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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 appropriate null hypothesis for performing a chi-square test is that (a) equal proportions of female and male teenagers are almost certain they will be married in 10 years. (b) there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers' opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population.

Short Answer

Expert verified
The correct answer is (e).

Step by step solution

01

Understand the Chi-Square Test Objective

The chi-square test of independence is used to determine if there is an association between two categorical variables. In this case, the variables are gender and opinion about future marriage.
02

Identify the Null and Alternative Hypotheses

The null hypothesis ( H_0 ) generally states that there is no association between the variables being studied. Based on the context of the chi-square test, we can identify the null hypothesis related to gender and marriage opinion in the population.
03

Interpret the Choices

Each choice provides a different interpretation of what the null hypothesis could be. We need to select the one that reflects no association between the variables in the context of the population as the chi-square test typically concerns population-level inferences.
04

Determine the Correct Answer

The correct null hypothesis that matches the scenario of a chi-square test for independence is option (e): "there is no association between gender and opinion about marriage in the population." This states that gender and opinion on marriage are independent in the population, which aligns with the objective of the chi-square test.

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Key Concepts

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

Categorical Variables
Categorical variables are variables that classify data into distinct categories or groups. These are typically qualitative in nature, rather than numerical. For example, in a survey about opinions on marriage, gender is a categorical variable, with categories like 'male' and 'female'. Another categorical variable in this context is the opinion on the likelihood of getting married, with categories like 'almost no chance', 'some chance', 'a 50-50 chance', 'a good chance', and 'almost certain'.

Understanding categorical variables is key because they allow us to segment data in ways that can reveal patterns and relationships between these segments. For instance, by examining responses separately for males and females, researchers can investigate if there's a gender-based difference in marriage expectations.
Null Hypothesis
In statistics, a null hypothesis is a kind of default position that indicates no association between the variables under study. It assumes that any kind of difference or significance seen in the data is due to chance. In the context of chi-square tests for independence, a typical null hypothesis would state that there is no association between the categorical variables being examined.

For instance, when considering the question of whether gender influences opinions about marriage, the null hypothesis would be that gender and opinion are independent, meaning one's gender does not affect or change one's opinion on marriage likelihood. This provides a starting point for statistical analysis, as the chi-square test will determine whether there is enough evidence to reject this assumption.
Statistical Independence
Statistical independence refers to a scenario where the occurrence of one event does not affect the probability of another event happening. In the context of a chi-square test for independence, two categorical variables are said to be statistically independent if the distribution of one variable is the same across the levels of the other variable.

In our example, we're testing whether gender and opinion about future marriage are independent. If they are, knowing someone's gender wouldn’t provide any additional information about their opinion on the likelihood of marriage. The chi-square test helps us assess this by comparing observed frequencies in a contingency table to expected frequencies, assuming independence. If observed and expected frequencies differ significantly, it suggests a possible association.
Population Inference
Population inference refers to the process of using data from a sample to make generalizations about a larger population. In statistical tests like the chi-square test, we often analyze data from a sample and then try to infer if the patterns observed apply to the larger group as a whole.

For example, by using responses from 4877 teens regarding marriage expectations, researchers can infer whether these patterns likely hold true for all teens in the same age range. The null hypothesis often frames these inferences; in this context, it would suggest that any observed association between gender and marriage likelihood in the sample is likely to apply to the larger population as well. The chi-square test helps validate whether we can confidently make such an inference or whether the pattern seen might be the result of sampling variability.

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Most popular questions from this chapter

No chi-square A school's principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night: $$ \begin{array}{lccccccc} \hline \text { Night: } & \text { Sunday } & \text { Monday } & \text { Tuesday } & \text { Wednesday } & \text { Thursday } & \text { Friday } & \text { Saturday } \\ \text { Average } & 130 & 108 & 115 & 104 & 99 & 37 & 62 \\ \text { time: } & & & & & & & \\ \hline \end{array} $$ Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

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 (10.2) Summary statistics for the two groups from Minitab are provided below. $$ \begin{array}{cccc} \text { Type of reader }\quad\mathrm{N} & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \text {Heavy}\quad47 & 3.640 & 0.324 & 0.047 \\ \text {Light}\quad 32 & 3.356 & 0.380 & 0.067 \end{array} $$ (a) Explain why it is acceptable to use two-sample \(t\) procedures in this setting. (b) Construct and interpret a \(95 \%\) confidence interval for the difference in the mean English grade for light and heavy readers. (c) Does the interval in part (b) provide convincing evidence that reading more causes a difference in students' English grades? Justify your answer.

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: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ (a) \(\frac{(18-25)^{2}}{25}+\frac{(22-25)^{2}}{25}+\frac{(39-25)^{2}}{25}+\frac{(21-25)^{2}}{25}\) (b) \(\frac{(25-18)^{2}}{18}+\frac{(25-22)^{2}}{22}+\frac{(25-39)^{2}}{39}+\frac{(25-21)^{2}}{21}\) (c) \(\frac{(18-25)}{25}+\frac{(22-25)}{25}+\frac{(39-25)}{25}+\frac{(21-25)}{25}\) (d) \(\frac{(18-25)^{2}}{100}+\frac{(22-25)^{2}}{100}+\frac{(39-25)^{2}}{100}+\frac{(21-25)^{2}}{100}\) (e) \(\frac{(0.18-0.25)^{2}}{0.25}+\frac{(0.22-0.25)^{2}}{0.25}+\frac{(0.39-0.25)^{2}}{0.25}\) \(+\frac{(0.21-0.25)^{2}}{0.25}\) The chi-square statistic is

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

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