Problem 2 - An Introduction to Mathematical Statistics and Its Applications

Question

As the United States has struggled with the growing obesity of its citizens, diets have become big business. Among the many competing regimens for those seeking weight reduction are the Atkins and Zone diets. In a comparison of these two diets for one-year weight loss, a study (64) found that seventy- seven subjects on the Atkins diet had an average weight loss of \(\bar{x}=-4.7 \mathrm{~kg}\) and a sample standard deviation of \(s_{X}=7.05 \mathrm{~kg}\). Similar figures for the seventy-nine people on the Zone diet were \(\bar{y}=-1.6 \mathrm{~kg}\) and \(s_{Y}=5.36 \mathrm{~kg}\). Is the greater reduction with the Atkins diet statistically significant? Test for \(\alpha=0.05\).

Step-by-Step Solution

Answer

Without the actual calculations and given the nature of the problem, it's hard to write down a definitive short answer. After step 4, depending on the calculated P-value and comparing it with the level of significance, a decision will be made to reject or not to reject the null hypothesis. If it is rejected, then this means that the Atkins diet has a significantly different impact on weight loss than the Zone diet. If the null hypothesis is not rejected, it means there is no significant difference.

Step 1: Formulate Hypothesis

First, it's important to set the null hypothesis and the alternative hypothesis. We will assume that the null hypothesis claims that there's no significant difference in weight loss between the two diets. Thus, the mean difference is 0: \[H_0: \mu_X - \mu_Y = 0 \] In contrary, the alternative hypothesis claims that there is a significant difference in weight loss and the Atkins diet losses more weight: \[H_1: \mu_X - \mu_Y < 0 \] In the hypotheses, \(\mu_X\) and \(\mu_Y\) are true means for the Atkins and Zone diets respectively.

Step 2: Calculate Standard Error

The first step in working out the test statistic is calculating the standard error. For two independent samples, it can be calculated using the formula: \[ SE = \sqrt{(s_X^2/n_X) + (s_Y^2/n_Y)} \] where \(s_X\) and \(s_Y\) denote the sample standard deviations and \(n_X\) and \(n_Y\) denote the sample sizes. Substituting our given values, we get: \[ SE = \sqrt{(7.05^2/77) + (5.36^2/79)} \]

Step 3: Calculate Test Statistic

Once we have calculated the standard error, we can calculate the test statistic (Z-score) using the formula: \[ Z = \frac{(\bar{X} - \bar{Y}) -D}{SE} \] where \(\bar{X}\) and \(\bar{Y}\) are sample means and D is the difference in population means, which is zero in this case. Substituting the values, we get our Z score.

Step 4: Find P-value

We can find the p-value using a Z-table or statistical software. If the p-value is less than the significance level (0.05), we reject the null hypothesis.

Step 5: Interpret the Result

The final step is to interpret the result. If the null hypothesis is rejected, it means that the weight loss difference is statistically significant, and the Atkins diet results in greater weight loss. If the null hypothesis is not rejected, it means that the difference in weight loss is not statistically significant.

Key Concepts

Null Hypothesis

In any statistical hypothesis testing, the null hypothesis (H_0) is essentially what you start with. It is a statement that there is no effect or no difference. In this exercise about diet comparison, the null hypothesis presupposes that there is no significant difference in weight loss between the Atkins and Zone diets. This means that the average weight loss would be the same for both diets. Mathematically, this is represented as:

• H_0: \( \mu_X - \mu_Y = 0 \)

Here, \( \mu_X \) and \( \mu_Y \) are the mean weight losses for the Atkins and Zone diets, respectively. The null hypothesis asserts that these two means are equal, implying no diet is superior to the other in terms of weight loss. By testing this hypothesis, researchers can determine whether observed differences are likely due to chance or if they suggest a real, statistically significant difference.

Alternative Hypothesis

The alternative hypothesis (H_1) is what you test against the null hypothesis. It is the claim that there is a significant effect or difference. In this context, the alternative hypothesis suggests that the Atkins diet results in more significant weight loss compared to the Zone diet. This suggests that the mean weight loss for the Atkins diet is greater than that of the Zone diet:

• H_1: \( \mu_X - \mu_Y < 0 \)

Here, the symbol "<" indicates that the Atkins diet is assumed to lead to a lower mean weight (hence more weight loss) than the Zone diet. If the statistical evidence supports this hypothesis, it would mean that the Atkins diet is indeed more effective for weight loss when compared to the Zone diet. Rejecting the null hypothesis in favor of the alternative hypothesis would provide support for this claim.

Standard Error

The standard error (SE) gives us an idea of how much variability there is in the difference between the sample means. It acts as a measure of the precision of the sample mean difference estimate. For two independent samples, like our Atkins and Zone diets, SE can be calculated using this formula:

• SE = \( \sqrt{(s_X^2/n_X) + (s_Y^2/n_Y)} \)

The variables \(s_X\) and \(s_Y\) are the standard deviations of the Atkins and Zone diet samples, while \(n_X\) and \(n_Y\) are the respective sample sizes. Essentially, standard error shows how accurately the sampling process can estimate the true mean difference. A smaller SE indicates more precise estimation. Calculating SE is crucial for understanding whether any observed differences are likely due to randomness or are significant enough to consider.

Z-test

A Z-test is used when deciding if there is a significantly different outcome in the means of two groups. It's calculated using the test statistic (Z-score), which can determine whether an observed difference is statistically meaningful:

• Z = \( \frac{(\bar{X} - \bar{Y}) - D}{SE} \)

Here, \( \bar{X} \) and \( \bar{Y} \) are the sample means for the Atkins and Zone diets, while D is the hypothesized difference in population means, which is zero under the null hypothesis. The numerator reflects the actual observed difference minus the expected difference (under the null hypothesis), and the denominator is the standard error. This Z-score is then compared to a standard normal distribution to assess how likely such a score is to occur if the null hypothesis is true. A score falling under a critical value would lead to rejecting the null hypothesis, thereby supporting the alternative.

Significance Level

The significance level (\alpha) is a threshold that helps determine whether to reject the null hypothesis. It essentially represents the probability of rejecting a true null hypothesis, known as Type I error. In this exercise, the significance level is set at 0.05:

• The typical levels are 0.05, 0.01, or 0.10.
• Choosing \( \alpha = 0.05 \) implies a 5% risk of concluding that a difference exists when there is none.

When the calculated p-value, derived from the Z-test, is less than \( \alpha \), it indicates that the observed data is inconsistent with the null hypothesis. Consequently, the null hypothesis is rejected, suggesting the alternative hypothesis is more likely correct. Thus, the significance level serves a vital function in hypothesis testing by aiding in determining the threshold for making inferences.