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Gaming and distracted eating, Part I. A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decrease in the amount of food consumption. To test this hypothesis, they monitored food intake for a group of 44 patients who were randomized into two equal groups. The treatment group ate lunch while playing solitaire, and the control group ate lunch without any added distractions. Patients in the treatment group ate 52.1 grams of biscuits, with a standard deviation of 45.1 grams, and patients in the control group ate 27.1 grams of biscuits, with a standard deviation of 26.4 grams. Do these data provide convincing evidence that the average food intake (measured in amount of biscuits consumed) is different for the patients in the treatment group? Assume that conditions for inference are satisfied. \(^{41}\)

Short Answer

Expert verified
There is convincing evidence of a difference in average biscuit consumption between the groups.

Step by step solution

01

Define Hypotheses

To determine if there's a significant difference in biscuit consumption, we start by defining the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that there is no difference in average biscuit consumption between the treatment and control groups, or \(\mu_1 = \mu_2\). The alternative hypothesis (\(H_a\)) is that there is a difference, or \(\mu_1 eq \mu_2\).
02

Identify the Sample Means and Standard Deviations

The sample mean for the treatment group \((\bar{x}_1)\) is 52.1 grams, with a standard deviation \((s_1)\) of 45.1 grams. The sample mean for the control group \((\bar{x}_2)\) is 27.1 grams, with a standard deviation \((s_2)\) of 26.4 grams. Both groups have \(n = 22\) participants.
03

Calculate the Test Statistic

To compare the means, we'll use a t-test for two independent samples. The test statistic is calculated using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the values: \[ t = \frac{52.1 - 27.1}{\sqrt{\frac{45.1^2}{22} + \frac{26.4^2}{22}}} \]
04

Compute the Degrees of Freedom

The degrees of freedom for this t-test can be approximated using the formula for unequal variances:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Solving this with the given values will give the approximate degrees of freedom.
05

Determine the p-value

Using the calculated t-statistic and degrees of freedom from the previous steps, refer to a t-distribution table or use statistical software to find the p-value. This p-value indicates whether the difference in means is statistically significant.
06

Decision Making

Compare the p-value to the significance level (commonly \(\alpha = 0.05\)). If the p-value is less than \(\alpha\), reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis (H_0) represents a default position. It is a statement asserting that there is no effect or no difference between groups. In this exercise, the null hypothesis is that there is no difference in the average biscuit consumption between the patients who are distracted and those who are not. This is symbolically represented as \(\mu_1 = \mu_2\), where \(\mu_1\) is the mean intake for the treatment group, and \(\mu_2\) is the mean for the control group.

The null hypothesis is always tested indirectly. We assume it is true and use statistical methods to challenge it. The aim is to determine if there's enough evidence to reject this hypothesis in favor of an alternative hypothesis (H_a), which suggests a significant difference exists. Understanding the null hypothesis is crucial because it forms the cornerstone of statistical hypothesis testing.
T-Test
The t-test is a statistical method used to compare the means of two groups. It's especially useful when the sample sizes are small and the variances of the groups might not be equal. In the context of our exercise, we utilize a t-test to determine if the difference in biscuit consumption between the treatment and control groups is significant.

There are different types of t-tests, but we focus on the independent samples t-test here. It is used when the two groups being compared are independent of one another. The test statistic is calculated using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where:
  • \(\bar{x}_1\) and \(\bar{x}_2\) are sample means,
  • \(s_1\) and \(s_2\) are standard deviations,
  • \(n_1\) and \(n_2\) are sample sizes.
The resulting t-value helps determine whether the observed differences are likely to be a result of random chance or if they indicate a real effect.
P-Value
The p-value is a probability measure that helps decide the strength and significance of the evidence against the null hypothesis. In hypothesis testing, after calculating the test statistic, the p-value is derived to determine the probability of observing a test result at least as extreme as the one observed, assuming that the null hypothesis is true.

A low p-value (usually less than a conventional threshold like \(\alpha = 0.05\)) indicates that the observed data is unlikely under the null hypothesis. Therefore, we might reject the null hypothesis in favor of the alternative hypothesis. A high p-value, however, suggests that the observed result is consistent with the null hypothesis.
  • If \(p \leq 0.05\): Evidence against the null hypothesis is strong enough to reject it.
  • If \(p > 0.05\): Not enough evidence to reject the null hypothesis.
Understanding p-values is crucial because they allow us to quantify the uncertainty and determine the significance of our results.
Degrees of Freedom
Degrees of freedom refer to the number of independent values or quantities that can be assigned to a statistical distribution. In essence, it is tied to the number of values in the final calculation of a statistic that are free to vary. In our t-test example, degrees of freedom are essential for determining the critical value from a t-distribution.

For a t-test involving two independent samples, calculating the degrees of freedom can be slightly complex, especially if variances are not equal. It involves the formula:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Understanding this concept is important as varying degrees of freedom affect the shape of the t-distribution and the resulting critical values, ultimately influencing hypothesis testing outcomes.

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Most popular questions from this chapter

Work hours and education, Part II. The General Social Survey described in Exercise 4.16 included random samples from two groups: US residents with a college degree and US residents without a college degree. For the 505 sampled US residents with a college degree, the average number of hours worked each week was 41.8 hours with a standard deviation of 15.1 hours. For those 667 without a degree, the mean was 39.4 hours with a standard deviation of 15.1 hours. Conduct a hypothesis test to check for a difference in the average number of hours worked for the two groups.

Body fat in women and men. The third National Health and Nutrition Examination Survey collected body fat percentage (BF) data from 13,601 subjects whose ages are 20 to \(80 .\) A summary table for these data is given below. Note that BF is given as mean \(\pm\) standard error. Construct a \(95 \%\) confidence interval for the difference in average body fat percentages between men and women, and explain the meaning of this interval. Tip: the standard error can be calculated as \(S E=\sqrt{S E_{M}^{2}+S E_{W}^{2}}\)

Math scores of 13 year olds, Part I. The National Assessment of Educational Progress tested a simple random sample of 1,000 thirteen year old students in both 2004 and 2008 (two separate simple random samples). The average and standard deviation in 2004 were 257 and 39 , respectively. In \(2008,\) the average and standard deviation were 260 and \(38,\) respectively. Calculate a \(90 \%\) confidence interval for the change in average scores from 2004 to 2008 , and interpret this interval in the context of the application. (Reminder: check conditions.) \(^{31}\)

Child care hours, Part I. The China Health and Nutrition Survey aims to examine the effects of the health, nutrition, and family planning policies and programs implemented by national and local governments. One of the variables collected on the survey is the number of hours parents spend taking care of children in their household under age 6 (feeding, bathing, dressing, holding, or watching them). In 2006,487 females and 312 males were surveyed for this question. On average, females reported spending 31 hours with a standard deviation of 31 hours, and males reported spending 16 hours with a standard deviation of 21 hours. Calculate a \(95 \%\) confidence interval for the difference between the average number of hours Chinese males and females spend taking care of their children under age 6 . Also comment on whether this interval suggests a significant difference between the two population parameters. You may assume that conditions for inference are satisfied. \(^{34}\)

\(t^{*}\) vs. \(z^{*}\). For a given confidence level, \(t_{d f}^{*}\) is larger than \(z^{*}\). Explain how \(t_{d f}^{*}\) being slightly larger than \(z^{*}\) affects the width of the confidence interval.

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