/*! 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 34 Weight change for controls A dis... [FREE SOLUTION] | 91影视

91影视

Weight change for controls A disadvantage of the experimental design in Example 8 on weight change in anorexic girls is that girls could change weight merely from participating in a study. In fact, girls were randomly assigned to receive a therapy or to serve in a control group, so it was possible to compare weight change for the therapy group to the control group. For the 26 girls in the control group, the weight change had \(\bar{x}=-0.5\) and \(s=8.0 .\) Repeat all five steps of the test of \(\mathrm{H}_{0}: \mu=0\) against \(\mathrm{H}_{a}: \mu \neq 0\) for this group, and interpret the P-value.

Short Answer

Expert verified
The test does not show a significant weight change in the control group.

Step by step solution

01

State the Hypotheses

We need to test the null hypothesis (H鈧) that there is no weight change, meaning the population mean is zero, \( H_0: \mu = 0 \). The alternative hypothesis (H鈧) is that the mean weight change is not zero, \( H_a: \mu eq 0 \).
02

Choose the Significance Level

Choose a significance level alpha (伪). Common choices are 0.05 or 0.01. We will use \( \alpha = 0.05 \) for this test.
03

Calculate the Test Statistic

The test statistic for a t-test is calculated using the formula:\[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]where \( \bar{x} = -0.5 \) (sample mean), \( \mu_0 = 0 \) (population mean under null hypothesis), \( s = 8.0 \) (sample standard deviation), and \( n = 26 \) (sample size). Substitute the values into the formula to calculate t:\[t = \frac{-0.5 - 0}{\frac{8.0}{\sqrt{26}}} \approx -0.340\]
04

Determine the Critical Value or Use P-value

Determine the critical value using a t-distribution table or software for \( n-1 = 25 \) degrees of freedom at \( \alpha = 0.05 \). Because this is a two-tailed test, divide \( \alpha \) by 2 for each tail. Otherwise, calculate the p-value using a t-distribution calculator. The critical t-value is approximately 卤2.060.
05

Make the Decision

Since the calculated t-statistic \( t \approx -0.340 \) is less extreme than the critical t-values (卤2.060) or if the p-value is greater than 0.05, we do not reject the null hypothesis.
06

Interpret the Result

The p-value indicates that there is not enough evidence to conclude that there is a significant weight change in the control group. Hence, we fail to reject \( H_0 \), suggesting no significant weight change for the control group.

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 91影视!

Key Concepts

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

Experimental Design
Experimental design is a critical part of conducting scientific research. It outlines how participants are allocated to various groups or conditions in a study. In our exercise, we see an example of a classic experimental design where participants, in this case, anorexic girls, were randomly assigned to either a therapy group or a control group.

This design helps control for variables that might otherwise skew the results, such as participants' psychological responses to simply being part of the study. By comparing changes in weight between the therapy and control groups, researchers can more accurately assess the effect of therapy itself, separate from other factors.
  • Random assignment balances out unknown factors across both groups.
  • Control groups serve to provide a baseline or standard for comparison.
  • This design enhances the validity of the test results, leading to more reliable and unbiased conclusions.
Understanding the role of experimental design is crucial in hypothesis testing, as it sets the stage for effective and meaningful data analysis.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold set by researchers to determine whether to reject a null hypothesis. It's a fundamental concept in hypothesis testing, serving as a measure of how extreme the data must be before we decide that our null hypothesis is unlikely enough to be rejected. In our case, \( \alpha \) is set to 0.05.

This means that there's a 5% risk of concluding a significant effect (weight change in this context) when there is none. It's a balance between making false positives and negatives.
  • A lower \( \alpha \) suggests stricter criteria for statistical significance.
  • Common alpha levels are 0.05 and 0.01, representing a 5% and 1% probability of error, respectively.
  • Choosing \( \alpha \) reflects researchers' tolerance for risk in their findings.
Setting the right significance level ensures that the results of the hypothesis test are both meaningful and appropriate for the context of the study.
T-Test
The T-Test is a statistical method used to determine whether there is a significant difference between the means of two groups. This approach is applied when the data is assumed to follow a normal distribution and is especially useful when dealing with small sample sizes. Our exercise involved a one-sample T-Test.

The T-Test formula used is:
\[t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\]Here, \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(s\) is the standard deviation, and \(n\) is the sample size.
  • The calculated t-value helps us decide if the observed data fits the null hypothesis.
  • A T-Test can be one-sample, two-sample, or paired, depending on the study design.
  • This test requires knowing or assuming the normality of the data for the results to be valid.
Using a T-Test properly ensures robust conclusions about the differences in means, essential for drawing scientifically sound conclusions.
P-Value
The P-Value is the probability of observing test results at least as extreme as the results actually obtained, under the assumption that the null hypothesis is correct. It's a measure that helps scientists determine the strength of their evidence.

In our problem, understanding the P-Value helps us interpret if the observed weight change is due to chance. A small P-Value (typically 鈮 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis.
  • A high P-Value suggests that the observed effect was likely due to random chance.
  • P-Values help in making informed decisions in hypothesis testing.
  • They should not be the lone metric in decision-making but rather part of a broader statistical analysis.
Proper interpretation of the P-Value allows researchers to draw meaningful conclusions about their hypotheses, strengthening the validity of their study outcomes.

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

Test and CI Results of \(99 \%\) confidence intervals are consistent with results of two-sided tests with which significance level? Explain the connection.

Men at work When the 636 male workers in the 2008 GSS were asked how many hours they worked in the previous week, the mean was 45.5 with a standard deviation of 15.16 . Does this suggest that the population mean work week for men exceeds 40 hours? Answer by: a. Identifying the relevant variable and parameter. b. Stating null and alternative hypotheses. c. Reporting and interpreting the P-value for the test statistic value of \(t=9.15\). d. Explaining how to make a decision for the significance level of 0.01

Young workers When the 127 workers aged \(18-25\) in the 2008 GSS were asked how many hours they worked in the previous week, the mean was 37.47 with a standard deviation of \(13.63 .\) Does this suggest that the population mean work week for this age group differs from 40 hours? Answer by: a. Identifying the relevant variable and parameter. b. Stating null and alternative hypotheses. c. Finding and interpreting the test statistic value. d. Reporting and interpreting the P-value and stating the conclusion in context.

Get P-value from \(z\) For a test of \(\mathrm{H}_{0}: p=0.50,\) the \(z\) test statistic equals 1.04 a. Find the P-value for \(\mathrm{H}_{a} \cdot p>0.50\). b. Find the P-value for \(\mathrm{H}_{a}: p \neq 0.50\). c. Find the P-value for \(\mathrm{H}_{c}: p<0.50 .\) (Hint: The P-values for the two possible one-sided tests must sum to \(1 .)\) d. Do any of the P-values in part a, part b, or part c give strong evidence against \(\mathrm{H}_{0}\) ? Explain.

Misleading summaries? Two researchers conduct separate studies to test \(\mathrm{H}_{0}: p=0.50\) against \(\mathrm{H}_{a}: p \neq 0.50\) each with \(n=400\). a. Researcher A gets 220 observations in the category of interest, and \(\hat{p}=220 / 400=0.550\) and test statistic \(z=2.00 .\) Show that the P-value \(=0.046\) for Researcher A's analysis. b. Researcher B gets 219 in the category of interest, and \(\hat{p}=219 / 400=0.5475\) and test statistic \(z=1.90 .\) Show that the P-value \(=0.057\) for Researcher B's analysis. c. Using \(\alpha=0.05,\) indicate in each case from part a and part b whether the result is "statistically significant." Interpret. d. From part a, part \(b\), and part \(c,\) explain why important information is lost by reporting the result of a test as "P-value \(\leq 0.05\) " versus "P-value \(>0.05\)," or as "reject \(\mathrm{H}_{0}\) " versus "do not reject \(\mathrm{H}_{0}\)," instead of reporting the actual P-value. e. Show that the \(95 \%\) confidence interval for \(p\) is (0.501,0.599) for Researcher \(\mathrm{A}\) and (0.499,0.596) for Researcher \(\mathrm{B}\). Explain how this method shows that, in practical terms, the two studies had very similar results.

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.