/*! 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 54 Refer to the following setting. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ The degrees of freedom for the chi-square test for this two-way table are (a) 4 . (c) 10 (e) 4876 . (b) 8 . (d) 20

Short Answer

Expert verified
The degrees of freedom are 4.

Step by step solution

01

Identify the Table Dimensions

First, record the number of categories for both variables in the table. In this case, there are 5 response categories (Almost no chance, Some chance, A 50-50 chance, A good chance, Almost certain) and 2 genders (Female and Male).
02

Use the Formula for Degrees of Freedom

The formula for calculating degrees of freedom in a chi-square test for a two-way table is \((r-1) \times (c-1)\), where \(r\) is the number of rows and \(c\) is the number of columns in the table.
03

Apply the Numbers to the Formula

Substitute \(r = 5\) (response categories) and \(c = 2\) (genders) into the formula: \((5-1) \times (2-1) = 4\times 1 = 4\).
04

Select the Correct Answer

Therefore, the degrees of freedom for the chi-square test in this two-way table is 4.

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Key Concepts

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

Degrees of Freedom
When conducting a chi-square test, the degrees of freedom play a crucial role in determining the test's validity and its result. Degrees of freedom, often abbreviated as dof, is like the number of values in your test that are free to vary. In a two-way table, it helps us understand how much variation is allowed when we are comparing observed data with expected data. In general terms, you can think of degrees of freedom as the number of choices you have minus the constraints placed on those choices. In the case of the chi-square test for a two-way table, it's important for accurately assessing the significance of the observed data patterns.For a two-way table with given rows and columns, the formula to calculate degrees of freedom is \[(r-1) \times (c-1)\] where \(r\) represents the number of response categories (rows) and \(c\) the number of subject characteristics or factors (columns). In simpler terms:
  • If there are more categories to compare, you'll need more freedom to see patterns.
  • The computed product provides a measure of how the separation of data could change if the data were different.
Thus, knowing the degrees of freedom gives you a better foundation on which to base your statistical conclusions.
Two-way Table
A two-way table is a handy tool used to explore the relationship between two categorical variables. It helps us visualize data by showing frequencies as counts across two facets. Imagine it as a grid where rows and columns intersect to highlight categorical data points. Each cell within the table displays information about frequencies or counts for each particular combination of categories. Here's how a typical two-way table setup looks:
  • Rows: These often represent each category or response option from one variable, such as survey responses.
  • Columns: These represent the different categories or groups of the other variable, like gender or grade level.
Using a two-way table, researchers can effortlessly identify patterns or unexpected results at a glance, such as whether one category consistently appears more than others. This makes it a powerful initial tool for any statistical analysis process.
Statistical Analysis
Statistical analysis allows us to interpret and draw conclusions from our data, lending structure and meaning to numerical observations. The chi-square test, often used with two-way tables, serves as a prime method in statistical analysis for understanding if there is a significant association between categorical variables. The basic principle involves comparing what is observed in the data to what would be expected if there were no relationship between the variables:
  • Observed Values: These are the actual results or frequencies obtained from the collected data.
  • Expected Values: These are the frequencies we would expect if the variables were truly independent of each other.
By assessing the differences between these observed and expected values, the chi-square test calculates a "chi-square statistic." This value can be referred to a chi-square distribution specific to its degrees of freedom to derive meaning about the relationship. A significant result may suggest that the association between your variables is not by random chance, but rather, there might be an inherent relationship worth exploring further.
Adolescent Health Study
The National Longitudinal Study of Adolescent Health provides rich data regarding the lives and development of adolescents. In this study, data is collected periodically to understand behavior, social relationships, and overall health trends among teens. Such studies are crucial as they offer insights into the social and health-related questions that play a role during adolescence, a pivotal time in human development:
  • It asks questions such as: What factors influence adolescent choices?
  • How do social environments impact adolescent decisions and health?
In the context of a two-way table and chi-square test, this study collected responses from teens about their perceptions of marriage in the next decade. By examining how responses vary between genders, researchers can uncover potential trends or differences in expectations or cultural norms. These findings open doors to discussions on how societal pressures and interpretations of adulthood vary across groups, further enhancing our understanding of adolescent development.

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Most popular questions from this chapter

Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.

Exercises 23 through 25 refer to the following setting. Do students who read more books for pleasure tend to earn higher grades in English? The boxplots below show data from a simple random sample of 79 students at a large high school. Students were classified as light readers if they read fewer than 3 books for pleasure per year. Otherwise, they were classified as heavy readers. Each student's average English grade for the previous two marking periods was converted to a GPA scale where \(A+=4.3\), \(A=4.0, A-=3.7, B+=3.3,\) and so on. Reading and grades (1.3) Write a few sentences comparing the distributions of English grades for light and heavy readers.

Birds in the trees Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, \(54 \%\) of the trees were Douglas firs, \(40 \%\) were ponderosa pines, and \(6 \%\) were other types of trees. At a randomly selected time during the day, the researchers observed 156 red-breasted nuthatches: 70 were seen in Douglas firs, 79 in ponderosa pines, and 7 in other types of trees. \(^{2}\) Do these data provide convincing evidence that nuthatches prefer particular types of trees when they're searching for seeds and insects?

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

Refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ The \(P\) -value for a chi-square test for goodness of fit is \(0.0129 .\) Which of the following is the most appropriate conclusion? (a) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is convincing evidence that the food choices are equally popular. (b) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is not convincing evidence that the food choices are equally popular. (c) Because 0.0129 is less than \(\alpha=0.05\), reject \(H_{0}\). There is convincing evidence that the food choices are not equally popular. (d) Because 0.0129 is less than \(\alpha=0.05,\) fail to reject \(H_{0}\). There is not convincing evidence that the food choices are equally popular. (e) Because 0.0129 is less than \(\alpha=0.05\), fail to reject \(H_{0}\). There is convincing evidence that the food choices are equally popular.

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