Independence Test Using TI-84: Videos & Practice Problems

Using a TI-84 calculator simplifies conducting a chi-square test of independence by entering categorical data into matrices. The null hypothesis assumes variables are independent, while the alternative suggests dependence. After inputting observed frequencies, the calculator computes expected frequencies and the p-value. Comparing the p-value to the significance level (alpha) determines if the null hypothesis is rejected, indicating dependence between variables. This method also applies to homogeneity tests with adjusted hypotheses and conclusions. Mastery of this process enhances understanding of categorical data analysis, hypothesis testing, and the chi-square distribution in inferential statistics.

Independence Test Using TI-84

Performing a chi-square independence test 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) posits that the variables are dependent, indicating a relationship between group membership and symptom improvement.

To conduct the 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 the observed frequency data into Matrix A. Each cell in the matrix should correspond exactly to the observed counts in the contingency table. After entering the data, initiate the chi-square test by pressing the STAT button, moving to the TESTS menu, and selecting the chi-square test option (usually 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:

\[E_{ij} = \frac{(R_i)(C_j)}{N}\]

where \(E_{ij}\) is the expected frequency for cell in row \(i\) and column \(j\), \(R_i\) is the total count for row \(i\), \(C_j\) is the total count for column \(j\), and \(N\) is the overall sample size.

After setting these matrices, select Calculate to obtain the test statistic and p-value. A p-value less than the significance level \(\alpha = 0.05\) indicates sufficient evidence to reject the null hypothesis, concluding that the variables are dependent. For example, a p-value of approximately \(1.7 \times 10^{-7}\) strongly suggests dependence between participant group and symptom improvement.

It is important to note that homogeneity tests share the same computational steps as independence tests on the TI-84, differing only in the wording of hypotheses and conclusions. Both tests rely on the chi-square distribution to assess relationships between categorical variables, making the TI-84 an efficient tool for these analyses once data is properly formatted as matrices.

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 two categorical variables are independent, such as the instruction delivery method and students' final grades, a chi-square test of independence is used. The null hypothesis assumes that the variables are independent, meaning the instruction delivery method does not affect the final grade. Conversely, the alternative hypothesis suggests that the variables are dependent, indicating a relationship between the delivery method and final grades.

To perform this test, data is organized into a contingency table and entered into a matrix format on a calculator, such as the TI-84. This matrix represents the observed frequencies of each category combination. The calculator then computes the expected frequencies based on the assumption of independence, using the formula:

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

The chi-square test statistic is calculated by summing the squared differences between observed and expected frequencies, divided by the expected frequencies, across all categories:

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

Here, \(O\) represents observed frequencies, and \(E\) represents expected frequencies. The resulting test statistic is then used to find the p-value, which 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 much 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 does not support a dependency between the instruction delivery method and final grades, so the college's claim that these variables are independent remains plausible.

Understanding how to conduct and interpret a chi-square test of independence is essential for analyzing relationships between categorical variables in educational research and beyond. It allows for informed decisions based on statistical evidence rather than assumptions.

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 commonly set at 0.05 for such tests.

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 for at least one book type, indicating a variation in sales distribution.

To perform the test, data is organized into a contingency table, and expected counts are calculated based on the assumption that the null hypothesis is true. The chi-square test statistic is computed by comparing observed counts to expected counts using the formula:

\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]

where \(O_i\) represents the observed frequency for each category, and \(E_i\) is the expected frequency under the null hypothesis. The degrees of freedom for the test are calculated as:

\[ df = (r - 1)(c - 1) \]

where r is the number of rows (book types) and c is the number of columns (subjects).

Using a statistical calculator or software, the observed data is input as a matrix, and the chi-square test function is selected. The calculator computes the expected counts and the chi-square statistic, then provides the p-value. If the p-value is less than the significance level (0.05), the null hypothesis is rejected, indicating that the proportions of subjects sold differ across book types.

In this scenario, the calculated p-value was approximately \(7.8 \times 10^{-7}\), which is significantly smaller than 0.05. This strong evidence leads to rejecting the null hypothesis, concluding that the distribution of subjects sold is not the same across digital, softcover, and hardcover books. This insight is crucial for understanding sales patterns and can guide marketing or inventory decisions for the textbook company.

Here’s what students ask on this topic:

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

To enter data as a matrix on the TI-84 for a chi-square test of independence, first press the 2nd button, then the x^-1 button to access the matrix menu. Navigate to the Edit tab and select a matrix (usually [A]). Enter the dimensions of your data table (rows and columns) and input the observed frequencies into the corresponding cells. Make sure the data matches the contingency table exactly. This step is crucial because the calculator uses this matrix as the observed data for the test. Once your data is entered, you can proceed to run the chi-square test using the calculator’s statistical test functions.

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 into a matrix as described earlier. (2) Press the STAT button, scroll right to the TESTS menu, and select option C: Chi-square test. (3) Set the Observed matrix to the one where you entered your data (e.g., [A]) and set the Expected matrix to a different matrix (e.g., [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 p-value. Compare the p-value to your significance level (usually 0.05) to decide whether to reject the null hypothesis of independence.

How do you interpret the results of a chi-square test of independence on the TI-84?

After running the chi-square test of independence on the TI-84, you will see the p-value displayed. To interpret the results, compare the p-value to your chosen significance level, typically $\alpha =\; 0.05$. If the p-value is less than $\alpha$, you reject the null hypothesis, indicating there is sufficient evidence to conclude that the variables are dependent or related. If the p-value is greater than $\alpha$, you fail to reject the null hypothesis, suggesting there is not enough evidence to say the variables are dependent, so they are considered independent. This conclusion helps determine if one categorical variable affects or is associated with another.

Can the TI-84 chi-square test function be used for homogeneity tests as well as independence tests?

Yes, the TI-84 chi-square test function can be used for both homogeneity tests and independence tests because the computational process is the same. The main difference lies in how you state your hypotheses and interpret the results. For homogeneity tests, the null hypothesis states that the distribution of a categorical variable is the same across different populations, while for independence tests, the null states that two variables are independent. When using the TI-84, you enter the data and run the chi-square test identically for both tests. Just remember to adjust your hypotheses and conclusions accordingly based on the context of your analysis.

What is the null hypothesis in a chi-square test of independence on the TI-84?

In a chi-square test of independence, the null hypothesis ($H\_0$) states that the two categorical variables are independent, meaning there is no association between them. For example, if you are testing whether the group a participant was in affects symptom improvement, the null hypothesis would be: "The group of participants is independent of symptom improvement." This means the group membership does not influence the likelihood of symptom improvement. The alternative hypothesis ($H\_a$) states that the variables are dependent, indicating an association exists. The TI-84 helps test this hypothesis by comparing observed and expected frequencies.