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Chapter 4: Problem 53

Multiple Sclerosis and Sunlight It is believed that sunlight offers some protection against multiple sclerosis (MS) since the disease is rare near the equator and more prevalent at high latitudes. What is it about sunlight that offers this protection? To find out, researchers \({ }^{15}\) injected mice with proteins that induce a condition in mice comparable to MS in humans. The control mice got only the injection, while a second group of mice were exposed to UV light before and after the injection, and a third group of mice received vitamin D supplements before and after the injection. In the test comparing UV light to the control group, evidence was found that the mice exposed to UV suppressed the MS-like disease significantly better than the control mice. In the test comparing mice getting vitamin D supplements to the control group, the mice given the vitamin \(\mathrm{D}\) did not fare significantly better than the control group. If the p-values for the two tests are 0.472 and 0.002 , which p-value goes with which test?

Short Answer

Expert verified
The p-values correspond to the following tests: The p-value 0.002 is for the test comparing UV light to the control group and the p-value 0.472 is for the test comparing vitamin D supplements to the control group.

Step by step solution

01

Interpret p-values

The p-value is a statistical concept that measures the strength of evidence in support of a null hypothesis or basically to deny the experiment hypothesis. A p-value less than 0.05 means there is strong evidence against the null hypothesis or that you reject the null hypothesis. A p-value higher than 0.05 means there is weak evidence against the null hypothesis or you fail to reject the null hypothesis.
02

Match the p-values to the tests

From the experiment it is known that the mice exposed to UV light had a significantly better result in contrast to the control group. Also, it is known that the mice given the vitamin D didn't fare significantly better than the control group. Hence, based on the p-value interpretations, the lower p-value (0.002) which equates to a significant result should be for the UV light group and the higher p-value (0.472) which equates to an insignificant result should be for the Vitamin D group.

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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 cornerstone concept in statistical hypothesis testing. It's a default position that suggests there is no effect or no difference in the context of scientific research. For instance, if researchers are studying the impact of sunlight on MS, the null hypothesis would be that sunlight does not affect the progression or suppression of MS.
In the experiment with mice, the null hypothesis might be stated as 'UV light or vitamin D supplements do not suppress MS-like conditions in mice.' Researchers then use statistical tests to challenge this hypothesis. The goal is to gather evidence from data to either support or reject the null hypothesis. When results significantly deviate from what the null hypothesis predicts, researchers may conclude that their findings provide evidence of an effect, subsequently leading to the rejection of the null hypothesis.
Understanding the null hypothesis is crucial for interpreting p-values. It serves as the basis for determining whether the experimental results are due to chance or if they reflect the actual effect of the treatment or condition under investigation.
Statistical Significance
Statistical significance is a term used to indicate that the result of a statistical test is unlikely to have occurred by random chance. This is determined by calculating a p-value, which is the probability of observing the results (or more extreme results) if the null hypothesis were true.
A commonly accepted threshold for declaring statistical significance is a p-value of 0.05 or less. This means there's less than a 5% probability that the results are due to random chance. In the context of the MS and sunlight study, the p-value of 0.002 suggests a statistically significant difference between the control group and the mice exposed to UV light. This would lead researchers to reject the null hypothesis and conclude that UV light has a protective effect against the development of an MS-like condition.

Importance of the 0.05 Threshold

The 0.05 threshold isn't a magical number but a convention that has been accepted in the scientific community. It's important to understand that 'statistical significance' doesn't necessarily equate to 'practical significance.' Even if a result is statistically significant, it's essential to consider whether the effect size is meaningful in a practical sense.
Research Methodology
Research methodology encompasses the techniques, strategies, and methods employed by researchers to systematically solve a research problem. It involves the whole process鈥攆rom defining research questions and hypotheses to data collection and analysis.
In our example with the sunlight and MS study, the research methodology included experimental design involving control groups and test groups (mice exposed to UV light and mice given vitamin D supplements), which is a common design in biomedical research. The dependability of the research findings relies heavily on the choice of methodology, including how the experiment is controlled, how samples are selected, and how data is analyzed.

Experimental Controls and Validity

Good research methodology is also key in maintaining the validity of the experiment. The use of control mice ensures that results can be attributed to the treatments rather than other confounding variables. For instance, the improved condition of the UV light group as compared to the control group strengthens the researchers' conclusions that UV light may have a protective effect against MS-like conditions.

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Most popular questions from this chapter

Exercise and the Brain It is well established that exercise is beneficial for our bodies. Recent studies appear to indicate that exercise can also do wonders for our brains, or, at least, the brains of mice. In a randomized experiment, one group of mice was given access to a running wheel while a second group of mice was kept sedentary. According to an article describing the study, "The brains of mice and rats that were allowed to run on wheels pulsed with vigorous, newly born neurons, and those animals then breezed through mazes and other tests of rodent IQ"10 compared to the sedentary mice. Studies are examining the reasons for these beneficial effects of exercise on rodent (and perhaps human) intelligence. High levels of BMP (bone- morphogenetic protein) in the brain seem to make stem cells less active, which makes the brain slower and less nimble. Exercise seems to reduce the level of BMP in the brain. Additionally, exercise increases a brain protein called noggin, which improves the brain's ability. Indeed, large doses of noggin turned mice into "little mouse geniuses," according to Dr. Kessler, one of the lead authors of the study. While research is ongoing in determining which effects are significant, all evidence points to the fact that exercise is good for the brain. Several tests involving these studies are described. In each case, define the relevant parameters and state the null and alternative hypotheses. (a) Testing to see if there is evidence that mice allowed to exercise have lower levels of BMP in the brain on average than sedentary mice (b) Testing to see if there is evidence that mice allowed to exercise have higher levels of noggin in the brain on average than sedentary mice (c) Testing to see if there is evidence of a negative correlation between the level of BMP and the level of noggin in the brains of mice

ADHD and Pesticides In Exercise 4.16 on page \(232,\) we describe an observational study investigating a possible relationship between exposure to organophosphate pesticides as measured in urinary metabolites (DAP) and diagnosis of ADHD (attention-deficit/hyperactivity disorder). In reporting the results of this study, the authors \({ }^{25}\) make the following statements: "The threshold for statistical significance was set at \(P<.05 . "\) "The odds of meeting the ... criteria for ADHD increased with the urinary concentrations of total DAP metabolites." \- "The association was statistically significant." (a) What can we conclude about the p-value obtained in analyzing the data? (b) Based on these statements, can we distinguish whether the evidence of association is very strong vs moderately strong? Why or why not? (c) Can we conclude that exposure to pesticides is related to the likelihood of an ADHD diagnosis? (d) Can we conclude that exposure to pesticides causes more cases of ADHD? Why or why not?

Exercise Hours Introductory statistics students fill out a survey on the first day of class. One of the questions asked is "How many hours of exercise do you typically get each week?" Responses for a sample of 50 students are introduced in Example 3.25 on page 207 and stored in the file ExerciseHours. The summary statistics are shown in the computer output. The mean hours of exercise for the combined sample of 50 students is 10.6 hours per week and the standard deviation is 8.04 . We are interested in whether these sample data provide evidence that the mean number of hours of exercise per week is different between male and female statistics students. Variable Gender N Mean StDev Minimum Maximum \(\begin{array}{lllllll}\text { Exercise } & \text { F } 30 & 9.40 & 7.41 & 0.00 & 34.00\end{array}\) \(\begin{array}{llll}20 & 12.40 & 8.80 & 2,00\end{array}\) Discuss whether or not the methods described below would be appropriate ways to generate randomization samples that are consistent with \(H_{0}: \mu_{F}=\mu_{M}\) vs \(H_{a}: \mu_{F} \neq \mu_{M} .\) Explain your reasoning in each case. (a) Randomly label 30 of the actual exercise values with " \(\mathrm{F}^{\prime \prime}\) for the female group and the remaining 20 exercise values with " \(\mathrm{M} "\) for the males. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) (b) Add 1.2 to every female exercise value to give a new mean of 10.6 and subtract 1.8 from each male exercise value to move their mean to 10.6 (and match the females). Sample 30 values (with replacement) from the shifted female values and 20 values (with replacement) from the shifted male values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\) - (c) Combine all 50 sample values into one set of data having a mean amount of 10.6 hours. Select 30 values (with replacement) to represent a sample of female exercise hours and 20 values (also with replacement) for a sample of male exercise values. Compute the difference in the sample means, \(\bar{x}_{F}-\bar{x}_{M}\)

Radiation from Cell Phones and Brain Activity Does heavy cell phone use affect brain activity? There is some concern about possible negative effects of radiofrequency signals delivered to the brain. In a randomized matched-pairs study, \(^{24}\) 47 healthy participants had cell phones placed on the left and right ears. Brain glucose metabolism (a measure of brain activity) was measured for all participants under two conditions: with one cell phone turned on for 50 minutes (the "on" condition) and with both cell phones off (the "off" condition). The amplitude of radiofrequency waves emitted by the cell phones during the "on" condition was also measured. (a) Is this an experiment or an observational study? Explain what it means to say that this was a "matched-pairs" study. (b) How was randomization likely used in the study? Why did participants have cell phones on their ears during the "off" condition? (c) The investigators were interested in seeing whether average brain glucose metabolism was different based on whether the cell phones were turned on or off. State the null and alternative hypotheses for this test. (d) The p-value for the test in part (c) is 0.004 . State the conclusion of this test in context. (e) The investigators were also interested in seeing if brain glucose metabolism was significantly correlated with the amplitude of the radiofrequency waves. What graph might we use to visualize this relationship? (f) State the null and alternative hypotheses for the test in part (e). (g) The article states that the p-value for the test in part (e) satisfies \(p<0.001\). State the conclusion of this test in context.

In Exercises 4.150 to \(4.152,\) a confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A \(95 \%\) confidence interval for \(p: 0.48\) to 0.57 (a) \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) (b) \(H_{0}: p=0.75\) vs \(H_{a}: p \neq 0.75\) (c) \(H_{0}: p=0.4\) vs \(H_{a}: p \neq 0.4\)
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