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Mating Choice and Offspring Fitness: MiniExperiments Exercise 4.153 explores the question of whether mate choice improves offspring fitness in fruit flies, and describes two seemingly identical experiments yielding conflicting results (one significant, one insignificant). In fact, the second source was actually a series of three different experiments, and each full experiment was comprised of 50 different mini-experiments (runs), 10 each on five different days. (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 \(50 \mathrm{p}\) -values would you expect to yield significant results at \(\alpha=0.05 ?\) (b) The 50 p-values, testing the alternative \(H_{a}\) : \(p_{C}>p_{N C}\) (proportion of flies surviving is higher in the mate choice group) are given below: $$ \begin{array}{lllllllllll} \text { Day 1: } & 0.96 & 0.85 & 0.14 & 0.54 & 0.76 & 0.98 & 0.33 & 0.84 & 0.21 & 0.89 \\ \text { Day 2: } & 0.89 & 0.66 & 0.67 & 0.88 & 1.00 & 0.01 & 1.00 & 0.77 & 0.95 & 0.27 \\ \text { Day 3: } & 0.58 & 0.11 & 0.02 & 0.00 & 0.62 & 0.01 & 0.79 & 0.08 & 0.96 & 0.00 \\ \text { Day 4: } & 0.89 & 0.13 & 0.34 & 0.18 & 0.11 & 0.66 & 0.01 & 0.31 & 0.69 & 0.19 \\ \text { Day 5: } & 0.42 & 0.06 & 0.31 & 0.24 & 0.24 & 0.16 & 0.17 & 0.03 & 0.02 & 0.11 \end{array} $$ How many are actually significant using \(\alpha=0.05 ?\) (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? Explain. (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, \(H_{a}:\) \(p_{C}

Short Answer

Expert verified
3 significant results might be expected if mating choice has no impact on offspring fitness, while there are actually 11 significant results in given data. It wouldn't be ethical to only report one very significant result. Reversed hypothesis wouldn't yield significant results with high p-values. Contradictory results could be due to multiple testing problem. Replication of studies is crucial, as it validates the original findings and brings up potential variations due to different settings or researchers.

Step by step solution

01

Determining Expected Significant Results

If the null hypothesis is true and mating choice has no impact on offspring fitness, at a significance level \(\alpha = 0.05\), we would expect 5\% of the p-values to be less than 0.05, therefore yielding a significant result. So from 50 experiments we would expect \(50 * 0.05 = 2.5\) significant results. Since we can't have half an experiment, it's typically rounded up to 3.
02

Count Actual Significant Results

Count the number of p-values from the given data that are less than 0.05, which makes them significant at \(\alpha = 0.05\). After counting, there are 11 p-values that are less than 0.05.
03

Discussing Reporting One Significant Result

Even though the p-value for one run is very low (0.0001), it wouldn't be appropriate or ethical to only report this one run and ignore the rest. This is because it presents a biased interpretation of the data and ignores the larger context of the experiments. Reporting only this result could lead to skewed interpretations and falsely report that mate choice always improves offspring fitness.
04

Testing Opposite Alternative

If the alternative hypothesis was \(H_a: p_C < p_NC\) (proportion surviving is lower in the mate choice group) and we consider p-values of 1 on day 2, these would not yield significant results. This is because a p-value of 1 indicates strong evidence in favor of the null hypothesis and against the alternative hypothesis. A low p-value (not a high one as in this case) would be needed to reject the null hypothesis.
05

Explaining Contradictory Results

The problem of multiple testing could explain contradictory results. The more tests we conduct, the higher the chance of obtaining a statistically significant result purely by chance, even if there's no real effect or difference. This issue generally inflates the Type I error rate (false positives).
06

Importance of Replication of Studies

Replication of studies is crucial because it helps validate the original findings. Different settings or researchers might introduce variations that could result in different outcomes, thereby providing a broader perspective on the phenomenon being studied. Replications can strengthen, refine or disprove the initial findings, making them an essential part of scientific inquiry.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
The null hypothesis is a fundamental concept in statistical testing. It posits that there is no effect or no difference between groups or variables; any observed effect is due to chance rather than a real effect. In the context of the mating choice in fruit flies, the null hypothesis would state that mate choice has no impact on offspring fitness.

Understanding the null hypothesis is crucial as it forms the basis for statistical significance testing. It's the default position that a researcher tries to reject with their data. Should the data provide sufficient evidence, the null hypothesis can be rejected, leading to accepting the alternative hypothesis, which suggests there is a meaningful effect or difference.

However, the decision to reject the null hypothesis should not be taken lightly. Researchers must look at the strength of the evidence, commonly measured by the p-value, and consider the potential for error.
P-value
The p-value is a statistical measure that helps researchers determine the strength of the evidence against the null hypothesis. It quantifies how likely it is to obtain the observed data, or something more extreme, if the null hypothesis were true.

A p-value is a probability, with a range from 0 to 1. A lower p-value indicates stronger evidence against the null hypothesis. Typically, a threshold (alpha level) is set to determine statistical significance, commonly 0.05. If the p-value is less than this threshold, the result is deemed statistically significant, and the null hypothesis is rejected in favor of the alternative hypothesis.

In the fruit fly exercise, a p-value below 0.05 for a mini-experiment would suggest a significant effect of mate choice on offspring fitness. Misinterpreting p-values can lead to false conclusions, so it's crucial to understand both what they represent and their limitations.
Multiple Testing
Multiple testing occurs when a researcher performs multiple statistical tests within the same study. Each test carries a risk of a Type I error – falsely finding evidence of an effect when there is none. When multiple tests are conducted, this risk compounds; we call this the problem of multiple testing.

Consider the mini-experiments with fruit flies. If each experiment is considered separately, the chance of observing at least one significant result by chance increases with the number of tests. Therefore, techniques such as the Bonferroni correction are often used to adjust significance levels and control the family-wise error rate. But even with such adjustments, the risk of spurious findings persists.

This phenomenon explains why the 50 mini-experiments yielded more statistically significant results (11) than expected by chance alone (approximately 3) if the null hypothesis were true.
Replication of Studies
Replication is the process of repeating research to verify its results. It serves a critical role in scientific progress since it can substantiate or refute the findings of a study. Replicating studies provides confidence in the reliability of the results and helps uncover any possible errors or biases in the original research.

In the case of the fruit fly experiments, replication across different types of flies and conditions brought varying results. This inconsistency emphasizes the importance of replication in different settings and by different researchers. Replications can reveal the nuances of the phenomenon under study and help the scientific community build a robust understanding of the subject. Ongoing replication efforts contribute to the self-correcting nature of science, allowing for comprehensive evaluation of scientific claims.

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Most popular questions from this chapter

In Exercise 3.129 on page \(254,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKeyor other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

Indicate whether it is best assessed by using a confidence interval or a hypothesis test or whether statistical inference is not relevant to answer it. (a) What percent of US voters support instituting a national kindergarten through \(12^{\text {th }}\) grade math curriculum? (b) Do basketball players hit a higher proportion of free throws when they are playing at home than when they are playing away? (c) Do a majority of adults riding a bicycle wear a helmet? (d) On average, were the 23 players on the 2014 Canadian Olympic hockey team older than the 23 players on the 2014 US Olympic hockey team?

Hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu=15\) vs \(H_{a}: \mu \neq 15\) (a) \(95 \%\) confidence interval for \(\mu: \quad 13.9\) to 16.2 (b) \(95 \%\) confidence interval for \(\mu: \quad 12.7\) to 14.8 (c) \(90 \%\) confidence interval for \(\mu: \quad 13.5\) to 16.5

Exercise 2.19 on page 58 introduces a study examining whether giving antibiotics in infancy increases the likelihood that the child will be overweight. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of a child's life, and each child was classified as overweight or not at age 12. (Exercise 2.19 looked at the results for age 9.) The researchers compared the proportion overweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likely to be overweight later in childhood compared with those who were unexposed \((32.4 \%\) versus \(18.2 \%\) at age 12 \(P=0.002) "\) (a) What is the explanatory variable? What is the response variable? Classify each as categorical or quantitative. (b) Is this an experiment or an observational study? (c) State the null and alternative hypotheses and define the parameters. (d) Give notation and the value of the relevant sample statistic. (e) Use the p-value to give the formal conclusion of the test (Reject \(H_{0}\) or Do not reject \(H_{0}\) ) and to give an indication of the strength of evidence for the result. (f) Can we conclude that whether or not children receive antibiotics in infancy causes the difference in proportion classified as overweight?

Give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p<0.5\) Sample data: \(\hat{p}=38 / 100=0.38\) with \(n=100\)

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