in each cell and E the expected frequency in each cell 2 2 2 2 2 2 2 12 172 28 228 14 16 23 21 24 168 15 222 172 228 16 21 168 2 862 4 Examine the table of critical values for this test Table 2 For the chisquare test the degrees of freedom formula is df rows 1 columns 1 21 31 2 We compare the test statistic to the critical value if it is bigger we reject the null hypothesis The calculated X2 is 862 which is greater than 599 The three groups of snails moved differently A second type of chisquare test is called the chisquare goodnessof t test In this case the experimenter tests to see how the data match expected values that were determined before the test was run For example in Mendelian genetics we can predict the outcome of different crosses the ratio of the different types of offspring is known in advance In this case we compare the observed values from the experiment with the expected values generated by theory The calculations are performed in exactly the same way as for the chisquare test of independence Test 6 The binomial test This test is useful for categorical data where we have only two categories and when we are interested in testing whether the data are equally likely to fall into either category Animal Behavior BIOLWFS 42305230 Spring 2008 Example You ve been using a coin to randomly assign treatments to your experimental animals but you are beginning to suspect that the coin is not fair and you decide you d better test this The null hypothesis is the coin is equally likely to come up tails as heads 1 Flip the coin 11 times Nine times it comes up heads and twice it comes up tails 2 Using Table 5 locate the value for N in this case 11 along the left side Compare your smallest number in this case 2 which is the number of tails to the smaller of the two numbers in the 2tailed test on 005 column If your number is equal to or less than this table value the result is signi cant Since 2 gt 1 we cannot reject the null hypothesis and so we must assume the coin is fair Animal Behavior BIOLWFS 42305230 Spring 2008 H amewark Problems Optional For further practice you may attempt the following problems In each an experiment is described Determine which statistical test is most appropriate and answer all questions posed Refer back to the worked examples to help you understand how to conduct each test 1 Elephants make lowfrequency sounds inaudible to humans Apparently these sounds are used in longdistance communication between individuals You are interested in the response of bull and female elephants to the sound of a female who is ready to mate You mount a giant speaker on top of your van and drive around the plains looking for elephants When you nd one you stop 15 m away play the sound and watch the elephant s response You discover 9 bull elephants approach the van 2 bull elephants do not approach the van 3 female elephants approach the van 11 female elephants do not approach the van Your experiment ends prematurely when one of the bull elephants apparently enraged by the absence of a female tips the van over and damages the speaker You hope that you have enough data to make a claim about males and females A What is the null hypothesis B What statistical test should you use C Calculate your test statistic Is your result statistically signi cant D What conclusion can you draw from this experiment 2 Male butter ies sometimes court females of other species with similar wing patterns You are interested in how long males persist in courting the wrong female You decide to test each male with a dead female to control for the effect of the female s behavior You use three types of test females one from the same species as the males one from a different species with a similar wing pattern and one from a different species with a different wing pattern Each pair is placed in a cage and you measure courtship time in seconds female of same species 23 20 17 25 28 female of different species similar pattern 18 27 24 21 female of different species different pattern 22 21 23 20 A Calculate the mean variance and standard deviation for each group B Qualitatively compare the means and standard deviations for each group Do they look very different Very similar C Which statistical test would you use to look for differences D Perform the test What is your test statistic Can you reject your null hypothesis E Give a biological reason why your test may have come out the way it did 3 Honeybees returning from foraging convey information to bees in the hive about the location of food resources One way they do this is through a waggle dance that other bees watch Another way they convey information is by regurgitating some of the food they have collected to other bees a process known as trophyllaxis You are interested in the speed at which bees nd a resource another bee tells them about You decide to compare bees that have only observed a dance with bees that have observed a dance M accepted regurgitated food You mark a lot of bees with bee tags little numbered discs that you glue to the back of the thorax This enables you to watch the same individual repeatedly One day you choose a lot of bees that have seen a waggle dance but not accepted food You measure in seconds how long it takes for them to nd the resource A week later you go back to the hive and nd the same individuals This time you watch until they see a dance and accept food and again measure how long it takes them to reach the resource Here are your data The numbers are seconds needed for the bee to reach the resource Animal Behavior BIOLWFS 42305230 Spring 2008 Bee Watch Only Watch and Accept Food 1 87 80 2 53 48 3 57 57 4 89 88 5 48 38 6 109 160 7 109 100 8 48 78 9 29 26 10 45 41 11 67 53 12 120 98 13 55 55 14 89 78 A What sort of data are these Which test should you choose B What is the test statistic The table statistic C You decide that a bee that has both watched and gotten food from another bee nds the resource faster than one that has just watched What other factor that is a result of your experimental protocol might also explain your results

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