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What impact does fast-food consumption have on various dietary and health characteristics? The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality among Children in a National Household Study" (Pediatrics, 2004: \(112-118\) ) reported the accompanying summary statistics on daily calorie intake for a representative sample of teens who do not typically eat fast food and a representative sample of teens who do eat fast food. Is there convincing evidence that the mean calorie intake for teens who typically eat fast food is greater than the mean intake for those who don't by more than 200 calories per day?

Short Answer

Expert verified
Based on the hypothesis test using the two-sample t-test at a significance level of \(\alpha = 0.05\), we compare the calculated test statistic to the critical value from the t-distribution with corresponding degrees of freedom. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is convincing evidence that teens who typically eat fast food have a mean calorie intake greater than those who don't by more than 200 calories per day. If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and cannot make a conclusion on the difference in calorie intake between the two groups.

Step by step solution

01

Define the hypotheses

We are testing if the mean calorie intake for teens who eat fast food is greater than the mean intake for teens who do not by more than 200 calories per day. Let's denote the mean calorie intake for teens who do not eat fast food as \(\mu_1\) and the mean calorie intake for teens who eat fast food as \(\mu_2\). The hypotheses are: \(H_0:\) \(\mu_2 - \mu_1 \leq 200\) \(H_a:\) \(\mu_2 - \mu_1 > 200\)
02

Set the significance level

We should choose a significance level (\(\alpha\)) for our hypothesis test. This is the probability of rejecting the null hypothesis when it is true. A common choice is \(\alpha = 0.05\).
03

Calculate the test statistic

Since we have summary statistics for both groups, we will use the two-sample t-test. The test statistic for this test is given by: \(t = \frac{(\bar{x}_2 - \bar{x}_1) - D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\) 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, and \(D_0\) is the hypothesized difference between the population means. Use the provided summary statistics to calculate the test statistic.
04

Determine the critical value

We will compare the test statistic to a critical value from the t-distribution with a given degree of freedom. To find the degree of freedom (\(df\)), we can use the conservative approximation: \(df = min(n_1 - 1, n_2 - 1)\) Find the critical value for the given significance level \(\alpha\) and degrees of freedom using the t-distribution table.
05

Compare the test statistic to the critical value and draw a conclusion

If the test statistic is greater than the critical value, we will reject the null hypothesis and conclude that there is convincing evidence that the mean calorie intake for teens who typically eat fast food is greater than the mean intake for those who don't by more than 200 calories per day. If the test statistic is less than or equal to the critical value, we fail to reject the null hypothesis and cannot conclude that there is a significant difference in calorie intake between the two groups. Compare the test statistic calculated in Step 3 to the critical value found in Step 4 and draw a conclusion.

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Key Concepts

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

Two-Sample T-Test
The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In the context of the exercise, this test helps us compare the average daily calorie intake between two groups of teens: those who frequently consume fast food and those who do not.

To perform a two-sample t-test, you need data on both groups which typically involves their means, standard deviations, and sample sizes. The hypothesis test assesses if any observed difference in sample means is statistically significant or if it could have occurred by random chance given the variability within each sample.
Null and Alternative Hypotheses
The null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \) are the foundational components of any hypothesis test. The null hypothesis posits that there is no effect or no difference between groups, while the alternative hypothesis suggests the presence of an effect or difference.

In the present exercise, the null hypothesis states that the mean difference in calorie intake is 200 calories or less, whereas the alternative hypothesis claims that the mean difference is greater than 200 calories. Formulating these hypotheses correctly is crucial as they guide the entire testing process.
Statistical Significance
Statistical significance is a determination of whether the outcome of a statistical test is unlikely to have occurred by chance. In hypothesis testing, we establish a significance level (\( \)alpha), usually set at 0.05, which represents a 5% risk of concluding that a difference exists when there is no actual difference.

When the p-value of the test is less than the chosen significance level, we reject the null hypothesis, indicating that the observed effect or difference is statistically significant. In our exercise, if the test statistic exceeds the critical value at the 5% significance level, we can claim with 95% confidence that the mean calorie intake for teens who eat fast food exceeds that of their counterparts by more than 200 calories.
Test Statistic Calculation
The test statistic for a two-sample t-test is calculated using the formula:
\[\begin{equation} t = \frac{(\bar{x}_2 - \bar{x}_1) - D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\end{equation}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means from each group, \(s_1^2\) and \(s_2^2\) are their respective sample variances, \(n_1\) and \(n_2\) are the sample sizes, and \(D_0\) is the hypothesized difference in population means.

The value of \(D_0\) is often zero when testing for any difference, but it can be a non-zero value when testing for a specific difference, as in our exercise where \(D_0\) is 200. This calculation results in a t-value which is then compared against a critical value from the t-distribution to determine if the difference between groups is statistically significant.

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Most popular questions from this chapter

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.

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.

The accompanying data on food intake (in Kcal) for 15 men on the day following two nights of only 4 hours of sleep each night and for 15 men on the day following two nights of 8 hours of sleep each night is consistent with summary quantities in the paper "Short-Term Sleep Loss Decreases Physical Activity Under Free-Living Conditions But Does Not Increase Food Intake Under Time- Deprived Laboratory Conditions in Healthy Men" (American Journal of Clinical Nutrition [2009]: \(1476-1482\) ). The men participating in this experiment were randomly assigned to one of the two sleep conditions.

The paper "Driving Performance While Using a Mobile Phone: A Simulation Study of Greek Professional Drivers" (Transportation Research Part F [2016]: \(164-170)\) describes a study in which 50 Greek male taxi drivers drove in a driving simulator. In the simulator, they were asked to drive following a lead car. On one drive, they had no distractions, and the average distance between the driver's car and the lead car was recorded. In a second drive, the drivers talked on a mobile phone while driving. The authors of the paper used a paired-samples \(t\) test to determine if the mean following distance is greater when the driver has no distractions than when the driver is talking on a mobile phone. The mean of the 50 sample differences (no distraction - talking on mobile phone) was 0.47 meters and the standard deviation of the sample differences was 1.22 meters. The authors concluded that there was evidence to support the claim that the mean following distance for Greek taxi drivers is greater when there are no distractions than when the driver is talking on a mobile phone. Do you agree with this conclusion? Carry out a hypothesis test to support your answer. You may assume that this sample of 50 drivers is representative of Greek taxi drivers.

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