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Red Wine and Weight Loss Resveratrol, an ingredient in red wine and grapes, has been shown to promote weight loss in rodents. A recent study \(^{19}\) investigates whether the same phenomenon holds true in primates. The grey mouse lemur, a primate, demonstrates seasonal spontaneous obesity in preparation for winter, doubling its body mass. A sample of six lemurs had their resting metabolic rate, body mass gain, food intake, and locomotor activity measured for one week prior to resveratrol supplementation (to serve as a baseline) and then the four indicators were measured again after treatment with a resveratrol supplement for four weeks. Some p-values for tests comparing the mean differences in these variables (before vs after treatment) are given below. In parts (a) to (d), state the conclusion of the test using a \(5 \%\) significance level, and interpret the conclusion in context. (a) In a test to see if mean resting metabolic rate is higher after treatment, \(p=0.013\). (b) In a test to see if mean body mass gain is lower after treatment, \(p=0.007\) (c) In a test to see if mean food intake is affected by the treatment, \(p=0.035\). (d) In a test to see if mean locomotor activity is affected by the treatment, \(p=0.980\) (e) In which test is the strongest evidence found? The weakest? (f) How do your answers to parts (a) to (d) change if the researchers make their conclusions using a stricter \(1 \%\) significance level? (g) For each p-value, give an informal conclusion in the context of the problem describing the level of evidence for the result. (h) The sample only included six lemurs. Do you think that we can generalize to the population of all lemurs that body mass gain is lower on average after four weeks of a resveratrol supplement? Why or why not?

Short Answer

Expert verified
The changes in resting metabolic rate, body mass gain and food intake are statistically significant with a \(5\%\) significance level. The strongest evidence is for body mass gain, while the weakest is for locomotor activity. At a \(1\%\) significance level, only the change in body mass gain is statistically significant. Informally, the resveratrol seems to noticeably impact all measurements except locomotor activity. It would be premature to generalize these results to all lemurs due to the small sample size.

Step by step solution

01

Analyze The P-Values

A p-value is a statistical measure of the probability that the result of your test occurred at random. If the p-value is less than your significance level (here, \(5\%\)), then the observed data is considered statistically significant. Below are the conclusions of test based on given p-values:
02

Understanding Results For Part (a), (b), (c) and (d)

(a) For resting metabolic rate, as the p-value \(0.013 < 0.05\), the change is considered significant, hence the mean metabolic rate is higher after treatment. (b) For body mass gain, as the p-value \(0.007 < 0.05\), the change is considered significant, hence the mean body mass gain is lower after treatment. (c) For food intake, as the p-value \(0.035 < 0.05\), the change is considered significant, hence the mean food intake is affected by the treatment. (d) For locomotor activity, as the p-value \(0.980 > 0.05\), the change is not considered significant, hence the mean locomotor activity is not significantly affected by the treatment.
03

Deciding Test With Strongest and Weakest Evidence

Since the p-value gives the evidence against the null hypothesis, a smaller p-value provides stronger evidence. Therefore, the test with the smallest p-value (body mass gain) shows the strongest evidence and the locomotor activity has the weakest evidence with the highest p-value.
04

Reanalyzing At \(1\%\) Significance Level

The main change is the level of significance, which is now lower at \(1\%\). This means that you are allowing a smaller possibility of error. If the p-value is smaller than the significance level of \(1\%\), (that is, if \(p < 0.01\)), the difference is significant. However, only the p-value of the body mass gain test is smaller than the \(1\%\) level. Hence, only the result of that test would be considered significant at the \(1\%\) significance level.
05

Informal Conclusion For Each P-Value

Informally, each p-value gives a sense of whether there is evidence to reject the null hypothesis. For the resting metabolic rate, body mass gain, and food intake tests, small p-values suggest that the resveratrol treatment has a significant impact. However, the high p-value for locomotor activity suggests that there is no significant evidence to suggest that resveratrol affects locomotor activity.
06

Discussing Sample Size

The results are based on a sample of just six lemurs. It’s risky to generalize this result to all lemurs because a bigger sample size usually provides a more accurate representation of the population. Therefore, while the study provides some evidence of resveratrol’s effect on weight loss, broader conclusions would require a larger sample size.

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Key Concepts

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

Statistical Significance
Statistical significance plays a pivotal role in research studies, including those that test the effectiveness of treatments or supplements like resveratrol. But what exactly is statistical significance? It's a measure that tells us whether the results observed in a study are likely due to genuine effects, rather than random chance.

When researchers set out to test a hypothesis, they decide on a 'significance level'—often 5%. This is represented as \(0.05\) in probability terms. If the p-value obtained from the study is less than this threshold, the result is deemed statistically significant. In the context of the resveratrol supplementation study, a p-value less than 0.05 implies that the observed changes in the lemurs (in terms of metabolic rate, body mass gain, etc.) are statistically significant and unlikely to be due to random variation.

For example, observed p-values like \(0.007\) for reduced body mass gain indicate a high likelihood that resveratrol truly affects weight. Conversely, a p-value of \(0.980\) suggests that changes in locomotor activity are likely to be random, not a true effect of the supplement. Interpreting these values correctly is crucial for drawing reliable conclusions about the effects of resveratrol supplementation in the study.
Null Hypothesis Testing
In statistical analysis, the concept of null hypothesis testing is central to interpreting study results. The null hypothesis—often denoted as \(H_0\)—is a default assumption that there is no effect or no difference; in this case, that resveratrol supplementation has no impact on the lemurs. Researchers collect data and use it to test this hypothesis.

To determine if the null hypothesis can be rejected, p-values are calculated. They represent the probability of observing data at least as extreme as what was found, assuming that the null hypothesis is true. If this probability (the p-value) is low enough, the null hypothesis is considered unlikely and is rejected in favor of an alternative hypothesis (such as \(H_1\): resveratrol supplementation does have an effect).

For instance, a low p-value \(0.007\) in the resveratrol study indicates strong evidence against the null hypothesis that body mass gain remains unchanged after treatment, suggesting that the supplement does indeed affect weight. Conversely, a high p-value, like the one observed for locomotor activity, would not lead to rejection of the null hypothesis, signifying a lack of evidence for the supplement's effect on that particular measure.
Resveratrol Supplementation Study
A study examining the impact of resveratrol supplementation on grey mouse lemurs is an intriguing example of how p-value interpretation is used to gauge the efficacy of a treatment. Resveratrol, a compound found in grapes and red wine, has been associated with weight loss in rodents, prompting scientists to investigate if similar benefits occur in primates.

Key aspects of the lemurs' health and behavior—like resting metabolic rate, body mass gain, food intake, and locomotor activity—were measured before and after administering resveratrol. P-values from these measurements were then used to decide if the changes were likely due to supplementation or just random fluctuations. For the lemurs, certain indicators (e.g., body mass gain, resting metabolic rate) showed statistical significance with low p-values, suggestive of a true effect from resveratrol. Other indicators did not show such a pattern, indicating no clear evidence of the supplement's effect.

The study, while limited by its small sample size of six lemurs, offers preliminary insights. Low p-values in some areas provide evidence that supports further investigation into resveratrol's potential benefits, reinforcing the need for larger studies to confirm these findings and generalize them to a broader population.

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Most popular questions from this chapter

Influencing Voters Exercise 4.38 on page 235 describes a possible study to see if there is evidence that a recorded phone call is more effective than a mailed flyer in getting voters to support a certain candidate. The study assumes a significance level of \(\alpha=0.05\) (a) What is the conclusion in the context of this study if the p-value for the test is \(0.027 ?\) (b) In the conclusion in part (a), which type of error are we possibly making: Type I or Type II? Describe what that type of error means in this situation. (c) What is the conclusion if the p-value for the test is \(0.18 ?\) (d) In the conclusion in part (c), which type of error are we possibly making: Type I or Type II? Describe what that type of error means in this situation.

Election Poll In October before the 2008 US presidential election, \(A B C\) News and the Washington Post jointly conducted a poll of "a random national sample" and asked people who they intended to vote for in the 2008 presidential election. \(^{37}\) Of the 1057 sampled people who answered either Barack Obama or John McCain, \(55.2 \%\) indicated that they would vote for Obama while \(44.8 \%\) indicated that they would vote for MeCain. While we now know the outcome of the election, at the time of the poll many people were very curious as to whether this significantly predicts a winner for the election. (While a candidate needs a majority of the electoral college vote to win an election, we'll simplify things and simply test whether the percentage of the popular vote for Obama is greater than \(50 \% .\) ) (a) State the null and alternative hypotheses for testing whether more people would vote for Obama than MeCain. (Hint: This is a test for a single proportion since there is a single variable with two possible outcomes.) (b) Describe in detail how you could create a randomization distribution to test this (if you had many more hours to do this homework and no access to technology).

Do iPads Help Kindergartners Learn: A Subtest The Auburn, Maine, school district conducted an early literacy experiment in the fall of \(2011 .\) In September, half of the kindergarten classes were randomly assigned iPads (the intervention group) while the other half of the classes got them in December (the control group.) Kids were tested in September and December and the study measures the average difference in score gains between the control and intervention group. \(^{46}\) The experimenters tested whether the mean score for the intervention group was higher on the HRSIW subtest (Hearing and Recording Sounds in Words) than the mean score for the control group. (a) State the null and alternative hypotheses of the test and define any relevant parameters. (b) The p-value for the test is 0.02 . State the conclusion of the test in context. Are the results statistically significant at the \(5 \%\) level? (c) The effect size was about two points, which means the mean score for the intervention group was approximately two points higher than the mean score for the control group on this subtest. A school board member argues, "While these results might be statistically significant, they may not be practically significant." What does she mean by this in this context?

Exercises 4.117 to 4.122 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: \(\hat{p}=30 / 50=0.60\) with \(n=50\)

State the null and alternative hvpotheses for the statistical test described. Testing to see if there is evidence that the correlation between two variables is negative

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