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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

Some people believe that talking on a cell phone while driving slows reaction time, increasing the risk of accidents. The study described in the paper "A Comparison of the Cell Phone Driver and the Drunk Driver" (Human Factors [2006]: \(381-391\) ) investigated the braking reaction time of people driving in a driving simulator. Drivers followed a pace car in the simulator, and when the pace car's brake lights came on, the drivers were supposed to step on the brake. The time between the pace car brake lights coming on and the driver stepping on the brake was measured. Two samples of 40 drivers participated in the study. The 40 people in one sample used a cell phone while driving. The 40 people in the second sample drank a mixture of orange juice and alcohol in an amount calculated to achieve a blood alcohol level of \(0.08 \%\) (a value considered legally drunk in most states). For the cell phone sample, the mean braking reaction time was 779 milliseconds and the standard deviation was 209 milliseconds. For the alcohol sample, the mean breaking reaction time was 849 milliseconds and the standard deviation was \(228 .\) Is there convincing evidence that the mean braking reaction time is different for the population of drivers talking on a cell phone and the population of drivers who have a blood alcohol level of \(0.08 \%\) ? For purposes of this exercise, you can assume that the two samples are representative of the two populations of interest.

The article "More Students Taking AP Tests" (San Luis Obispo Tribune, January 10,2003 ) provided the following information on the percentage of students in grades 11 and 12 taking one or more AP exams and the percentage of exams that earned credit in 1997 and 2002 for seven high schools on the central coast of California. $$ \begin{array}{cccccc} & {\begin{array}{c} \text { Percentage of } \\ \text { Students Taking } \\ \text { One or More } \\ \text { AP Exams } \end{array}} & & {\begin{array}{c} \text { Percentage of } \\ \text { Exams That } \\ \text { Earned College } \end{array}} \\ { 2 - 3 } { 5 - 6 } & & & & {\text { Credit }} \\ { 2 - 3 } { 5 - 6 } \text { School } & 1997 & 2002 & & 1997 & 2002 \\ & 1 & 13.6 & 18.4 & & 61.4 & 52.8 \\ 2 & 20.7 & 25.9 & & 65.3 & 74.5 \\ 3 & 8.9 & 13.7 & & 65.1 & 72.4 \\ 4 & 17.2 & 22.4 & & 65.9 & 61.9 \\ 5 & 18.3 & 43.5 & & 42.3 & 62.7 \\ 6 & 9.8 & 11.4 & & 60.4 & 53.5 \\ 7 & 15.7 & 17.2 & & 42.9 & 62.2 \\ \hline \end{array} $$ a. Assuming that it is reasonable to regard these seven schools as a random sample of high schools located on the central coast of California, carry out an appropriate test to determine if there is convincing evidence that the mean percentage of exams earning college credit at central coast high schools in 1997 and in 2002 were different. b. Do you think it is reasonable to generalize the conclusion of the test in Part (a) to all California high schools? Explain. c. Would it be appropriate to use the paired-samples \(t\) test with the data on percentage of students taking one or more AP exams? Explain.

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The paper "Ladies First?" A Field Study of Discrimination in Coffee Shops" (Applied Economics [2008]: 1-19) describes a study in which researchers observed wait times in coffee shops in Boston. Both wait time and gender of the customer were observed. The mean wait time for a sample of 145 male customers was 85.2 seconds. The mean wait time for a sample of 141 female customers was 113.7 seconds. The sample standard deviations (estimated from graphs in the paper) were 50 seconds for the sample of males and 75 seconds for the sample of females. Suppose that these two samples are representative of the populations of wait times for female coffee shop customers and for male coffee shop customers. Is there convincing evidence that the mean wait time differs for males and females? Test the relevant hypotheses using a significance level of 0.05

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 of independent samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean hip range of motion was 75.6 degrees and the sample standard deviation was 5.9 degrees, whereas the sample mean and sample standard deviation for position players were 79.6 degrees and 7.6 degrees, respectively. Assuming that the two samples are representative of professional baseball pitchers and position players, test hypotheses appropriate for determining if mean range of motion for pitchers is less than the mean for position players.

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