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Coffee, depression, and physical activity. Caffeine is the world's most widely used stimulant, with approximately \(80 \%\) consumed in the form of coffee. Participants in a study investigating the relationship between coffee consumption and exercise were asked to report the number of hours they spent per week on moderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on these data the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value always greater than \(0 .\) The table below gives summary statistics of MET for women in this study based on the amount of coffee consumed. 3 (a) Write the hypotheses for evaluating if the average physical activity level varies among the different levels of coffee consumption. (b) Check conditions and describe any assumptions you must make to proceed with the test. (c) Below is part of the output associated with this test. Fill in the empty cells. (d) What is the conclusion of the test?

Short Answer

Expert verified
Evaluate hypotheses with ANOVA; conclude based on p-value. If the p-value < 0.05, averages differ; otherwise, they don't.

Step by step solution

01

Formulate Hypotheses

Let's define the hypothesis for this scenario. - Null Hypothesis ( H_0 ): The average physical activity (MET hours per week) does not vary significantly among different levels of coffee consumption. - Alternative Hypothesis ( H_a ): The average physical activity (MET hours per week) does vary significantly among different levels of coffee consumption.
02

Check Conditions and Assumptions

To conduct an ANOVA test, we must check the following assumptions: 1. **Independence**: The MET measures are independent of one another. This is assumed to be true as we are dealing with individual responses. 2. **Normality**: The MET values for each group should follow a normal distribution. This can be checked using a normal probability plot or Shapiro-Wilk test, but we assume it is met due to the central limit theorem if sample sizes are large. 3. **Homogeneity of variances**: The variance within each group should be approximately equal. This can be tested using Levene's Test. Without specific data, assume that these conditions hold true based on typical study characteristics.
03

Analyze the Output

To fill in the missing cells of the output, we typically need to compute the F-statistic and compare it with a critical value from an F-distribution table or obtain the p-value: - Use the ANOVA table with variances within and between groups to calculate an F-statistic. - The degrees of freedom for between groups is (k - 1) , where k is the number of coffee consumption categories. - The degrees of freedom for within groups is (n - k) , where n is the total number of observations. - Use these to compute the F-value and find the corresponding p-value.
04

Interpret the Results

Based on the completed ANOVA table: - If the p-value is less than the significance level (commonly 0.05), reject the null hypothesis (H_0 ). This would mean there's statistically significant evidence that MET levels vary with coffee consumption. - If the p-value is greater, do not reject the null hypothesis, as there is no evidence of variation in MET levels among the different coffee consumption levels.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions based on data analysis. In the context of ANOVA, it helps us determine if there are any statistically significant differences between the means of several groups.
In this exercise, the task is to evaluate if the average physical activity level, measured in MET hours per week, varies among different coffee consumption levels.
We start by formulating two hypotheses:
  • **Null Hypothesis ( H_0 ):** Assumes no difference in MET levels between coffee consumption groups.
  • **Alternative Hypothesis ( H_a ):** Assumes a significant difference exists among these groups.
Through careful analysis, including checking conditions and calculating statistics, we determine if we can reject or fail to reject the null hypothesis.
Normality Assumption
The normality assumption is crucial in hypothesis testing, particularly for methods like ANOVA that assess group differences. This assumption suggests that the data for each group should be approximately normally distributed.
For smaller sample sizes, checking normality becomes essential, typically through visual inspections like normal probability plots or statistical tests like the Shapiro-Wilk test.
For larger samples, the central limit theorem often justifies normality, allowing us to assume it holds true unless proven otherwise. In the context of MET measurements and coffee consumption, assuming normality helps ensure the ANOVA results are accurate and reliable.
Homogeneity of Variances
Homogeneity of variances is another important assumption in the ANOVA test, often referred to as homoscedasticity. This implies that the variance within each group should be roughly similar.
Why is this important? If one group has much higher variance, it could unduly influence the test results, leading to incorrect conclusions.
To check this assumption, tests like Levene's Test can be applied. This analytical tool helps verify if the variances are similar across groups. If the variances are homogeneous, ANOVA remains a valid analysis for comparing group means.
F-statistic
The F-statistic is a vital component of the ANOVA test, representing the ratio of variances: specifically, the variance among group means to the variance within groups.
Mathematically, it's given by \[ F = \frac{\text{Variance between groups}}{\text{Variance within groups}} \]Calculating the F-statistic involves degrees of freedom both between groups and within groups:
  • **Degrees of freedom between groups:** \(k - 1\), where \(k\) is the number of groups.
  • **Degrees of freedom within groups:** \(n - k\), where \(n\) is the total number of observations.
A higher F-statistic indicates a more significant difference between group means. This F-value is then compared against a critical value from the F-distribution table or used to determine the p-value, guiding us in accepting or rejecting the null hypothesis.

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Most popular questions from this chapter

4.36 True or false, Part I. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. (a) When comparing means of two samples where \(n_{1}=20\) and \(n_{2}=40,\) we can use the normal model for the difference in means since \(n_{2} \geq 30\). (b) As the degrees of freedom increases, the T distribution approaches normality. (c) We use a pooled standard error for calculating the standard error of the difference between means when sample sizes of groups are equal to each other.

Diamonds, Part I. Prices of diamonds are determined by what is known as the 4 Cs: cut, clarity, color, and carat weight. The prices of diamonds go up as the carat weight increases, but the increase is not smooth. For example, the difference between the size of a 0.99 carat diamond and a 1 carat diamond is undetectable to the naked human eye, but the price of a 1 carat diamond tends to be much higher than the price of a 0.99 diamond. In this question we use two random samples of diamonds, 0.99 carats and 1 carat, each sample of size \(23,\) and compare the average prices of the diamonds. In order to be able to compare equivalent units, we first divide the price for each diamond by 100 times its weight in carats. That is, for a 0.99 carat diamond, we divide the price by 99. For a 1 carat diamond, we divide the price by \(100 .\) The distributions and some sample statistics are shown below. \(^{37}\) Conduct a hypothesis test to evaluate if there is a difference between the average standardized prices of 0.99 and 1 carat diamonds. Make sure to state your hypotheses \begin{tabular}{|l|} \hline \\ \hline \end{tabular} clearly, check relevant conditions, and interpret your results in context of the data.

Find the p-value. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given set of hypotheses and \(T\) test statistic. Also determine if the null hypothesis would be rejected at \(\alpha=0.05\). (a) \(H_{A}: \mu>\mu_{0}, n=11, T=1.91\) (c) \(H_{A}: \mu \neq \mu_{0}, n=7, T=0.83\) (b) \(H_{A}: \mu<\mu_{0}, n=17, T=-3.45\) (d) \(H_{A}: \mu>\mu_{0}, n=28, T=2.13\)

True or false, Part II. Determine if the following statements are true or false in ANOVA, and explain your reasoning for statements you identify as false. (a) As the number of groups increases, the modified significance level for pairwise tests increases as well. (b) As the total sample size increases, the degrees of freedom for the residuals increases as well. (c) The constant variance condition can be somewhat relaxed when the sample sizes are relatively consistent across groups. (d) The independence assumption can be relaxed when the total sample size is large.

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