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The paper "Overweight Among Low-Income Preschool Children Associated with the Consumption of Sweet Drinks" (Pediatrics [2005]: 223-229) described a study of children who were underweight or normal weight at age 2 . Children in the sample were classified according to the number of sweet drinks consumed per day and whether or not the child was overweight one year after the study began. Is there evidence of an association between whether or not children are overweight after one year and the number of sweet drinks consumed? Assume that the sample of children in this study is representative of 2 - to 3 -year-old children, Test the appropriate hypotheses using a 0.05 significance level.

Short Answer

Expert verified
The solution will be that the null hypothesis is either rejected or failed to be rejected based on the Chi-Square test of independence with the 0.05 significance level. However, without exact data, the definite outcome can't be stated.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (H0) will be: there is no association between whether or not children are overweight after one year and the number of sweet drinks consumed. The alternative hypothesis (H1) is: there is an association between whether or not children are overweight after one year and the number of sweet drinks consumed.
02

Identify a Test Statistic

Since we are trying to determine a relationship between two variables, a Chi-Square test of independence would be fitting. This means that the test statistic will be a Chi-Square statistic, determined by the data provided.
03

Determine the Significance Level

The significance level for this test has been given as 0.05. This will determine the rejection region of our hypotheses test.
04

Calculate the Test Statistic

To calculate the test statistic, use the data about the number of drinks consumed and the weight condition of children. Input these data into the Chi-Square formula, which includes observed and expected frequencies in its calculations.
05

Make a Decision

Compare the calculated Chi-Square test statistic with the Chi-Square critical value corresponding to the 0.05 significance level. If the calculated test statistic is greater than the critical value, reject the null hypothesis.
06

Interpret the Results

If the null hypothesis were rejected due to a large Chi-Square statistic, it would mean that there is evidence suggesting an association between the number of sweet drinks consumed and children being overweight after a year. If the null hypothesis is not rejected, there may not be enough evidence to support the association.

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Most popular questions from this chapter

Think about how you would answer the following question. Next Wednesday's meeting has been moved forward two days. What day is the meeting now that it has been rescheduled? This question is ambiguous, as "moved forward" can be interpreted in two different ways. Would you have answered Monday or Friday? The authors of the paper "Even Abstract Motion Influences the Understanding of Time" (Metaphor and Symbol [2011]\(: 260-271)\) wondered if the answers Monday and Friday would be provided an equal proportion of the time. Each student in a random sample of students at Stanford University was asked this question, and the responses are summarized in the following table. The authors of the paper used a chi-square goodness-of-fit test to test the null hypothesis \(H_{0}: p_{1}=0.50, p_{2}=0.50\), where \(p_{1}\) is the proportion who would respond Monday, and \(p_{2}\) is the proportion who would respond Friday. They reported \(X^{2}=11.00\) and \(P\) -value \(<0.001\). What conclusion can be drawn from this test?

Give an example of a situation where it would be appropriate to use a chi- square test of independence. Describe the population that would be sampled and the two variables that would be recorded.

What is the approximate \(P\) -value for the following values of \(X^{2}\) and \(\mathrm{df} ?\) a. \(X^{2}=6.62, \mathrm{df}=3\) b. \(X^{2}=16.97, \mathrm{df}=10\) c. \(X^{2}=30.19, \mathrm{df}=17\)

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.

A certain genetic characteristic of a particular plant can appear in one of three forms (phenotypes). A researcher has developed a theory, according to which the hypothesized proportions are \(p_{1}=0.25, p_{2}=0.50,\) and \(p_{3}=0.25 .\) A random sample of 200 plants yields \(X^{2}=4.63\). a. Carry out a test of the null hypothesis that the theory is correct, using level of significance \(\alpha=0.05\). b. Suppose that a random sample of 300 plants had resulted in the same value of \(X^{2}\). How would your analysis and conclusion differ from those in Part (a)?

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