91Ó°ÊÓ

The paper referenced in the Preview Example of this chapter ("Mood Food: Chocolate and Depressive Symptoms in a Cross-Sectional Analysis," Archives of Internal Medicine [2010]: \(699-703\) ) describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on their test score. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings per month of chocolate for the sample of people that screened positive for depression was \(8.39,\) and the sample standard deviation was \(14.83 .\) For the sample of people who did not screen positive for depression, the mean was \(5.39,\) and the standard deviation was \(8.76 .\) The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants included 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is greater than the mean number of chocolate servings per month for people who would not screcn positive.

Step by step solution

01

State the Hypotheses

To perform a hypothesis test, we first need to state the null and alternative hypotheses. In this case, they are: - Null hypothesis (H0): There is no significant difference in the mean number of servings of chocolate between the two groups, i.e., μ1 = μ2. - Alternative hypothesis (H1): The mean number of servings of chocolate per month for people who would screen positive for depression is greater than the mean number of chocolate servings per month for people who would not screen positive, i.e., μ1 > μ2.

02

Determine the Significance Level

The significance level, denoted by α, is the probability of rejecting the null hypothesis when it is true. It is not given in the problem, but a common choice is 0.05.

03

Calculate Test Statistic

To carry out the hypothesis test, we will use the two-sample t-test. The test statistic is calculated as: \[t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] Where: - \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means. - \(s_1^2\) and \(s_2^2\) are the sample variances. - \(n_1\) and \(n_2\) are the sample sizes. Using the given data: \[\bar{x}_1 = 8.39, \quad s_1 = 14.83, \quad n_1 = 311\] \[\bar{x}_2 = 5.39, \quad s_2 = 8.76, \quad n_2 = 620\] Substitute the values into the formula: \[t = \frac{(8.39 - 5.39) - 0}{\sqrt{\frac{14.83^2}{311} + \frac{8.76^2}{620}}} \approx 3.72\]

04

Calculate the Critical Value and p-value

Since we are conducting a one-tailed test with a significance level of 0.05, we need to find the critical t value and compare it to our test statistic. We will use the degrees of freedom from the smaller sample size, which is 311-1 = 310. Using a t-distribution table or a calculator, we find the critical t value to be approximately 1.65. Now, we need to find the p-value for our test statistic. Using a t-distribution calculator, we find that the p-value for a t-score of 3.72 with 310 degrees of freedom is approximately 0.0001.

05

Make a Decision

Since the p-value is less than the significance level (0.0001 < 0.05) and the test statistic is greater than the critical value (3.72 > 1.65), we reject the null hypothesis. This provides evidence that the mean number of servings of chocolate per month for people who would screen positive for depression is greater than the mean number of chocolate servings per month for people who would not screen positive.

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Key Concepts

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

Two-Sample t-Test

A two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two groups. This is particularly useful when comparing two independent groups, like in our example with individuals who screen positive for depression versus those who do not.

The main idea is to see if the observed difference in means is due to random chance, or if it reflects a real difference in the populations. In essence, it helps answer questions like "Is there really a difference, or are we seeing patterns in the numbers which aren’t significant?"

In the exercise, we calculate a test statistic using the t-test formula:

• \[t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
• This compares the means (\(\bar{x}_1\) and \(\bar{x}_2\)) of two samples.

It takes into account the variances (\(s_1^2\), \(s_2^2\)) and sizes (\(n_1\), \(n_2\)) of the samples too. The calculated test value helps in determining whether to accept or reject the null hypothesis.

Null Hypothesis

In hypothesis testing, the null hypothesis (\( H_0 \)) is a starting assumption that there is no effect or difference. It's like saying, "Everything is the same—there's no storm brewing beneath the surface."

For the chocolate consumption study, the null hypothesis would be:

• There is no difference in the mean chocolate servings consumed by those who screen positive or negative for depression.
• Symbolically, we say \( \mu_1 = \mu_2 \).

Rejecting the null hypothesis suggests that there truly might be a difference worth noting.

It forms the foundation of scientific testing because we try to gather enough evidence to reject this idea. If we fail to reject the null hypothesis, it simply means our data was not convincing enough to see a difference. However, it does not prove that the null hypothesis is true.

p-value

The p-value is a critical concept in hypothesis testing. It tells us the probability of observing the results given that the null hypothesis is true. Think of it as a measure of surprise!

If the p-value is very low, it suggests that such an extreme result would be surprising if the null hypothesis were true. In our example, a p-value of 0.0001 implies that our observed difference in chocolate consumption is very unlikely under the null hypothesis.

• A low p-value (< significance level) provides evidence to reject the null hypothesis.
• The smaller the p-value, the stronger the evidence against the null hypothesis.

Essentially, a p-value helps quantify evidence against the null hypothesis. A common threshold is 0.05, below which results are typically considered statistically significant.

Significance Level

The significance level, often denoted by \( \alpha \), is a threshold set by researchers to determine how strong the evidence must be to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it's actually true (type I error).

For most scientific studies, a standard value used is \( \alpha = 0.05 \). This means there's a 5% risk of saying there's an effect when there isn't one.

• Our example uses \( \alpha = 0.05 \) to evaluate the test results.
• If p-value < \( \alpha \), we reject the null hypothesis.

Choosing the right significance level is crucial as it balances the chances of making errors in conclusions. It ultimately reflects how much risk one is willing to take in declaring a significant result.