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The article “Regional Differences in Attitudes Toward Corporal Punishment" (Journal of Marriage and Family [1994]: 314-324) presents data resulting from a random sample of 978 adults. Each individual in the sample was asked whether he or she agreed with the following statement: "Sometimes it is necessary to discipline a child with a good, hard spanking." Respondents were also classified according to the region of the United States in which they lived. The resulting data are summarized in the accompanying table. Is there an association between response (agree, disagree) and region of residence? Use \(\alpha=.01\). $$ \begin{array}{l|cc} & \text { Response } \\ \hline \text { Region } & \text { Agree } & \text { Disagree } \\ \hline \text { Northeast } & 130 & 59 \\ \text { West } & 146 & 42 \\ \text { Midwest } & 211 & 52 \\ \text { South } & 291 & 47 \\ \hline \end{array} $$

Short Answer

Expert verified
After performing the Chi-Square Test for Independence, if the Chi-Square Test Statistic is greater than the critical value, then we reject the null hypothesis and conclude that there is an association between response and region of residence, otherwise, we don't reject the null hypothesis.

Step by step solution

01

State the Hypotheses

The null hypothesis, \(H_0\), is that there is no association between response and region of residence. The alternative hypothesis, \(H_A\), is that there is an association between response and region of residence.
02

Calculate Expected Frequencies

Each expected frequency can be calculated as (row total × column total) / grand total. For instance, the expected frequency for Northeast Agrees is \((130+59) \times (130+146+211+291) / 978 = 144.4\). In similar way calculate expected frequencies for others.
03

Compute the Test Statistic

The Chi-Square Test Statistic has the formula: \(\chi^2 = \sum (O-E)^2 / E\) where O represents the observed frequency and E represents the expected frequency. The Chi-Square Statistic for each cell is computed then added up to get the total Chi-square.
04

Make Decision

Find the critical value of \(\chi^2\) from the table of \(\chi^2\) distribution for \(0.01\) significance level with 3 degrees of freedom. Compare the test statistic with the critical value. If the test statistic is greater than the critical value, reject the null hypothesis.

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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 decide whether there is enough evidence in a sample to infer that a certain condition holds for the entire population. In the context of the Chi-Square test, it helps us determine the presence of an association between categorical variables. For the given exercise, the two categorical variables are "region of residence" and "response toward spanking."

The process begins with formulating two hypotheses. The null hypothesis (\( H_0 \) ) states that there is no association between the variables, implying that any observed relationship might just be due to random chance. On the other hand, the alternative hypothesis (\( H_A \) ) suggests there is a genuine association between the variables.

In hypothesis testing, your goal is to evaluate whether the sample data provides enough evidence to reject the null hypothesis. The decision is based on the calculated test statistic and its comparison with a critical value from the statistical distribution. If the calculated statistic is greater than the critical value, the null hypothesis is rejected, providing evidence in favor of the alternative hypothesis.
Statistical Significance
Statistical significance is a key concept used to interpret the results of a hypothesis test. It determines whether the observed effect in your sample data is likely to be true for the entire population, given a specified significance level (\( \alpha \) ). This significance level reflects the probability of rejecting the null hypothesis when it is indeed true, also known as the Type I error rate.

In this exercise, \( \alpha \) is set at 0.01, which means you are willing to accept a 1% chance of incorrectly concluding that there is an association between region and response when in fact, there isn't one. This is a stringent level, providing strong evidence if the null hypothesis is rejected.

Statistical significance does not measure the size or importance of the effect, just the likelihood that any observed effect is not due to sampling variation. Thus, it’s essential to interpret the size and practical impact of the association alongside the statistical significance.
Expected Frequencies
Expected frequencies are crucial in a Chi-Square test as they represent the frequencies you would expect if the null hypothesis were true. They serve as a baseline against which the observed frequencies can be compared. Calculating expected frequencies involves using the proportions of the total sample that belong to each category.

The formula for calculating an expected frequency is \( (\text{row total} \times \text{column total}) / \text{grand total} \). For the Northeast region agreeing with the statement, the expected frequency is \( \frac{( ext{total agree} \times ext{Northeast total})}{ ext{grand total}} \).

Comparing these expected frequencies against the observed frequencies allows you to see where and how much they differ. Large discrepancies contribute to a higher Chi-Square statistic, which indicates that the observed distribution is unlikely under the null hypothesis. Thus, calculating expected frequencies is a foundational step in assessing whether the sample data displays a significant association.
Degree of Freedom
Degrees of freedom (df) refer to the number of values in the final calculation of a statistic that are free to vary. In the Chi-Square test, they are critical for interpreting the test statistic by determining which distribution to use for significance testing.

The degrees of freedom in a Chi-Square test for independence are calculated as \( (r-1)(c-1) \), where \( r \) is the number of rows and \( c \) is the number of columns in your contingency table. For this exercise, with 4 regions and 2 responses, the degrees of freedom are \( (4-1)(2-1) = 3 \).

Knowing the degrees of freedom helps you find the critical value in a Chi-Square distribution table. It's this value against which you compare your calculated statistic to make a decision about the null hypothesis. A high or low degree of freedom can influence the variability of the data and thus the outcome of the test.

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Most popular questions from this chapter

Packages of mixed nuts made by a certain company contain four types of nuts. The percentages of nuts of Types \(1,2,3\), and 4 are supposed to be \(40 \%, 30 \%\), \(20 \%\), and \(10 \%\), respectively. A random sample of nuts is selected, and each one is categorized by type. a. If the sample size is 200 and the resulting test statistic value is \(X^{2}=19.0\), what conclusion would be appropriate for a significance level of .001? b. If the random sample had consisted of only 40 nuts, would you use the chi- square test here? Explain your reasoning.

The press release titled "Nap Time" (pewresearch.org, July 2009 ) described results from a nationally representative survey of 1488 adult Americans. The survey asked several demographic questions (such as gender, age, and income) and also included a question asking respondents if they had taken a nap in the past 24 hours. The press release stated that \(38 \%\) of the men surveyed and \(31 \%\) of the women surveyed reported that they had napped in the past 24 hours. For purposes of this exercise, suppose that men and women were equally represented in the sample. a. Use the given information to fill in observed cell counts for the following table: $$ \begin{array}{l|ll} & \text { Napped } & \text { Did Not Nap } & \text { Row Total } \\ \hline \text { Men } & & & 744 \\ \text { Women } & & & 744 \\ \hline \end{array} $$ b. Use the data in the table from Part (a) to carry out a hypothesis test to determine if there is an association between gender and napping. c. The press release states that more men than women nap. Although this is true for the people in the sample, based on the result of your test in Part (b), is it reasonable to conclude that this holds for adult Americans in general? Explain.

The article "Cooperative Hunting in Lions: The Role of the Individual" (Behavioral Ecology and Sociobiology [1992]: \(445-454\) ) discusses the different roles taken by lionesses as they attack and capture prey. The authors were interested in the effect of the position in line as stalking occurs; an individual lioness may be in the center of the line or on the wing (end of the line) as they advance toward their prey. In addition to position, the role of the lioness was also considered. A lioness could initiate a chase (be the first one to charge the prey), or she could participate and join the chase after it has been initiated. Data from the article are summarized in the accompanying table. $$ \begin{array}{l|cc} & \text { Role } \\ \hline \text { Position } & \text { Initiate Chase } & \text { Participate in Chase } \\ \hline \text { Center } & 28 & 48 \\ \text { Wing } & 66 & 41 \\ \hline \end{array} $$ Is there evidence of an association between position and role? Test the relevant hypotheses using \(\alpha=.01\). What assumptions about how the data were collected must be true for the chi-square test to be an appropriate way to analyze these data?

The report "Fatality Facts 2004: Bicydes" (Insurance Institute, 2004) included the following table classifying 715 fatal bicycle accidents according to time of day the accident occurred. $$ \begin{array}{lc} \text { Time of Day } & \text { Number of Accidents } \\ \hline \text { Midnight to } 3 \text { A.M. } & 38 \\ \text { 3 A.M. to } 6 \text { A.M. } & 29 \\ \text { 6 A.M. to } 9 \text { A.M. } & 66 \\ \text { 9A.M. to Noon } & 77 \\ \text { Noon to } 3 \text { P.M. } & 99 \\ \text { 3 P.M. to } 6 \text { r.M. } & 127 \\ \text { 6 P.M. to 9 p.M. } & 166 \\ \text { 9 P.M. to Midnight } & 113 \\ \hline \end{array} $$ a. Assume it is reasonable to regard the 715 bicycle accidents summarized in the table as a random sample of fatal bicycle accidents in 2004 . Do these data support the hypothesis that fatal bicycle accidents are not equally likely to occur in each of the 3 -hour time periods used to construct the table? Test the relevant hypotheses using a significance level of \(.05\). b. Suppose a safety office proposes that bicycle fatalities are twice as likely to occur between noon and midnight as during midnight to noon and suggests the following hypothesis: \(H_{0}: p_{1}=1 / 3, p_{2}=2 / 3\), where \(p_{1}\) is the proportion of accidents occurring between midnight and noon and \(p_{2}\) is the proportion occurring between noon and midnight. Do the given data provide evidence against this hypothesis, or are the data consistent with it? Justify your answer with an appropriate test. (Hint: Use the data to construct a one-way table with just two time categories.)

The article "In Bronx, Hitting Home Runs Is A Breeze" (USA Today, June 2, 2009) included a classification of 87 home runs hit at the new Yankee Stadium according to direction that the ball was hit, resulting in the accompanying data. $$ \begin{array}{l|ccccc} \hline \text { Direction } & \begin{array}{l} \text { Left } \\ \text { Field } \end{array} & \begin{array}{l} \text { Left } \\ \text { Center } \end{array} & \text { Center } & \begin{array}{l} \text { Right } \\ \text { Center } \end{array} & \begin{array}{l} \text { Right } \\ \text { Field } \end{array} \\ \hline \begin{array}{c} \text { Number of } \\ \text { Home Runs } \end{array} & 18 & 10 & 7 & 18 & 34 \\ \hline \end{array} $$ a. Assuming that it is reasonable to regard this sample of 87 home runs as representative of home runs hit at Yankee Stadium, carry out a hypothesis test to determine if there is convincing evidence that the proportion of home runs hit is not the same for all five directions. b. Write a few sentences describing how the observed counts for the five directions differ from what would have been expected if the proportion of home runs is the same for all five directions.

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