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Chapter 13: Problem 79

The paper "Supervised Exercise Versus Non-Supervised Exercise for Reducing Weight in Obese Adults" (The Journal of Sports Medicine and Physical Fitness [2009]: \(85-90\) ) describes an experiment in which participants were randomly assigned either to a supervised exercise program or a control group. Those in the control group were told that they should take measures to lose weight. Those in the supervised exercise group were told they should take measures to lose weight, but they also participated in regular supervised exercise sessions. Weight loss at the end of four months was recorded. Data consistent with summary quantities given in the paper are shown in the accompanying table.Use the given information to construct and interpret a \(95 \%\) confidence interval for the difference in mean weight loss for the two treatments.

Short Answer

Expert verified
The 95% confidence interval for the difference in mean weight loss between the supervised exercise group and the control group can be calculated by first finding the sample means and standard deviations for both groups. With degrees of freedom calculated as \(df = n_1 + n_2 - 2\), find the t-score corresponding to a 95% confidence level. The standard error of the difference in means can be calculated using \(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\). The margin of error is found by multiplying the t-score by the standard error and the confidence interval is \((\bar{x}_1 - \bar{x}_2) \pm\) Margin of Error. If the interval is completely above zero, the supervised exercise program is significantly more effective. If the interval contains zero, there is no significant difference between the two treatments.

Step by step solution

01

Identify the given information

In the problem statement, we are given the following information: - The objective is to find the 95% confidence interval of the difference in mean weight loss between supervised exercise and control group. - The sample data of weight loss at the end of four months is provided.
02

Calculate the sample means and sample standard deviations

Calculate the sample means (\(\bar{x}_1\) and \(\bar{x}_2\)) and sample standard deviations (\(s_1\) and \(s_2\)) for both groups using the provided data.
03

Determine the degrees of freedom and t-score for a 95% confidence interval

The degrees of freedom for this t-distribution can be calculated using the formula: \(df = n_1 + n_2 - 2\) Where \(n_1\) and \(n_2\) are the sample sizes of both groups. Using a table of t-distribution critical values or a calculator, find the corresponding t-score for a 95% confidence interval with the calculated degrees of freedom.
04

Calculate the standard error of the difference in means

To calculate the standard error of the difference in means, use the following formula: \(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\) Where \(s_1\), \(s_2\), \(n_1\), and \(n_2\) are the sample standard deviations and sample sizes of both groups respectively.
05

Calculate the margin of error and confidence interval

To calculate the margin of error, multiply the standard error by the corresponding t-score obtained in Step 3. Margin of Error = \(t \times SE\) Now, calculate the confidence interval for the difference in mean weight loss for the two treatments using the following formula: Confidence Interval = \((\bar{x}_1 - \bar{x}_2) \pm\) Margin of Error
06

Interpret the confidence interval

If the confidence interval obtained in the previous step is completely above zero, it suggests a significant difference in mean weight loss, with the supervised exercise program being more effective. If the interval contains zero, it indicates that there is no significant difference between the two treatments.

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Key Concepts

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

t-distribution
The t-distribution is a way to understand variability and confidence intervals, especially when working with small sample sizes. Unlike the normal distribution, the t-distribution adjusts for sample size. It has fatter tails, which means more variability or uncertainty, particularly when sample sizes are small. This distribution becomes important when we don't know the population standard deviation and instead use the sample standard deviation, as is often the case in real-world situations.
  • Useful for small sample sizes and unknown standard deviations.
  • Becomes closer to the normal distribution as the sample size increases.
In our context, using the t-distribution helps to accurately estimate the confidence interval for mean weight loss between the two groups in the exercise study.
standard error
Standard error is a statistical term that measures the accuracy with which a sample represents a population. More specifically, it's the estimated standard deviation of the sample mean. For calculating the confidence interval, the standard error helps quantify the amount of variation or "error" in the sample's mean.
  • Calculated using sample standard deviations and sample sizes.
  • Reflects how much the sample mean will differ from the population mean.
In the exercise study, the standard error of the difference in means is calculated to understand the variability between the two groups' mean weight losses, providing a clearer picture of how significant the weight changes are.
mean weight loss
Mean weight loss is the average amount of weight lost by participants in the study. To find the mean weight loss for each group, you add up all the individual weight loss figures and divide by the number of participants in each group. This value provides an insight into how effective each treatment is.
  • Mean = Total weight loss / Number of participants.
  • Gives a central value of the weight loss data distribution.
In the case of this study, comparing the mean weight loss of the supervised exercise to the control group helps in understanding the effectiveness of the supervised program.
degrees of freedom
Degrees of freedom (df) is a concept that's vital when using the t-distribution to analyze two samples. It is determined by the number of independent values in the calculation minus the number of estimated parameters. For comparing two samples, the formula applied is:\[ df = n_1 + n_2 - 2 \]where \( n_1 \) and \( n_2 \) are the sizes of the two groups being compared.
  • Degrees of freedom impact the shape of the t-distribution.
  • Important for finding the correct t-score from the t-table.
In the weight-loss study, knowing the degrees of freedom allows us to locate the correct t-score to use when calculating the 95% confidence interval, ensuring that the interval is accurate and valid for inference about the mean weight loss difference.

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Most popular questions from this chapter

The paper "Sodium content of Lunchtime Fast Food Purchases at Major U.S. Chains" (Archives of Internal Medicine [2010]: \(732-734\) ) reported that for a random sample of 850 meal purchases made at Burger King, the mean sodium content was \(1685 \mathrm{mg},\) and the standard deviation was \(828 \mathrm{mg}\). For a random sample of 2107 meal purchases made at Mel)onald's, the mean sodium content was \(1477 \mathrm{mg},\) and the standard deviation was \(812 \mathrm{mg}\). Based on these data, is it reasonable to conclude that there is a difference in mean sodium content for meal purchases at Burger King and meal purchases at McDonald's? Use \(\alpha=0.05\).

The article "Dieters Should Use a Bigger Fork" (Food Network Magazine, January/February 2012) described an experiment conducted by researchers at the University of Utah. The article reported that when people were randomly assigned to cither cat with a small fork or to eat with a large fork, the mean amount of food consumed was significantly less for the group that ate with the large fork. a. What are the two treatments in this experiment? b. In the context of this experiment, explain what it means to say that the mean amount of food consumed was significantly less for the group that ate with the large fork.

The humorous puper "Will Humans Swim Faster or Slower in Syrup?" (American Institute of Chemical Engineers Journal [2004]: \(2646-2647\) ) investigated the fluid mechanics of swimming. Twenty swimmers each swam a specified distance in a water-filled pool and in a pool in which the water was thickened with food grade guar gum to create a syruplike consistency. Velocity, in meters per second, was recorded. Values estimated from a graph in the paper are given. The authors of the paper concluded that swimming in guar syrup does not change mean swimming speed. Are the given data consistent with this conclusion? Carry out a hypothesis test using a 0.01 significance level.

Do children diagnosed with attention deficit/ hyperactivity disorder (ADHD) have smaller brains than children without this condition? This question was the topic of a research study described in the paper "Developmental Trajectories of Brain Volume Abnormalities in Children and Adolescents with Attention Deficit/Hyperactivity Disorder" (Journal of the American Medical Association [2002]: \(1740-1747\) ). Brain scans were completed for a representative sample of 152 children with ADHI) and a representative sample of 139 children without \(A D H D\). Summary values for total cerebral volume (in cubic milliliters) are given in the following table: Is there convincing evidence that the mean brain volume for children with ADHD is smuller than the mean for children without ADHD? Test the relevant hypotheses using a 0.05 level of significance.

Use the information in the previous exercise to construct a \(95 \%\) bootstrap confidence interval to estimate the difference in mean Personal Meaning scores for patients with cancer in the high-dose and low-dose psilocybin groups. Interpret the interval in context. You can use make use of the Shiny apps in the collection at statistics.cengage.com/Peck2e/Apps.html.
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