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91Ó°ÊÓ

When is chi-squared not valid? Give an example of a contingency table for which the chi-squared test of independence should not be used.

Short Answer

Expert verified
Chi-squared is not valid if any expected frequency is less than 5, as in the provided example table.

Step by step solution

01

Understand Chi-Squared Validity

The chi-squared test of independence is sensitive to sample size and expected frequencies. It is generally not valid if the expected frequency in any cell of the contingency table is less than 5, as this may lead to inaccurate results.
02

Create a Contingency Table

Let's create a contingency table where some cells have expected frequencies less than 5. Assume we have a table for favorite fruit preference (Apples, Bananas) among two groups (Group A, Group B) with this observed data: | | Apples | Bananas | |----------|--------|---------| | Group A | 2 | 3 | | Group B | 1 | 4 |
03

Calculate Expected Frequencies

To calculate expected frequencies, use the formula: \[ E_{ij} = \frac{(\text{Row Total}_i) \times (\text{Column Total}_j)}{\text{Grand Total}} \]For Apples and Group A, \[ E_{11} = \frac{(5) \times (3)}{10} = 1.5 \]Here, the expected frequency of 1.5 for Apples in Group A is less than 5.
04

Identify Invalidation

Since one of the expected frequencies here (1.5) is less than 5, this violates a chi-squared test assumption making it invalid to use the test for this dataset.

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Key Concepts

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

Contingency Table
A contingency table is a tabular representation that displays the frequency distribution of certain variables. Typically, these tables are used in statistics to show the relationship between two categorical variables. Rows often represent one variable and columns another, with the table cells filled with frequency counts.

### Understanding Contingency Tables
To better grasp contingency tables, let's take an example involving fruit preference. Suppose we want to explore how different groups, say Group A and Group B, choose between Apples and Bananas. The observed frequencies might look something like this:
  • Group A prefers Apples: 2 times
  • Group A prefers Bananas: 3 times
  • Group B prefers Apples: 1 time
  • Group B prefers Bananas: 4 times
This table organizes our data and helps us see if there's an apparent preference pattern or difference between the groups.

Such contingency tables are crucial for applying chi-squared tests, which help determine if the observed relationship is due to chance or an actual association.
Expected Frequency
Expected frequency is a fundamental concept in hypothesis testing using chi-squared tests. It involves determining the expected count of occurrences in each cell of a contingency table if there was no association between the variables.

### Calculating Expected Frequencies
The formula to calculate expected frequency (for cell at the intersection of row i and column j) is:\[E_{ij} = \frac{(\text{Row Total}_i) \times (\text{Column Total}_j)}{\text{Grand Total}}\]

Continuing with our fruit preference example, let's calculate the expected frequency for Group A's preference for Apples. If:
  • Row Total for Group A is 5
  • Column Total for Apples is 3
  • Grand Total of all observations is 10
Then, the expected frequency for Apples and Group A would be:\[E_{11} = \frac{(5) \times (3)}{10} = 1.5\]
This result means that, under the assumption of independence, we would expect Group A to choose Apples 1.5 times. If expected frequencies are too low (less than 5), the chi-squared test results can become unreliable.
Hypothesis Testing
Hypothesis testing is a methodical approach used to make statistical decisions using experimental data. In the context of a chi-squared test, it’s used to examine if there is an association between the categorical variables in a contingency table.

### Steps in Hypothesis Testing with Chi-Squared
Here's how you typically conduct a chi-squared test:
  • Define Null Hypothesis ( H_0 ): There is no association between the variables.
  • Define Alternative Hypothesis ( H_A ): There is an association between the variables.
  • Calculate the Chi-Squared Statistic: This involves summing up the squared difference between observed and expected frequencies, divided by expected frequencies, for each table cell.
  • Determine the p-value: Compare the calculated statistic against the chi-squared distribution to find the p-value.
  • Make a Decision: If the p-value is less than the significance level (commonly 0.05), reject the null hypothesis in favor of the alternative hypothesis. Otherwise, do not reject the null hypothesis.

In our example, if one of the expected frequencies is less than 5, the test might not be valid. This means it wouldn’t be appropriate to draw conclusions about the association based on this test. Hypothesis testing requires careful consideration of assumptions to ensure findings are reliable.

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Most popular questions from this chapter

Women's role A recent GSS presented the statement, "Women should take care of running their homes and leave running the country up to men," and \(14.8 \%\) of the male respondents agreed. Of the female respondents, \(15.9 \%\) agreed. Of respondents having less than a high school education, \(39.0 \%\) agreed. Of respondents having at least a high school education, \(11.7 \%\) agreed. a. Report the difference between the proportion of males and the proportion of females who agree. b. Report the difference between the proportion at the low education level and the proportion at the high education level who agree. c. Which variable, gender or educational level, seems to have the stronger association with opinion? Explain your reasoning.

Standardized residuals for \(2 \times 2\) tables The table that follows shows the standardized residuals in parentheses for GSS data about the statement, "Women should take care of running their homes and leave running the country up to men." The absolute value of the standardized residual is 13.2 in every cell. For chi-squared tests with \(2 \times 2\) tables, since \(d f=1,\) only one nonredundant piece of information exists about whether an association exists. If observed count \(>\) expected count in one cell, observed count \(<\) expected count in the other cell in that row or column. Explain why this is true, using the fact that observed and expected counts have the same row and column totals. (In fact, in \(2 \times 2\) tables, all four standardized residuals have absolute value equal to the square root of the \(X^{2}\) test statistic.)

Life after death and gender In the \(2008 \mathrm{GSS}, 620\) of 809 males and 835 of 978 females indicated a belief in life after death. (Source: Data from CSM, UC Berkeley.) a. Construct a \(2 \times 2\) contingency table relating gender of respondent (SEX, categories male and female) as the rows to belief about life after death (POSTLIFE, categories yes and no) as the columns. b. Find the four expected cell counts for the chisquared test. Compare them to the observed cell counts, identifying cells having more observations than expected. c. The data have \(X^{2}=22.36 .\) Set up its calculation by showing how to substitute the observed and expected cell counts you found into its formula.

Happiness and sex \(\quad\) A contingency table from the 2008 GSS relating happiness to number of sex partners in the previous year \((0,1,\) at least 2\()\) had standardized residuals as shown in the table. Interpret the highlighted standardized residuals. $$ \begin{aligned} &\text { Results on Happiness and Sex }\\\ &\begin{array}{lccc} \text { Rows: } & \text { partners } & \text { Columns: happy } & \\ & \text { not } & \text { pretty } & \text { very } \\ 0 & 84 & 235 & 95 \\ & (3.1) & (0.7) & (-3.3) \\ 1 & 130 & 578 & 381 \\ & (-5.2) & (-2.3) & (6.6) \\ 2 & 58 & 160 & 41 \\ & (3.4) & (2.3) & (-5.2) \end{array} \end{aligned} $$

Prison and gender \(\quad\) According to the U.S. Department of Justice, in 2009 the incarceration rate in the nation's prisons was 949 per 100,000 male residents, and 67 per 100,000 female residents. a. Find the relative risk of being incarcerated, comparing males to females. Interpret. b. Find the difference of proportions of being incarcerated. Interpret. c. Which measure do you think is more appropriate for these data? Why?

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