Independence Test Using TI-84: Videos & Practice Problems

Using a TI-84 calculator simplifies performing chi-square tests of independence and homogeneity by entering data as matrices and utilizing the chi-square test function. The null hypothesis assumes variables are independent, while the alternative suggests dependence. The calculator computes expected values and the chi-square statistic, providing a p-value to compare against the alpha level (e.g., 0.05). A p-value less than alpha leads to rejecting the null hypothesis, indicating a significant association between categorical variables. This method streamlines hypothesis testing for categorical data, reinforcing concepts like contingency tables, expected values, and significance in statistical inference.

Independence Test Using TI-84

Performing a chi-square test of independence using a TI-84 calculator streamlines the process of determining whether two categorical variables are related. This test evaluates if the distribution of one variable is independent of the other by comparing observed frequencies to expected frequencies under the assumption of independence. The null hypothesis (H₀) states that the variables are independent, meaning the group a participant belongs to does not affect symptom improvement. Conversely, the alternative hypothesis (H_a) asserts that the variables are dependent, indicating a relationship between group membership and symptom improvement.

To conduct this test on a TI-84, data must first be entered as a matrix. Access the matrix menu by pressing the 2nd button followed by the x^-1 button, then navigate to the Edit tab to input your observed frequency data into Matrix A. Each cell in the matrix should correspond exactly to the values in your contingency table. After entering the data, press the STAT button, move to the TESTS menu, and select the chi-square test function (option C). Ensure that the observed matrix is set to Matrix A and designate a different matrix, such as Matrix B, to store the expected frequencies calculated by the calculator using the formula for expected counts:

\( E_{ij} = \frac{( \text{row total}_i ) ( \text{column total}_j )}{\text{grand total}} \)

After confirming these settings, select Calculate to obtain the test statistic and p-value. The p-value indicates the probability of observing the data assuming the null hypothesis is true. If the p-value is less than the significance level α (commonly 0.05), reject the null hypothesis, concluding that the variables are dependent. For example, a p-value of approximately \(1.7 \times 10^{-7}\) is significantly less than 0.05, providing strong evidence that symptom improvement depends on the participant group.

It is important to note that chi-square tests of homogeneity are closely related to independence tests. Both use the same computational steps and formulas, but differ slightly in hypothesis wording and interpretation. In homogeneity tests, the focus is on whether different populations have the same distribution of a categorical variable, whereas independence tests assess the relationship between two variables within a single population. When using the TI-84, the same matrix input and chi-square test function apply to both tests; only the hypotheses and conclusions are adjusted accordingly.

Mastering the use of matrices and the chi-square test function on the TI-84 enhances efficiency in statistical analysis, allowing for quick and accurate testing of relationships between categorical variables. This skill is essential for interpreting data in research settings, such as clinical trials, where understanding dependencies between factors can inform decision-making and further study design.

A student performs an independence test using technology to see if pet ownership is independent of relationship status. They get the following results: ${\chi }^2=0.545$ & $p=0.7614$. What can they conclude about pet ownership and relationship status?

• A. Reject $H_0$ since there is enough evidence to suggest that pet ownership and relationship status are dependent.
• B. Fail to reject $H_0$ since there is not enough evidence to suggest that pet ownership and relationship status are dependent.
• C. Reject $H_0$ since there is not enough evidence to suggest that pet ownership and relationship status are dependent.
• D. Fail to reject $H_0$ since there is enough evidence to suggest that pet ownership and relationship status are dependent.

Answer: (B)

Independence Test Using TI-84 Example 1

When testing whether customer satisfaction ratings are independent of the restaurant location, a chi-square test of independence is an effective statistical method. The null hypothesis assumes that the two categorical variables—location and customer satisfaction rating—are independent, meaning the satisfaction rating does not depend on which of the three locations a customer dined at. Conversely, the alternative hypothesis posits that these variables are dependent, indicating a relationship between location and satisfaction.

To perform this test, data must be organized into a contingency table, which can be input as a matrix in a statistical calculator such as the TI-84. Each cell in the matrix corresponds to the frequency count of customers for a specific satisfaction rating at a particular location. After entering the observed data matrix, the chi-square test function calculates the expected frequencies under the assumption of independence using the formula:

Expected frequency = \(\frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}\)

The test statistic is computed by summing over all cells the squared difference between observed and expected frequencies divided by the expected frequency:

\(\chi^2 = \sum \frac{(O - E)^2}{E}\)

where \(O\) represents observed frequencies and \(E\) represents expected frequencies. The resulting p-value indicates the probability of observing the data assuming the null hypothesis is true.

In this scenario, the p-value was approximately 0.99, which is significantly greater than the significance level \(\alpha = 0.05\). Since the p-value exceeds \(\alpha\), there is insufficient evidence to reject the null hypothesis. This means the data do not support a dependency between customer satisfaction and restaurant location, allowing the restaurant to maintain the assumption that satisfaction ratings are independent of location.

Understanding how to conduct and interpret a chi-square test of independence is crucial for analyzing relationships between categorical variables in real-world contexts, such as customer feedback across different service locations.

Independence Test Using TI-84 Example 2

When testing whether the proportion of books sold for each subject is consistent across different book types, a chi-square test for homogeneity is appropriate. This test evaluates if the distribution of categories (such as subjects like statistics, calculus, algebra, and other math books) is the same across different groups (digital, softcover, and hardcover books). The significance level, denoted by α, is set at 0.05 for this analysis.

The hypotheses for a homogeneity test are framed to compare proportions across groups. The null hypothesis (H₀) states that the proportions of subjects sold are equal across all book types, implying no difference in the distribution of subjects between digital, softcover, and hardcover books. The alternative hypothesis (H_a) asserts that at least one subject’s proportion differs among the book types, indicating a variation in sales distribution.

To perform the test, data is organized into a matrix representing observed frequencies of book sales by subject and type. Using a TI-84 calculator, the data is entered into Matrix A. The chi-square test function is then accessed through the calculator’s statistics menu, selecting the chi-square test option. The observed data matrix is input as Matrix A, and the calculator computes the expected frequencies, typically stored in Matrix B, based on the assumption that proportions are equal across groups.

The chi-square test statistic is calculated using the formula:

$$\chi ²=\sum _{i=1}^r\sum _{j=1}^c\frac{{\mathrm{(O_ij\; -\; E_ij)}}^{}}{2}\mathrm{E_ij}$$

where O_ij represents the observed frequency for the ith row and jth column, and E_ij is the expected frequency under the null hypothesis.

After calculation, the p-value is obtained, which quantifies the probability of observing the data assuming the null hypothesis is true. In this case, the p-value is approximately $7.8\times {10}^{-7}$, which is significantly less than the alpha level of 0.05. This leads to rejecting the null hypothesis, providing strong evidence that the proportions of subjects sold differ across book types.

In summary, the chi-square test for homogeneity effectively determines whether categorical proportions are consistent across different groups. By carefully entering observed data, calculating expected frequencies, and interpreting the p-value relative to the significance level, one can conclude whether sales distributions vary by book type. This method is essential for analyzing categorical data in contexts such as market research and product sales analysis.

Here’s what students ask on this topic:

How do you enter data as a matrix on the TI-84 for a chi-square independence test?

To enter data as a matrix on the TI-84 for a chi-square independence test, first press the 2nd button, then the x^-1 button to access the matrix menu. Scroll right to the Edit tab and select Matrix A (option 1). You will then input your observed frequency data into the matrix cells, making sure each value corresponds exactly to the cells in your contingency table. After entering all data, press 2nd then Mode to quit. This step is crucial because the chi-square test function uses this matrix as the observed data for calculations.

What are the steps to perform a chi-square test of independence on the TI-84?

To perform a chi-square test of independence on the TI-84, follow these steps: (1) Enter your observed data as a matrix (usually Matrix A). (2) Press the Stat button, scroll right to the Tests menu, and select option C: χ²-Test. (3) Set the Observed matrix to the one where you entered your data (e.g., Matrix A) and set the Expected matrix to a different matrix (e.g., Matrix B) where the calculator will store expected frequencies. (4) Scroll down to Calculate and press Enter. The calculator will display the test statistic, degrees of freedom, and the p-value. Compare the p-value to your alpha level (commonly 0.05) to decide whether to reject the null hypothesis of independence.

How do you interpret the p-value from a chi-square independence test on the TI-84?

The p-value from a chi-square independence test on the TI-84 tells you the probability of observing your data, or something more extreme, assuming the null hypothesis is true. The null hypothesis states that the two categorical variables are independent. If the p-value is less than your significance level $\alpha =0.05$, you reject the null hypothesis, concluding there is evidence that the variables are dependent (associated). If the p-value is greater than $\alpha$, you fail to reject the null, meaning there is not enough evidence to say the variables are dependent, so you assume independence. For example, a p-value of 1.7 × 10^-7 is much smaller than 0.05, so you reject the null hypothesis.

What is the difference between a chi-square test of independence and a homogeneity test on the TI-84?

The chi-square test of independence and the homogeneity test are very similar and use the same procedure on the TI-84. The main differences lie in the wording of the hypotheses and conclusions. For an independence test, the null hypothesis states that the two categorical variables are independent, while the alternative states they are dependent. For a homogeneity test, the null hypothesis states that different populations have the same distribution of a categorical variable, and the alternative states they differ. On the TI-84, you perform the test identically by entering data as a matrix and using the χ²-Test function. Just be sure to adjust your hypotheses and interpret your conclusions accordingly.

How do you set up hypotheses for a chi-square test of independence?

For a chi-square test of independence, the hypotheses are set up as follows: The null hypothesis ($H{0}_{}$) states that the two categorical variables are independent, meaning one variable does not affect the other. The alternative hypothesis ($H{a}_{}$) states that the variables are dependent, meaning there is an association between them. For example, if testing whether patient group affects symptom improvement, $H{0}_{}$ would say group and improvement are independent, and $H{a}_{}$ would say they are dependent. These hypotheses guide the interpretation of the test results.

Why is it important to use a different matrix for expected values in the TI-84 chi-square test?

When performing a chi-square test on the TI-84, it is important to use a different matrix for expected values than the one used for observed data because the calculator needs a separate space to calculate and store the expected frequencies. The observed matrix contains your actual data, while the expected matrix will be automatically filled by the calculator using the formula for expected counts based on marginal totals. Using the same matrix for both would overwrite your observed data and cause errors. Typically, you enter observed data in Matrix A and specify Matrix B for expected values. This separation ensures the test runs correctly and results are accurate.