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The article "The Effects of a Low-Fat, Plant-Based Dietary Intervention on Body Weight, Metabolism, and Insulin Sensitivity in Postmenopausal Women" (Amer. J. of Med., \(2005: 991-997\) ) reported on the results of an experiment in which half of the individuals in a group of 64 postmenopausal overweight women were randomly assigned to a particular vegan diet, and the other half received a diet based on National Cholesterol Education Program guidelines. The sample mean decrease in body weight for those on the vegan diet was \(5.8\) \(\mathrm{kg}\), and the sample SD was \(3.2\), whereas for those on the control diet, the sample mean weight loss and standard deviation were \(3.8\) and \(2.8\), respectively. Does it appear the true average weight loss for the vegan diet exceeds that for the control diet by more than \(1 \mathrm{~kg}\) ? Carry out an appropriate test of hypotheses at significance level \(.05\) based on calculating a \(P\)-value.

Short Answer

Expert verified
The true average weight loss for the vegan diet does not exceed the control diet by more than 1 kg at a 0.05 significance level.

Step by step solution

01

Set Up Hypotheses

We need to test if the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg. This is a one-sided test.- Null Hypothesis \( H_0: \mu_1 - \mu_2 = 1 \)- Alternative Hypothesis \( H_a: \mu_1 - \mu_2 > 1 \)where \( \mu_1 \) is the true mean weight loss for the vegan diet and \( \mu_2 \) for the control diet.
02

Calculate the Test Statistic

To perform the hypothesis test, we use the formula for the test statistic of two independent samples:\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - D_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where:- \( \bar{x}_1 = 5.8 \text{ kg} \) and \( \bar{x}_2 = 3.8 \text{ kg} \) are the sample means,- \( s_1 = 3.2 \text{ kg} \) and \( s_2 = 2.8 \text{ kg} \) are the sample standard deviations,- \( n_1 = n_2 = 32 \) are the sample sizes,- \( D_0 = 1 \text{ kg} \) is the hypothesized difference.Plugging in the values:\[ t = \frac{(5.8 - 3.8) - 1}{\sqrt{\frac{3.2^2}{32} + \frac{2.8^2}{32}}} \]
03

Compute the Test Statistic Value

First, calculate the standard error:\[ SE = \sqrt{\frac{3.2^2}{32} + \frac{2.8^2}{32}} = \sqrt{\frac{10.24}{32} + \frac{7.84}{32}} = \sqrt{0.32 + 0.245} = \sqrt{0.565} \approx 0.751 \]Then, compute the test statistic:\[ t = \frac{2 - 1}{0.751} \approx \frac{1}{0.751} \approx 1.331 \]
04

Find the P-Value

Using a t-distribution with \( n_1 + n_2 - 2 = 62 \) degrees of freedom, we find the p-value corresponding to our test statistic \( t = 1.331 \). This requires consulting a t-table or using a statistical software:- \( P(T > 1.331) \) where \( T \) follows a t-distribution with 62 degrees of freedom.
05

Compare P-Value and Decision

Assume, upon consultation, the p-value is approximately 0.095. At a significance level of \( \alpha = 0.05 \), the p-value \( 0.095 \) is greater than \( 0.05 \). Therefore, we do not reject the null hypothesis.This suggests that there isn't strong enough evidence to state that the true average weight loss for the vegan diet exceeds the control diet by more than 1 kg.

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Key Concepts

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

Two-Sample t-Test
In hypothesis testing, a two-sample t-test is often used to determine if there are statistically significant differences between the means of two independent samples. It helps to test whether the average of one group is significantly different from the average of another. In this scenario, the two samples are the group of women following the vegan diet and the group following the control diet.
Two-sample t-tests assume that the data follows a normal distribution and that the variances are equal in both samples. The formula for the test statistic involves calculating the difference between the sample means, adjusting for any hypothesized difference, and comparing it to the variability of the data. This variability, or standard error, accounts for the spread of individual measurements within each group.
  • The goal is to see if the obtained difference between the sample means is greater than what would be expected by chance.
  • The calculated t-statistic helps us determine how extreme our results are, assuming the null hypothesis is true.
Understanding the assumptions and calculations of a two-sample t-test is crucial for precise data analysis and helps verify if any observed differences have real-world implications.
P-Value Calculation
In hypothesis testing, the p-value is a key measure that helps us determine the significance of our test results. It represents the probability of obtaining a test statistic at least as extreme as the one observed, under the assumption that the null hypothesis is true. Here, we are testing whether the average difference between the two diets exceeds 1 kg by more than just chance.
The p-value is calculated using the t-distribution, based on the calculated t-statistic and the degrees of freedom, which in this case is the total number of women minus 2. This step usually involves a statistical software or reference to a t-table.
  • If the p-value is less than the chosen significance level (e.g., 0.05), we reject the null hypothesis, suggesting the results are statistically significant.
  • If it is greater, as in this example with a p-value of 0.095, it suggests that the evidence is insufficient to prove that the vegan diet leads to an average weight loss greater than 1 kg compared to the control diet.
Understanding how to calculate and interpret the p-value is essential for making informed conclusions based on statistical data.
Statistical Significance
Statistical significance is a determination of whether the results of a study are likely to be true and not just due to random chance. It hinges on the concept of a p-value and a predetermined significance level, often denoted by \( \alpha \). A common threshold is 0.05, meaning there's a 5% probability of the results occurring by chance under the null hypothesis.
When a p-value is less than \( \alpha \), as per our criteria, the result is considered statistically significant, indicating that it is unlikely to have occurred by random variation alone. This would lead to rejecting the null hypothesis in favor of the alternative. However, when, as in this exercise, the p-value is 0.095, which is greater than 0.05, it lacks statistical significance.
  • This outcome means we cannot confidently claim that the vegan diet results in an average weight loss more than 1 kg greater than the control diet.
  • The concept underscores the importance of thoroughly interpreting statistical results rather than solely relying on numerical outputs.
By understanding statistical significance, researchers can make better decisions about hypothesis testing and draw meaningful conclusions from their data.

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Most popular questions from this chapter

Wait staff at restaurants have employed various strategies to increase tips. An article in the Sept. 5, 2005, New Yorker reported that "In one study a waitress received \(50 \%\) more in tips when she introduced herself by name than when she didn't." Consider the following (fictitious) data on tip amount as a percentage of the bill: \(\begin{array}{lrll}\text { Introduction: } & m=50 & \bar{x}=22.63 & s_{1}=7.82 \\ \text { No introduction: } & n=50 & \bar{y}=14.15 & s_{2}=6.10\end{array}\) Does this data suggest that an introduction increases tips on average by more than \(50 \%\) ? State and test the relevant hypotheses. [Hint: Consider the parameter \(\theta=\mu_{1}-1.5 \mu_{2}\).]

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