/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A writer in a medical journal sa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A writer in a medical journal says: "An uncontrolled experiment in 37 women found a significantly improved mean clinical symptom score after treatment. Methodologic flaws make it difficult to interpret the results of this study." The writer is skeptical about the significant improvement because (a) there is no control group, so the improvement might be due to the placebo effect or to the fact that many medical conditions improve over time. (b) the \(P\)-value given was \(P=0.048\), which is too large to be convincing. (c) the response variable might not have an exactly Normal distribution in the population.

Short Answer

Expert verified
(a) Lack of a control group affects the study's validity.

Step by step solution

01

Identify the Key Points

We are given a statement about an experiment with 37 women that showed significant improvement after treatment. The problem highlights methodologic flaws, and we must evaluate potential reasons for skepticism.
02

Understand the Role of a Control Group

A control group is crucial in experiments to ensure that the observed effect is due to the treatment and not other factors like the placebo effect or natural improvement over time. Option (a) questions the lack of a control group, suggesting this could be a reason for skepticism.
03

Evaluate Given P-Value

The given P-value is 0.048, indicating statistical significance at the 5% level. Option (b) claims this P-value is too large to be convincing, but at 5%, it is traditionally considered significant. Hence, a P-value of 0.048 is not typically a skeptical point in itself.
04

Check Assumptions of Statistical Tests

Normality of distribution is an assumption in many parametric tests. Option (c) raises the possibility that the response variable might not follow a Normal distribution, which could affect the validity of the results. However, moderate deviations from normality often have limited impact, especially in large samples.
05

Conclude the Reason for Skepticism

Upon analyzing each option, the lack of a control group is a significant methodological flaw that could lead to skepticism about the study's findings. This aligns with option (a) as the main reason for questioning the results.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Control Group Significance
In scientific research, a control group serves as a critical component in validating experimental results. Without a control group, it's hard to determine whether the treatment itself causes any observed effects.
Control groups help distinguish between genuine treatment effects and other influences, such as the placebo effect or natural improvements over time. This is why researchers often stress the importance of including a control group in any experiment.
When the experiment on 37 women reported improved mean clinical symptom scores, the lack of a control group raised red flags. Improvements could potentially result from reasons unrelated to the treatment, like spontaneous recovery or psychological factors.
To ensure robust findings, a well-designed experiment generally includes a control group. This comparison group doesn't receive the experimental treatment, serving as a baseline. Thus, researchers can more reliably attribute observed changes to the treatment itself, rather than other variables. Such a methodological framework provides more confidence in the experimental results.
P-value Interpretation
The P-value represents the probability that the observed results would occur by chance if there were no actual effect. A common threshold for statistical significance is 0.05, meaning if the P-value is below this level, the results are generally considered statistically significant.
In the case of the experiment with 37 women, a P-value of 0.048 suggests statistical significance, just under the conventional threshold. However, it's critical not to overly rely on the P-value as a definitive measure of the effect.
Cautious interpretation is needed here: while a P-value of 0.048 indicates that there's only a 4.8% chance the results happened by random chance, it doesn't confirm the result is practically significant. The smaller the P-value, the stronger the evidence against the null hypothesis, but this does not guarantee that the effect is meaningful or solely due to the treatment.
Critics argue that P-values can be misinterpreted, and relying solely on them may overlook other important aspects of study design or data analysis. It's a piece of the puzzle and should be considered alongside other study factors and statistical indicators.
Normal Distribution Assumption
Statistical tests often assume that data have a normal distribution, particularly parametric tests like t-tests or ANOVAs. This assumption simplifies calculations and ensures the validity of the test results.
In experiments like the one with 37 women, questioning normality is important. A response variable not following a normal distribution might skew results, leading to potentially incorrect conclusions. However, small deviations from normality might have limited impact,
especially in larger sample sizes where the Central Limit Theorem suggests that the sampling distribution of the mean approaches normality, given a sufficiently large sample.
When the population data do not meet the normal distribution assumption, non-parametric tests may offer a better alternative. These tests don't require normality, providing a more flexible approach to analyzing non-normally distributed data.
Therefore, while the assumption of normal distribution is a foundational aspect of many tests, researchers should assess their data and choose appropriate statistical methods accordingly to ensure accurate and reliable findings.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Finding power by hand. Even though software is used in practice to calculate power, doing the work by hand builds your understanding. Return to the test in Example 18.6. There are \(n=10\) observations from a population with standard deviation \(\sigma=1\) and unknown mean \(\mu\). We will test $$ \begin{aligned} &H_{0}: \mu=0 \\ &H_{a}: \mu>0 \end{aligned} $$ with fixed significance level \(\alpha=0.05\). Find the power against the alternative \(\mu=0.8\) by following these steps. (a) The \(z\) test statistic is $$ \mathrm{z}=\mathrm{x}^{-}-\mu 0 \sigma / \mathrm{n}=\mathrm{x}^{-}-01 / 10=3.162 \mathrm{x}^{-}=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{n}}=\frac{\bar{x}-0}{1 / \sqrt{10}}=3.162 \bar{x} $$ (Remember that you won't know the numerical value of \(x^{-} \bar{x}\) until you have data.) What values of \(z\) lead to rejecting \(H_{0}\) at the \(5 \%\) significance level? (b) Starting from your result in part (a), what values of \(x^{-} \bar{x}\) lead to rejecting \(H_{0}\) ? The area above these values is shaded under the top curve in Figure 18.1. (c) The power is the probability that you observe any of these values of \(\mathrm{x}^{-} \bar{x}\) when \(\mu=0.8\). This is the shaded area under the bottom curve in Figure 18.1. What is this probability?

A college administrator questions the first 50 students he meets on campus the day after final exams are over. He asks them whether they had positive, neutral, or negative overall feelings about the term that had just ended. Suggest some reasons it may be risky to act as if the first 50 students at this particular time are an SRS of all students at this college.

Software can generate samples from (almost) exactly Normal distributions. Here is a random sample of size 5 from the Normal distribution with mean 12 and standard deviation 2.5: $$ \begin{array}{lllll} 14.94 & 9.04 & 9.58 & 12.96 & 11.29 \end{array} $$ These data match the conditions for a \(z\) test better than real data will: the population is very close to Normal and has known standard deviation \(\sigma=2.5\), and the population mean is \(\mu=12\). Although we know the true value of \(\mu\), suppose we pretend that we do not and we test the hypotheses $$ \begin{aligned} &H_{0}: \mu=10 \\ &H_{a}: \mu \neq 10 \end{aligned} $$ (a) What are the \(z\) statistic and its \(P\)-value? Is the test significant at the \(5 \%\) level? (b) We know that the null hypothesis does not hold, but the test failed to give strong evidence against \(H_{0}\). Explain why this is not surprising.

Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample (or the number of subjects in an experiment) on each of the following: (a) The \(P\)-value of a test, when \(H_{0}\) is false and all facts about the population remain unchanged as \(n\) increases. (b) (Optional) The power of a fixed level \(\alpha\) test, when \(\alpha\), the alternative hypothesis, and all facts about the population remain unchanged.

Mortality rates vary from city to city in the United States. We have lots of data on many U.S. cities. Statistical software makes it easy to perform dozens of significance tests on dozens of variables to see which ones best predict mortality rate. One interesting finding is that those cities with major league ballparks tend to have significantly higher mortality rates than other cities. To improve your chances of a long life, should you use this "significant" variable to decide where to live? Explain your answer.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.