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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 "The Effect of Multitasking on the Grade Performance of Business Students" (Research in Higher Education Journal [2010]: 1-10) describes an experiment in which 62 undergraduate business students were randomly assigned to one of two experimental groups. Students in one group were asked to listen to a lecture but were told that they were permitted to use cell phones to send text messages during the lecture. Students in the second group listened to the same lecture but were not permitted to send text messages during the lecture. Afterwards, students in both groups took a quiz on material covered in the lecture. The researchers reported that the mean quiz score for students in the texting group was significantly lower than the mean quiz score for students in the no-texting group. In the context of this experiment, explain what it means to say that the texting group mean was significantly lower than the no-text group mean. (Hint: See discussion on page \(662 .\) )

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in two population means. If not, explain why not.

The National Sleep Foundation surveyed representative samples of adults in six different countries to ask questions about sleeping habits ("2013 International Bedroom Poll Summary of Findings," www.sleepfoundation.org/sites /default/files/RPT495a.pdf, retrieved May 20,2017 ). Each person in a representative sample of 250 adults in each of these countries was asked how much sleep they get on a typical work night. For the United States, the sample mean was 391 minutes, and for Mexico the sample mean was 426 minutes. Suppose that the sample standard deviations were 30 minutes for the U.S. sample and 40 minutes for the Mexico sample. The report concludes that on average, adults in the United States get less sleep on work nights than adults in Mexico. Is this a reasonable conclusion? Support your answer with an appropriate hypothesis test.

To determine if chocolate milk is as effective as other carbohydrate replacement drinks, nine male cyclists performed an intense workout followed by a drink and a rest period. At the end of the rest period, each cyclist performed an endurance trial in which he exercised until exhausted, and the time to exhaustion was measured. Each cyclist completed the entire regimen on two different days. On one day, the drink provided was chocolate milk, and on the other day the drink provided was a carbohydrate replacement drink. Data consistent with summary quantities in the paper "The Efficacy of Chocolate Milk as a Recovery Aid" (Medicine and Science in Sports and Exercise [2004]: S126) are given in the table at the bottom of the page. Is there evidence that the mean time to exhaustion is greater after chocolate milk than after a carbohydrate replacement drink? Use a significance level of \(\alpha=0.05\).

Can moving their hands help children learn math? This question was investigated in the paper "Gesturing Gives Children New Ideas About Math" (Psychological Science [2009]: \(267-272\) ). Eighty-five children in the third and fourth grades who did not answer any questions correctly on a test with six problems of the form \(3+2+8=-8\) were participants in an experiment. The children were randomly assigned to either a no-gesture group or a gesture group. All the children were given a lesson on how to solve problems of this form using the stratcgy of trying to make both sides of the equation equal. Children in the gesture group were also taught to point to the first two numbers on the left side of the equation with the index and middle linger of one hand and then to point at the blank on the right side ol the equation. This gesture was supposed to emphasize that grouping is involved in solving the problem. The children then practiced udditional problems of this type. All children were then given a test with six problems to solve, and the number of correct answers was recorded for each child. Summary statistics are given below. Is there evidence that learning the gesturing approach to solving problems of this type results in a significantly higher mean number of correct responses? Test the relevant hypotheses using \(\alpha=0.05\).

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