/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The paper referenced in the Prev... [FREE SOLUTION] | 91Ó°ÊÓ

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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 higher than the mean number of chocolate servings per month for people who would not screen positive.

Short Answer

Expert verified
Based on the p-value obtained, if the p-value is less than or equal to a significance level of 0.05, the conclusion would be that the mean number of servings of chocolate per month is significantly higher in individuals who screen positive for depression. If the p-value is larger, then the conclusion is that there is no significant difference between the two groups.

Step by step solution

01

State the Hypotheses

The null hypothesis (`H_0`) is our status quo, where there's no difference between the means of the two groups. While the alternative hypothesis (`H_1`) is what we want to prove. So, `H_0: µ_1=µ_2` and `H_1: µ_1>µ_2`, where µ_1 and µ_2 are the means of chocolate consumption per month for people who screen positive and negative for depression, respectively.
02

Test Statistic

Since the population standard deviations are unknown, we use the sample standard deviations to compute a t-statistic. The formula for the t-statistic in an independent two-sample t-test with unequal variances is: \( t = \frac{\(X_1 - X_2\)}{\sqrt{\(\frac{s_1^2}{n_1}\) + \(\frac{s_2^2}{n_2}\)}\), where `X_1` and `X_2` are the sample means, `s_1` and `s_2` are the sample standard deviations, and `n_1` and `n_2` are the sample sizes. Plugging the given values into this formula, the t-statistic is calculated.
03

Degree of Freedom

The degrees of freedom for this t-test is calculated using the formula: \((s1^2/n1 + s2^2/n2)^2 / [ ((s1^2/n1)^2/(n1-1)) + ((s2^2/n2)^2/(n2-1)) ]\) and substituting the given values.
04

Compute P-value

After calculating the t-statistic and degrees of freedom, we can use these values to compute the p-value using a T-distribution table or with software. The p-value is the probability that you would observe a test statistic that extreme assuming the null hypothesis is true.
05

Conclusion

If the p-value is less than or equal to a significance level (normally 0.05), then we reject the null hypothesis in favor of the alternative hypothesis. That would suggest that the mean number of servings of chocolate per month is significantly higher for people who would screen positive for depression. If the p-value is greater than the significance level, then we fail to 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 and Alternative Hypotheses
Hypothesis testing in statistics starts with the formulation of two opposing statements: the null hypothesis (ull_hypothesis{H_0}) and the alternative hypothesis (ull_hypothesis{H_1}). The null hypothesis represents the default position that there is no effect or no difference, effectively serving as a starting point for the test. In the context of the chocolate consumption study, the null hypothesis suggests that there is no difference in chocolate consumption between those who screened positive and negative for depression.

Conversely, the alternative hypothesis proposes the presence of an effect or a difference. In our example, it asserts that individuals who screened positive for depression consume more chocolate than those who screened negative. The ultimate goal of hypothesis testing is to assess the evidence against the null hypothesis, and, if sufficient evidence exists, to reject the null hypothesis in favor of the alternative hypothesis.
t-statistic
In hypothesis testing, the t-statistic is a ratio that compares the difference between two sample means relative to the variability of the samples. It forms part of the t-test, which is used to determine if there is a significant difference between the means of two groups. The t-statistic is calculated from the sample data and represents the number of standard errors that the observed difference in means is away from the null hypothesis.For example, a higher t-statistic indicates that the means are further apart compared to the spread or variability of their respective distributions. The formula for the t-statistic incorporates the sample means, standard deviations, and sizes, allowing researchers to effectively gauge the magnitude of the difference with an understanding of how much natural variation exists in the data.
p-value
The p-value is a fundamental concept in hypothesis testing, representing the likelihood of observing test results as extreme as those actually observed, assuming the null hypothesis is true. A small p-value suggests that the observed data is highly unlikely under the null hypothesis and hence provides evidence against it.

In practice, researchers often use a significance level (often 0.05) to determine if the p-value warrants rejecting the null hypothesis. If the p-value is less than the significance level, the results are considered statistically significant, leading to the rejection of the null hypothesis. Conversely, if the p-value is higher, we fail to reject the null hypothesis, meaning that there is not enough evidence to support the alternative hypothesis.
Degrees of Freedom
Degrees of freedom (df) in statistics refer to the number of independent values or quantities that can freely vary in the calculation of a statistic without violating given constraints. It often represents the number of values that are free to vary in a dataset.

In the context of the t-test, the degrees of freedom are important for determining the exact distribution to refer to when interpreting the t-statistic. The calculation for the df in an independent two-sample t-test where sample variances are unequal (as in the chocolate consumption study) is more complex than the simple difference in the sample sizes. The df accounts for both sample sizes and their variances and helps ensure that the t-test remains reliable even when the underlying assumptions are not perfectly met.
Independent Two-sample t-test
The independent two-sample t-test is a statistical procedure used to compare the means of two independent groups. This test is applicable when the samples do not have a paired or matched relationship and are drawn from populations with approximately normal distributions.

Key assumptions of this test include the independence of samples, normally distributed populations, and equal or unequal population variances. When the variances are unequal, as assumed in the chocolate consumption study, the test is modified to account for this by adjusting the degrees of freedom and modifying the formula for the t-statistic. The independent two-sample t-test analyzes the null hypothesis that the population means are equal, producing a t-statistic and corresponding p-value, which are used to make a conclusion about the statistical significance of the observed difference in means.

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Most popular questions from this chapter

In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: \(2024-2033\) ). The following statement appeared in the paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004)\) a. What hypotheses did the researchers test to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(\quad t=5.00\) iii. \(t=2.33\) ii. \(t=2.65\) iv. \(t=1.47\)

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be for a difference in population means. If not, explain why not. Scenario 1: The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]: e953-e958) studied independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. You would like to determine if there is convincing evidence that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls. Scenario 2: The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 is \(\$ 48,722\). A random sample of 50 accounting graduates at a large university resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). You would like to determine if there is strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722\). Scenario 3: Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25,2006 ). The sample mean and standard deviation were 15.1 hours and 11.4 hours for males and 14.1 and 11.8 for females. You would like to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers.

Example 13.1 looked at a study comparing students who use Facebook and students who do not use Facebook ("Facebook and Academic Performance," Computers in Human Behavior [2010]: \(1237-1245\) ). In addition to asking the students in the samples about GPA, each student was also asked how many hours he or she spent studying each day. The two samples (141 students who were Facebook users and 68 students who were not Facebook users) were independently selected from students at a large, public Midwestern university. Although the samples were not selected at random, they were selected to be representative of the two populations. For the sample of Facebook users, the mean number of hours studied per day was 1.47 hours and the standard deviation was 0.83 hours. For the sample of students who do not use Facebook, the mean was 2.76 hours and the standard deviation was 0.99 hours. Do these sample data provide convincing evidence that the mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook? Use a significance level of 0.01 .

The article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids ages 8 to 18 . One survey was conducted in 1999 and the other was conducted in 2009 . Data on number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. $$ \begin{array}{lllllllllllll} \mathbf{2 0 0 9} & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 \\ & 10 & 9 & 8 & & & & & & & & & \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 \\ & 5 & 6 & 6 & & & & & & & & & \\ & & & & & & & & & & & \end{array} $$ a. Because the given sample sizes are small, what assumption must be made about the distributions of electronic media use times for the two-sample \(t\) test to be appropriate? Use the given data to construct graphical displays that would be useful in determining whether this assumption is reasonable. Do you think it is reasonable to use these data to carry out a two-sample \(t\) test? b. Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in \(1999 ?\) Test the relevant hypotheses using a significance level of 0.01 .

Research has shown that for baseball players, good hip range of motion results in improved performance and decreased body stress. The article "Functional Hip Characteristics of Baseball Pitchers and Position Players" (The American journal of Sports Medicine, \(2010: 383-388\) ) reported on a study involving independent samples of 40 professional pitchers and 40 professional position players. For the sample of pitchers, the mean hip range of motion was 75.6 degrees and the standard deviation was 5.9 degrees, whereas the mean and standard deviation for the sample of position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that these two samples are representative of professional baseball pitchers and position players, estimate the difference in mean hip range of motion for pitchers and position players using a \(90 \%\) confidence interval.

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