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Exercise and the Brain Exercise 4.19 on page 232 describes a study investigating the effects of exercise on cognitive function. \(^{28}\) Separate groups of mice were exposed to running wheels for \(0,2,4,7,\) or 10 days. Cognitive function was measured by \(Y\) maze performance. The study was testing whether exercise improves brain function, whether exercise reduces levels of BMP (a protein which makes the brain slower and less nimble), and whether exercise increases the levels of noggin (which improves the brain's ability). For each of the results quoted in parts (a), (b), and (c), interpret the information about the p-value in terms of evidence for the effect. (a) "Exercise improved Y-maze performance in most mice by the 7 th day of exposure, with further increases after 10 days for all mice tested \((p<.01)^{*}\) (b) "After only two days of running, BMP ... was reduced ... and it remained decreased for all subsequent time-points \((p<.01) . "\) (c) "Levels of noggin ... did not change until 4 days, but had increased 1.5 -fold by \(7-10\) days of exercise \((p<.001)\)." (d) Which of the tests appears to show the strongest statistical effect? (e) What (if anything) can we conclude about the effects of exercise on mice?

Short Answer

Expert verified
The strongest statistical effect was shown in the noggin test (p<0.001), pointing towards the conclusion that regular exercise can increase noggin levels significantly. In general terms, this study provides evidence that exercise can improve cognitive function, decrease BMP levels and increase noggin levels in mice.

Step by step solution

01

Interpret Result (a)

Result (a) saw exercise improving Y-maze performance by the 7th day of exposure, with further increases after 10 days for all mice tested \(p<.01\). A p-value of less than 0.01 indicates that the probability of observing these results (or more extreme), given that the null hypothesis is true, is less than 1%. Thus, there is strong evidence against the null hypothesis that exercise has no effect on Y-maze performance, and in favor of the alternate hypothesis that exercise does improve Y-maze performance.
02

Interpret Result (b)

Result (b) saw levels of BMP reduced after only two days of running and it remained decreased for all subsequent time points \((p<.01)\). A p-value of less than 0.01 suggests that the probability of seeing these results, assuming the null hypothesis that exercise has no effect on BMP levels, is less than 1%. This provides strong evidence against the null hypothesis and in favor of the alternate hypothesis that exercise does decrease BMP levels.
03

Interpret Result (c)

Result (c) saw levels of noggin didn't change until 4 days, but had increased 1.5 fold by 7-10 days of exercise \((p<.001)\). The p-value less than 0.001 indicates that the probability of observing these results, given the null hypothesis is true (i.e., exercise has no effect on noggin levels), is less than 0.1%. Hence, there is overwhelming evidence against the null hypothesis and in favor of the alternate hypothesis that exercise significantly increases noggin levels.
04

Identify Strongest Statistical Effect

Comparing the p-values from the three tests, the noggin test shows the strongest statistical effect because it has the smallest p-value. This suggests that the observed difference is very unlikely to have occurred by chance, therefore, it provides strong evidence for the alternate hypothesis that exercise increases noggin levels.
05

Conclude Effects of Exercise

Based on these results, it can be concluded that exercising has beneficial effects on mice, particularly in enhancing cognitive function (improved Y-maze performance), reducing levels of BMP (a protein that hinders brain function), and increasing levels of noggin (a protein that boosts brain ability).

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Key Concepts

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

Exercise and Brain Function
Regular physical activity has been shown to have numerous effects on the brain and its functions. When it comes to cognitive skills, exercise can be a real game-changer. How so? Well, the brain's functioning is enhanced through exercise as it can improve memory, learning, and even mental agility.
Among the molecules influencing cognition is BMP, which exercise reduces. This protein's presence is generally associated with slower brain responses. By decreasing BMP levels, exercise accelerates brain activity, making it more efficient.
Additionally, exercise in mice increased noggin, a protein renowned for enhancing brain capabilities. By interacting with brain cells, noggin supports the formation and maturation of neurons. This increase was particularly noted after prolonged physical activity, as highlighted in the study's results after 7 to 10 days of exercise.
Overall, regular exercise can improve the speed of learning and problem-solving capabilities, thus boosting cognitive function.
Statistical Significance
Statistical significance is a critical concept in research and science. It helps determine whether the findings of a study are likely due to chance or if there is indeed an important effect present.
In this exercise, the researchers used different p-values to assess the significance of changes due to exercise in mice. A result is typically considered statistically significant if its p-value is below a certain threshold, often 0.05. This threshold represents a 5% probability that the results observed happened by random chance, assuming the null hypothesis is true.
  • If a result is statistically significant, it means it is probably not just a fluke.
  • The lower the p-value, the stronger the evidence against the null hypothesis, indicating a true effect exists.
In this study, BMP and Y-maze performance improvements showed a p-value of less than 0.01, while noggin levels had an even smaller p-value of less than 0.001. These low p-values suggest strong statistical significance, supporting the hypothesis that exercise affects these areas.
P-Value Interpretation
Understanding p-values is vital in interpreting scientific results. In this exercise, p-values help determine the strength and validity of the results obtained through testing. A p-value measures how likely it is to observe results at least as extreme as studied, assuming the null hypothesis is true.
Here's how to interpret p-values:
  • A p-value less than 0.05 usually indicates statistical significance, suggesting real effects beyond chance.
  • Lower p-values mean stronger evidence against the null hypothesis. Thus, a p-value less than 0.01, as in this study for BMP reduction and Y-maze performance, shows very strong evidence of a true effect.
  • An even smaller p-value, like the one indicating noggin increases ( <0.001), suggests overwhelming evidence of an effect.
Thus, interpreting these p-values, the results consistently demonstrate that exercise impacts brain function in mice significantly. Understanding and correctly interpreting p-values is crucial for drawing valid conclusions and guiding further scientific inquiry.

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Most popular questions from this chapter

Influencing Voters: Is a Phone Call More Effective? Suppose, as in Exercise \(4.37,\) that we wish to compare methods of influencing voters to support a particular candidate, but in this case we are specifically interested in testing whether a phone call is more effective than a flyer. Suppose also that our random sample consists of only 200 voters, with 100 chosen at random to get the flyer and the rest getting a phone call. (a) State the null and alternative hypotheses in this situation. (b) Display in a two-way table possible sample results that would offer clear evidence that the phone call is more effective. (c) Display in a two-way table possible sample results that offer no evidence at all that the phone call is more effective. (d) Display in a two-way table possible sample results for which the outcome is not clear: There is some evidence in the sample that the phone call is more effective but it is possibly only due to random chance and likely not strong enough to generalize to the population.

In Exercises 4.14 and \(4.15,\) determine whether the sets of hypotheses given are valid hypotheses. State whether each set of hypotheses is valid for a statistical test. If not valid, explain why not. (a) \(H_{0}: \mu=15 \quad\) vs \(\quad H_{a}: \mu \neq 15\) (b) \(H_{0}: p \neq 0.5 \quad\) vs \(\quad H_{a}: p=0.5\) (c) \(H_{0}: p_{1}p_{2}\) (d) \(H_{0}: \bar{x}_{1}=\bar{x}_{2} \quad\) vs \(\quad H_{a}: \bar{x}_{1} \neq \bar{x}_{2}\)

Car Window Skin Cancer? A new study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{43}\) The study reviewed the records of all 1050 skin cancer patients referred to the St. Louis University Cancer Center in 2004 . Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

Arsenic in Chicken Data 4.5 on page 228 discusses a test to determine if the mean level of arsenic in chicken meat is above 80 ppb. If a restaurant chain finds significant evidence that the mean arsenic level is above \(80,\) the chain will stop using that supplier of chicken meat. The hypotheses are $$ \begin{array}{ll} H_{0}: & \mu=80 \\ H_{a}: & \mu>80 \end{array} $$ where \(\mu\) represents the mean arsenic level in all chicken meat from that supplier. Samples from two different suppliers are analyzed, and the resulting p-values are given: Sample from Supplier A: p-value is 0.0003 Sample from Supplier B: p-value is 0.3500 (a) Interpret each p-value in terms of the probability of the results happening by random chance. (b) Which p-value shows stronger evidence for the alternative hypothesis? What does this mean in terms of arsenic and chickens? (c) Which supplier, \(\mathrm{A}\) or \(\mathrm{B}\), should the chain get chickens from in order to avoid too high a level of arsenic?

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 \(H_{a}: p \neq 0.5\) Sample data: \(\hat{p}=42 / 100=0.42\) with \(n=100\)

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