Question

Mating Choice and Offspring Fitness 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 study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), 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, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

Short answer

To provide a short answer without doing the calculations: (a) and (b) Perform a hypothesis test for two proportions. (c) and (d) Perform a hypothesis test for a single mean of differences. (e) If the truth is mate choice does have a positive effect, but the follow-up study fails to confirm that, it is a Type II error. (f) If the truth is mate choice doesn't have a positive effect, but the original study claims that, it is a Type I error.

Step-by-step solution

Hypothesis testing for two proportions

Let \(p_1\) be the true proportion of the mate choice group and \(p_2\) the true proportion of the no mate choice group. Null hypothesis: \(H_0: p_1 - p_2 = 0\). Alternative hypothesis: \(H_a: p_1 - p_2 > 0\). Calculate the test statistic. This will follow a normal distribution under the null hypothesis.

Calculate p-value for hypothesis testing in Step 1

The p-value is given by \(P(Z > z)\) where \(Z\) is a standard normal random variable and \(z\) is the calculated test statistic in Step 1. Use standard normal table or statistical software to obtain the p-value.

Make conclusion for hypothesis testing in Step 1

If p-value \(< 0.05\), reject \(H_0\). If p-value \(> 0.05\) fail to reject \(H_0\). For the later case, we do not have enough evidence to support the claim that mate choice increases offspring survival rate.

Hypothesis testing for a single mean

This is the paired data case where the counts of survival larvae from two groups are treated as a pair for each run. We are interested in the variable of differences. Let \(µ_d\) be the true mean difference in larvae survival rates between the two group. Null hypothesis: \(H_0: µ_d <= 0\). Alternative hypothesis: \(H_a: µ_d > 0\). Calculate the test statistic. This will follow a t-distribution under the null hypothesis.

Calculate p-value for hypothesis testing in Step 4

The p-value is given by \(P(T > t)\) where \(T\) is a random variable following the t-distribution with degree of freedom \(n - 1\) where \(n\) is the number of differences (50 in this case) and \(t\) is the test statistic calculated in Step 4. Use t-distribution table or statistical software to obtain the p-value.

Make conclusion for hypothesis testing in Step 4

If p-value \(< 0.05\), reject \(H_0\). If p-value \(> 0.05\) fail to reject \(H_0\). For the later case, we do not have enough evidence to support the claim that mate choice increases offspring survival rate even after considering different conditions for each run.

Discuss potential error for the follow-up study

If the follow-up study fails to reject \(H_0\), but the truth is mate choice does improve offspring fitness. Then Type II error may have been made in this study (false negative).

Discuss potential error for the original study

If the original study rejects \(H_0\), but the truth is mate choice does not improve offspring fitness. Then Type I error may have been made in this original study (false positive).

Key concepts

Mating Choice and Offspring Fitness

Understanding the role of mating choice in offspring fitness is crucial in the study of biology and evolutionary sciences. It's hypothesized that allowing female fruit flies to choose their mate could result in offspring with higher survival rates. This hypothesis is rooted in the idea that females may select males with traits advantageous for the survival of their progeny. After a study published in Nature stirred controversy by supporting this claim, replication studies were necessary to validate these findings.

Exercise Improvement Advice:

When exploring this topic or conducting an experiment, it's key to clearly define what 'offspring fitness' means. Measurements should be accurate, replicable, and the experimental design should control external variables to ensure the reliability of the results.

Survival Rates in Fruit Flies

The survival rate of fruit fly larvae is an important measure of success in the studies of genetics and environmental effects on new generations. In experiments assessing the impact of mate choice, this rate is used as an indicator of offspring fitness, with a higher survival rate in the mate choice group as a sign of improved fitness. These studies often involve comparison between controlled groups where female fruit flies are either permitted or not permitted to choose their mates, examining the resultant effect on their larvae.

Exercise Improvement Advice:

To ensure statistical relevance, the sample size must be sufficiently large. Additionally, the study must be designed to take into consideration the possibility of variations across different experimental runs, which could otherwise skew the results.

P-Value Calculation

The p-value is a fundamental concept in statistics, primarily used to determine the significance of results in hypothesis testing. It is the probability of obtaining test results at least as extreme as the observed results, under the assumption that the null hypothesis is correct. In biological contexts, such as the study of mating choice effects on offspring fitness, the calculation of the p-value informs researchers whether there is enough evidence to reject the null hypothesis. For a given significance level \( \alpha \), if the p-value is less than \( \alpha \), the null hypothesis can be rejected, suggesting that the observed data is not likely under the null hypothesis and thus supports the alternate hypothesis.

Exercise Improvement Advice:

While computing the p-value, it is essential to use appropriate statistical formulas or software and to correctly interpret the resulting p-value within the context of the studied phenomena.

Statistical Significance

Statistical significance is used to determine if the finding of a study is unlikely to have occurred by chance. This concept hinges on a significance level, denoted as \( \alpha \), which is the threshold for p-values below which findings are considered statistically significant. Commonly set at 0.05, it implies a 5% risk of concluding that an effect exists when there is no actual effect. In biological studies, establishing the statistical significance of results lends credibility to the conclusions drawn regarding phenomena such as mating choice and offspring fitness.

Exercise Improvement Advice:

Students should note that statistical significance does not necessarily imply biological importance, and they must consider the practical relevance and biological context of their findings.

Type I and Type II Errors

In the context of hypothesis testing, errors can occur: Type I and Type II. A Type I error, also known as a 'false positive,' happens when the null hypothesis is incorrectly rejected when it is actually true. Conversely, a Type II error, or 'false negative,' occurs when the null hypothesis is not rejected when it is false. Understanding these errors is crucial in the interpretation of results, such as those from the mating choice experiments in fruit flies, as incorrect conclusions can have significant implications in the understanding of evolutionary and genetic theories.

Exercise Improvement Advice:

Researchers and students should be aware of the risks of these errors and should carefully design experiments and analyze data to minimize their occurrence and to correctly interpret the results of hypothesis testing.