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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 \(.05\).

Step by step solution

01

Set Up Hypotheses

To determine if the true mean weight loss for the vegan diet exceeds the control diet by more than 1 kg, we first set up the null and alternative hypotheses. The null hypothesis is \( H_0: \mu_1 - \mu_2 \leq 1 \), and the alternative hypothesis is \( H_a: \mu_1 - \mu_2 > 1 \) where \( \mu_1 \) is the mean weight loss for the vegan diet, and \( \mu_2 \) is the mean weight loss for the control diet.

02

Calculate Sample Mean Difference and Standard Error

Calculate the observed difference in sample means: \( \bar{x}_1 - \bar{x}_2 = 5.8 - 3.8 = 2.0 \) kg. Next, calculate the standard error (SE) for the difference of sample means: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3.2^2}{32} + \frac{2.8^2}{32}} \approx 0.825 \]

03

Find Test Statistic

Using the sample mean difference and SE calculated, compute the test statistic \( t \): \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - 1}{SE} = \frac{2.0 - 1}{0.825} \approx 1.212 \]

04

Determine Critical Value or P-value

The test involves a t-distribution with \( n_1 + n_2 - 2 = 64 - 2 = 62 \) degrees of freedom. At a significance level of \( \alpha = 0.05 \), using a t-table or calculator, we find the critical t-value for a one-tailed test is approximately 1.670 (for 62 degrees of freedom). Or compute the p-value.

05

Make Decision

Since our test statistic (1.212) is less than the critical t-value (1.670), or the p-value is greater than 0.05, we fail to reject the null hypothesis.

06

Conclusion

There is not enough statistical evidence to conclude that the true average weight loss for the vegan diet exceeds that for the control diet by more than 1 kg at the 0.05 significance level.

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Key Concepts

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

t-test

The t-test is a statistical test used to compare the means of two groups. In the context of the exercise, it helps determine if the average weight loss among women on the vegan diet is significantly greater than that on the control diet.
The t-test specifically checks if the observed difference in sample means is unlikely to have occurred due to chance alone. Here, the sample mean difference of weights between the two diets is compared to an expected difference set by the null hypothesis (e.g., 1 kg difference).

• The t-test assumes that the data is normally distributed and that the variances of the two samples are equal.
• It's appropriate for small to medium sample sizes, which applies in this experiment with 32 participants per group.

The test calculates a t-statistic, which indicates how far the observed data is from the expected mean difference under the null hypothesis in terms of standard error units. This value is then used to determine whether we reject or fail to reject the null hypothesis.

significance level

The significance level, denoted by \( \alpha \), is a threshold that researchers set to decide when to reject the null hypothesis. In this exercise, a significance level of 0.05 is used. This means there is a 5% risk of concluding that there is a true difference in mean weight loss between the two diets when there actually isn't.
A lower significance level means more rigorous criteria for rejecting the null hypothesis, reducing the likelihood of false positives.

• The choice of \(\alpha = 0.05\) is a standard in many scientific fields, balancing false positive risk and sensitivity of the test.
• It essentially indicates our tolerance for a Type I error, which occurs when we incorrectly reject a true null hypothesis.

In our context, since the calculated test statistic doesn’t surpass the critical value, the conclusion is to "fail to reject" the null hypothesis, as weight loss difference isn't significant enough based on the set threshold.

sample mean difference

The sample mean difference is the observed difference between the average results of the two sample groups. In this case, it's the difference in average weight loss between women on a vegan diet and those on the control diet.
The calculated mean difference is 2.0 kg, which is obtained by subtracting the average weight loss of the control group (3.8 kg) from that of the vegan diet group (5.8 kg).

• This number tells us how much more weight, on average, the vegan diet group lost compared to the control group.
• It serves as the primary statistic of interest in deciding whether there is a significant difference in outcomes between the diet groups.

However, in hypothesis testing, this observed difference has to be evaluated in the context of variability (captured by the standard error) to test its statistical significance.

standard error

The standard error (SE) measures the accuracy with which a sample represents a population. It is particularly crucial in hypothesis testing because it reflects the sampling variability of the sample mean difference.
In this example, the standard error for the difference of sample means is calculated to be approximately 0.825.
This is obtained using the standard deviations and sample sizes of both groups:
\[ SE = \sqrt{\frac{3.2^2}{32} + \frac{2.8^2}{32}} \approx 0.825 \]

• Lower SE suggests that the sample mean difference is more reliable as an estimate of the true population mean difference.
• The SE provides the denominator in the formula for the t-statistic, which indicates how much the sample mean difference deviates from the expected mean difference under the null hypothesis.

The smaller the SE, the more confident one can be that the sample mean difference is an accurate reflection of the true difference in the population means.