/*! 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 48 The following passage is from th... [FREE SOLUTION] | 91Ó°ÊÓ

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The following passage is from the paper "Gender Differences in Food Selections of Students at a Historically Black College and University" (College Student Journal \([2009]: 800-806):\) Also significant was the proportion of males and their water consumption ( 8 oz. servings) compared to females \(\left(X^{2}=8.166, P=.086\right) .\) Males came closest to meeting recommended daily water intake \((64\) oz. or more) than females ( \(29.8 \%\) vs. \(20.9 \%\) ). This statement was based on carrying out a chi-square test of homogeneity using data in a two-way table where rows corresponded to sex (male, female) and columns corresponded to number of servings of water consumed per day, with categories none, one, two to three, four to five, and six or more. a. What hypotheses did the researchers test? What is the number of degrees of freedom associated with the reported value of the \(X^{2}\) statistic? b. The researchers based their statement on a test with a significance level of 0.10 . Would they have reached the same conclusion if a significance level of 0.05 had been used? Explain.

Short Answer

Expert verified
The researchers tested the null hypothesis \((H_0)\) that there was no difference in proportions between males and females in their daily water consumption, and the alternative hypothesis \((H_a)\) that there was a difference in proportions. The degrees of freedom for the reported chi-square statistic was 4. The researchers would have reached a different conclusion if they had used a significance level of 0.05 instead of 0.10, as the p-value of 0.086 is greater than 0.05.

Step by step solution

01

Understand the hypotheses

The chi-square test of homogeneity is performed to see if there are any significant differences between groups in terms of their proportions in various categories. In this case, the groups are male and female students, and the categories are the number of servings of water consumed daily. The null hypothesis states that there is no difference in proportions across groups, while the alternative hypothesis states that there is a difference in proportions between groups. In this case, the null hypothesis \((H_0)\) and alternative hypothesis \((H_a)\) are: \(H_0\): There is no difference in proportions between males and females in their daily water consumption. \(H_a\): There is a difference in proportions between males and females in their daily water consumption.
02

Find the degrees of freedom for the reported chi-square statistic

Given that there are 5 columns and 2 rows in the two-way table, the degrees of freedom \((df)\) would be: \[df = (number\_of\_columns - 1) * (number\_of\_rows - 1)\] \[df = (5 - 1) * (2 - 1) = 4\] The degrees of freedom for the reported value of the \(X^{2}\) statistic is 4.
03

Determine if the conclusion would change with a different significance level

The researchers used a significance level of 0.10 and obtained a p-value of 0.086. Since the p-value is less than the chosen significance level, they rejected the null hypothesis and concluded that there is a difference in proportions between males and females in their daily water consumption. Now let us check if the conclusion would change if the researchers used a significance level of 0.05 instead of 0.10. Since the p-value of 0.086 is greater than the new significance level of 0.05, they would not reject the null hypothesis and would not conclude that there is a difference in proportions between males and females in their daily water consumption. Thus, the researchers would have reached a different conclusion with a significance level of 0.05.

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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 an essential statistical tool used to infer the properties of an entire population based on sample data. In the context of a chi-square test of homogeneity, we aim to determine whether different groups show similar characteristics across several categories. Here, male and female students are compared based on their water consumption rates.

The test begins with two hypotheses:
  • The **null hypothesis** ( \(H_0\) ) suggests that there is no difference in the proportion of water consumption across genders, implying that any observed differences are just due to random chance.
  • On the other hand, the **alternative hypothesis** ( \(H_a\)) posits that there is, in fact, a difference in the proportions of water consumption between males and females.
Once the hypotheses are established, the test analyzes the data to see if the observed differences could plausibly occur if the null hypothesis were true. The strength of hypothesis testing lies in its ability to handle data variability and help researchers make confident decisions about their hypotheses.
Degrees of Freedom
The concept of degrees of freedom is crucial in statistical tests as it determines the number of independent values that can vary in the data. Specifically, for the chi-square test, it reflects the number of independent comparisons we can make.

In our example, we have a two-way table with rows for males and females, and columns for different categories of water servings consumed. The formula for finding degrees of freedom in a chi-square test is: \[df = (\text{number of columns} - 1) \times (\text{number of rows} - 1) \]
Using this formula:
  • Our table has 5 columns (for different amounts of water servings) and 2 rows (for male and female).
  • Thus, the degrees of freedom calculation is \((5 - 1) \times (2 - 1) = 4\).
Understanding degrees of freedom helps in determining the proper statistical distribution to assess the chi-square statistic, thereby enabling accurate hypothesis testing.
Significance Level
The significance level is a threshold used in hypothesis tests to determine when we should reject the null hypothesis. It represents the probability of incorrectly rejecting the null hypothesis when it is true (also known as a Type I error).

In many studies, a standard significance level is set at 0.05, meaning a 5% risk of concluding a difference exists when there may be none. However, the choice of significance level affects the test outcomes. For the study discussed, a significance level of 0.10 was used.

When the p-value (0.086) obtained from the chi-square test is less than the significance level (0.10), the null hypothesis is rejected, suggesting a significant difference exists. However, if the study had used a lower significance level like 0.05, the same p-value would have been above this threshold, leading to different conclusions: not rejecting the null hypothesis.
  • **Significance level of 0.10:** Reject the null hypothesis (p-value = 0.086).
  • **Significance level of 0.05:** Do not reject the null hypothesis (p-value = 0.086).

This highlights how different significance levels can influence research conclusions and the importance of choosing an appropriate threshold reflecting the study's context and desired rigor.

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Most popular questions from this chapter

The authors of the paper "Risk of Malnutrition Is an Independent Predictor of Mortality, Length of Hospital Stay, and Hospitalization Costs in Stroke Patients" (Journal of Stroke and Cerebrovascular Diseases [2016]: \(799-806\) ) describe a sample of patients admitted to a hospital after suffering a stroke. Each of 537 patients was classified according to a measure of risk of malnutrition (with possible categories low, medium, and high) and whether or not the patient was alive at 6 months following the stroke. The authors concluded that there was an association between survival and the risk of malnutrition. Do you agree? Support your answer with evidence based on that the given data. For purposes of this exercise, you may assume that the sample of 537 patients is representative of stroke patients.

Give an example of a situation where it would be appropriate to use a chi- square test of homogeneity. Describe the populations that would be sampled and the variable that would be recorded.

The authors of the paper "Talking Smack: Verbal Aggression in Professional Wrestling" (Communication Studies [2008]: 242-258) analyzed the content of 36 hours of televised professional wrestling. Each instance of verbal aggression was classified according to the type of aggression, resulting in the frequency table at the bottom of the page. Assume that this sample of 804 verbal aggression incidents is representative of all verbal aggression incidents in televised professional wrestling. Carry out a hypothesis test to determine if there is convincing evidence that the proportions of verbal aggression incidents are not the same for all five types of aggression. Use a significance level of 0.01 .

Explain the difference between situations that would lead to a chi-square test for homogeneity and those that would lead to a chi-square test for independence.

The accompanying data on degree of spirituality for a random sample of natural scientists and a random sample of social scientists working at research universities appeared in the paper "Conflict Between Religion and Science Among Academic Scientists" (Journal for the Scientific Study of Religion [2009]: 276-292). Is there evidence that the spirituality category proportions are not the same for natural and social scientists? Test the relevant hypotheses using a significance level of \(\alpha=0.01\)

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