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Chapter 11: Problem 19

Explain how the expected frequencies for cells of a contingency table are calculated in a test of independence or homogeneity.

Short Answer

Expert verified
Expected frequencies in a contingency table are calculated by multiplying the row total for each cell by its column total, then dividing by the total number of observations. This is vital for the Chi-square test of independence or homogeneity which sees if observed frequencies significantly vary from expected frequencies.

Step by step solution

01

Understanding Contingency Tables

In statistics, a contingency table is a type of table in a matrix format that displays the multivariate frequency distribution of variables. They provide a basic picture of the interrelation between two variables and can help find interactions between them.
02

Identifying Total Values, Row & Column Totals

First, identify the total value (N), which is the total number of observations. This is found by summing all cell frequencies in the table. You also need to know the row and column totals.
03

Calculating Expected Frequencies

To calculate the expected frequency for any cell in the table, multiply the row total for that cell by the column total for that cell, then divide by the total number of observations. Repeat this process for each cell in the table.
04

Applying to a Test of Independence or Homogeneity

The expected frequencies are essential in a Chi-square test of independence or homogeneity. If the observed frequencies differ significantly from the expected frequencies, you would reject the null hypothesis of independence or homogeneity.

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Key Concepts

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

Expected Frequencies
Expected frequencies are a crucial part of analyzing data in a contingency table. They help us determine what we would expect the data to look like if two variables were independent of each other. To calculate them, follow a simple formula:
  • Identify the total number of observations in the table, denoted as \( N \).
  • For any cell, multiply the total of the row containing that cell by the total of the column.
  • Finally, divide this product by \( N \) to get the expected frequency for that specific cell.
This calculation ensures that expected frequencies account for both the row and column distributions, giving us a fair benchmark to compare our observed data against.
Chi-square Test
The Chi-square test is a statistical method used to check the association between two categorical variables in a contingency table. It evaluates whether the observed frequency distribution differs significantly from the expected distribution. When performing the test, the major steps include:
  • Calculate the expected frequencies for all cells in the table.
  • For each cell, find the difference between observed and expected frequencies.
  • Square these differences, then divide by the expected frequency for each cell.
  • Sum all these values to obtain the Chi-square statistic.
The result of this test helps determine if the association between variables is statistically significant, allowing researchers to infer dependencies or independencies.
Test of Independence
A test of independence looks to explore whether two variables in a contingency table are truly independent. The null hypothesis assumes that the distribution of one variable is independent of the other. To conduct this test:
  • Set up the null and alternative hypotheses. The null usually states independence between variables.
  • Calculate the Chi-square statistic using expected and observed frequencies.
  • Compare this statistic to a critical value from the Chi-square distribution table.
If your calculated statistic is higher than the table value for the chosen significance level, you reject the null hypothesis, suggesting a possible relationship between the variables.
Test of Homogeneity
The test of homogeneity is used to compare the distribution of a categorical variable across different populations. It's quite similar to the test of independence, but with a key difference in focus: rather than checking if variables are related, it examines if two or more populations are identical in distribution.
  • Formulate the null hypothesis stating that population distributions are homogeneous.
  • Use the same Chi-square method to calculate the statistic based on expected and observed data.
  • Comparison of this statistic against a critical value will suggest whether population distributions deviate significantly.
This testing method is important in research fields like biology and social sciences where comparisons across different groups can inform major conclusions.

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Most popular questions from this chapter

A random sample of 100 persons was selected from each of four regions in the United States. These people were asked whether or not they support a certain farm subsidy program. The results of the survey are summarized in the following table. $$ \begin{array}{lccc} \hline & \text { Favor } & \text { Oppose } & \text { Uncertain } \\ \hline \text { Northeast } & 56 & 33 & 11 \\ \text { Midwest } & 73 & 23 & 4 \\ \text { South } & 67 & 28 & 5 \\ \text { West } & 59 & 35 & 6 \\ \hline \end{array} $$ Using a \(1 \%\) significance level, test the null hypothesis that the percentages of people with different opinions are similar for all four regions.

You have collected data on a variable, and you want to determine if a normal distribution is a reasonable model for these data. The following table shows how many of the values fall within certain ranges of \(z\) values for these data. $$ \begin{array}{lr} \hline \text { Category } & \text { Count } \\ \hline z \text { score below }-2 & 48 \\ z \text { score from }-2 \text { to less than }-1.5 & 67 \\ z \text { score from }-1.5 \text { to less than }-1 & 146 \\ z \text { score from }-1 \text { to less than }-0.5 & 248 \\ z \text { score from }-0.5 \text { to less than } 0 & 187 \\ z \text { score from } 0 \text { to less than } 0.5 & 125 \\ z \text { score from } 0.5 \text { to less than } 1 & 88 \\ z \text { score from } 1 \text { to less than } 1.5 & 47 \\ z \text { score from } 1.5 \text { to less than } 2 & 25 \\ z \text { score of } 2 \text { or above } & 19 \\ \hline \text { Total } & 1000 \\ \hline \end{array} $$ Perform a hypothesis test to determine if a normal distribution is an appropriate model for these data. Use a significance level of \(5 \%\).

In a Harris Poll conducted October \(15-20,2014\), American adults were asked "to think ahead 2 to 5 years and assess if they feel solar energy will contribute to meeting our energy needs." Of the respondents, \(31 \%\) said solar energy will make a major contribution to meeting our energy needs within the next 2 to 5 years, \(53 \%\) felt it will make a minor contribution, and \(16 \%\) expected that it will make hardly any contribution at all (www.harrisinteractive.com). Assume that these results are true for the 2014 population of adults. Recently a random sample of 2000 American adults was selected and these adults were asked the same question. The results of the poll are presented in the following table. $$ \begin{array}{l|ccc} \hline \text { Response } & \begin{array}{c} \text { Major } \\ \text { Contribution } \end{array} & \begin{array}{c} \text { Minor } \\ \text { Contribution } \end{array} & \begin{array}{c} \text { Hardly Any } \\ \text { Contribution } \end{array} \\ \hline \text { Frequency } & 820 & 920 & 260 \\ \hline \end{array} $$ Test at a \(2.5 \%\) significance level whether the current distribution of opinions to the said question is significantly different from that for the 2014 opinions.

The following table gives the distributions of grades for three professors for a few randomly selected classes that each of them taught during the last 2 years. $$ \begin{array}{l|lccc} \hline & & \multicolumn{3}{c} {\text { Professor }} \\ \cline { 3 - 5 } & & \text { Miller } & \text { Smith } & \text { Moore } \\ \hline \multirow{4}{*} {\text { Grade }} & \text { A } & 18 & 36 & 20 \\ & \text { B } & 25 & 44 & 15 \\ & \text { C } & 85 & 73 & 82 \\ & \text { D and F } & 17 & 12 & 8 \\ \hline \end{array} $$ Using a \(2.5 \%\) significance level, test the null hypothesis that the grade distributions are homogeneous for these three professors.

The following table lists the frequency distribution for 60 rolls of a die. $$ \begin{array}{l|cccccc} \hline \text { Outcome } & 1 \text { -spot } & 2 \text { -spot } & \text { 3-spot } & 4 \text { -spot } & \text { 5-spot } & \text { 6-spot } \\ \hline \text { Frequency } & 7 & 12 & 8 & 15 & 11 & 7 \\ \hline \end{array} $$ Test at a \(5 \%\) significance level whether the null hypothesis that the given die is fair is true.
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