Chapter 4, Section 4, Exercise 154 Mating Choice and Offspring Fitness: Mini-Exp
ID: 3312077 • Letter: C
Question
Chapter 4, Section 4, Exercise 154 Mating Choice and Offspring Fitness: Mini-Experiments Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study1 published in Nature found that mate choice does increase offspring fitness in fruit flies, yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments with data provided in MateChoice. The second study was actually a series of three different experiments, and each full experiment was comprised of days. Click here for the dataset associated with this question. (a) Suppose each of the 50 mini-experiments from the first study were analyzed individually. If mating choice has no impact on offspring fitness, about how many of these 50 p-values would you expect to yield significant results at -0.05? Round your answer to one decimal place. (b) The 50 p-values, testing the alternative Ha Pc>Pre (proportion of flies surviving is higher in the mate choice group) are given below Day 1: 0.96 0.85 0.14 0.54 0.76 0.98 0.33 0.84 0.21 0.89 Day 2: 0.89 0.66 0.67 0.88 1.00 0.01 1.00 0.77 0.95 0.27 Day 3: 0.58 0.11 0.02 0.00 0.62 0.01 0.79 0.08 0.96 0.00 Day 4: 0.89 0.13 0.34 0.18 0.11 0.66 0.01 0.31 0.69 0.19 Day 5: 0.42 0.06 0.31 0.24 0.24 0.16 0.17 0.03 0.02 0.11 How many are actually significant using -0.05? (c) You may notice that two p-values (the fourth and last run on day 3) are 0.00 when rounded to two decimal places. The second of these is actually 0.0001 if we report more decimal places. This is very significant! Would it be appropriate and/or ethical to just report this one run, yielding highly statistically significant evidence that mate choice improves offspring fitness? d) You may also notice that two of the p-values on day 2 are 1 (rounded to two decimal places). If we had been testing the opposite alternative, Ha PcPvc (proportion surviving is lower in the mate choice group), would these two runs yield significant results that preventing mate choice actually improves offspring fitness? (e) Is this an example of the problem of multiple tests?Explanation / Answer
(a) We are testing at alpha=0.05. which means that the probability of falsely rejecting the null hypotheses is 5%. Here the null hypotheses is mating choice has no impact on offspring fitness. Hence there is a 5% probability of these experiments getting the null hypotheses rejected. So out of 50, we expect 0.05*50 = 2.5 , approx 3 experiments to reject the null hypotheses. So out of 50 p-values we expect only 2.5 p-values to give significant result.
(b) We have to count the number of p-value which are lower than 0.05 to be significant. There are 8 values in the 50 experimental run which are less than 0.05. So answer is 8
(c) We should not choose which run to report, becasue then we violate the assumption of randomness. We must report a significant number of runs and their outcome to represent the true scenario. Here we find that out of 50 runs, 8 are significant, so we can see that there is hardly any evidence of offspring fitness varying according to mating choice.
(d) If we fix the alternative to be opposite of the alternative mentioned in part (b) and (c) then the p-value will change direction and the 2 valies of 1 will be replaced by very small values in order of 0.01. Hence the result will become significant in favour of the alternative Pc < Pnc
(e) There is an embedded problem of multiple tests. We have fixed the alpha alue at 0.05 for each test. However we must actually fix the alpha value for the family of tests at 0.05. Then we must use the bonferroni method to compare each p-value against 0.05/50 and then comment on the significance.
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