Independence Tests: Videos & Practice Problems

Independence tests assess whether two categorical variables, such as students' heights and grade levels, are related. The null hypothesis assumes independence, while the alternative suggests dependence. The chi-square test statistic is calculated using observed and expected frequencies, with expected frequencies derived from row and column totals. Degrees of freedom are determined by the formula: $(r-1)(c-1)$, where r is rows and c is columns. A p-value comparison with alpha determines the conclusion.

Independence Test

Understanding the concept of independence between two variables is crucial in statistics, particularly when analyzing categorical data. Independence implies that the occurrence of one variable does not influence the occurrence of another. For instance, when examining students' heights in relation to their grade levels, we may want to determine if these two variables are related. To assess this relationship, we can utilize an independence test, which is conceptually similar to a goodness of fit test.

In an independence test, we start by formulating hypotheses. The null hypothesis (H₀) posits that the two variables are independent, meaning they do not affect each other. Conversely, the alternative hypothesis (H₁) suggests that the variables are dependent. For example, we might state that students' heights are unaffected by their grade levels.

The test statistic used in an independence test is the chi-squared statistic, calculated using observed and expected frequencies. The expected frequencies (E) are determined by the formula:

\[E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}\]

This formula allows us to create a contingency table that reflects the relationship between the two categorical variables. The chi-squared statistic is then computed using the formula:

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

where O represents the observed frequencies. In our example, a chi-squared value of 3.32 was calculated.

Next, we determine the degrees of freedom (df) for the test, which is calculated as:

\[df = (\text{Number of Rows} - 1) \times (\text{Number of Columns} - 1)\]

For a dataset with 2 rows and 3 columns, the degrees of freedom would be (2-1)(3-1) = 2. Using the chi-squared value and the degrees of freedom, we can find the p-value, which in this case is 0.19.

To draw a conclusion, we compare the p-value to our significance level (α), which is typically set at 0.05. Since 0.19 is greater than 0.05, we fail to reject the null hypothesis. This indicates that there is insufficient evidence to conclude that the variables are dependent, suggesting that students' heights and grade levels are indeed independent.

Before finalizing our results, it is essential to verify that the conditions for conducting an independence test are met. These include having random samples, ensuring that all categories have observed frequencies, and confirming that expected frequencies are at least 5 for each category. Meeting these criteria strengthens the validity of our test results.

In summary, an independence test allows us to explore the relationship between two categorical variables, using the chi-squared statistic to assess whether they are independent or dependent. This process involves hypothesis formulation, calculation of test statistics, and careful consideration of conditions to ensure accurate conclusions.

Independence Test Example 1

In this example, we explore the relationship between symptom improvement in ADHD patients and whether they received a placebo or not, using a chi-squared independence test. The goal is to determine if symptom improvement is independent of the treatment type. We start by establishing our null hypothesis, which states that symptom improvement is independent of whether a patient received a placebo. Conversely, the alternative hypothesis posits that symptom improvement is dependent on the treatment received.

Before proceeding, we ensure that our data meets the necessary criteria for the test: we have random samples, observed frequencies for all categories, and expected frequencies greater than five. The expected frequencies are calculated using the formula:

\[ E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}} \]

For our observed frequencies, we have values such as 18 (with an expected frequency of 26.4), 37 (expected 28.6), 30 (expected 21.6), and 15 (expected 23.4). With these values, we compute the chi-squared test statistic using the formula:

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

Calculating each term, we find:

• For 18: \[ \frac{(18 - 26.4)^2}{26.4} \]
• For 37: \[ \frac{(37 - 28.6)^2}{28.6} \]
• For 30: \[ \frac{(30 - 21.6)^2}{21.6} \]
• For 15: \[ \frac{(15 - 23.4)^2}{23.4} \]

Summing these values gives us a chi-squared statistic of 11.42. Next, we determine the degrees of freedom, calculated as:

\[ \text{Degrees of Freedom} = (r - 1)(c - 1) = (2 - 1)(2 - 1) = 1 \]

Using the chi-squared value and degrees of freedom, we find the p-value to be 0.0007. Since this p-value is significantly lower than our significance level of 0.01, we reject the null hypothesis. This indicates that there is sufficient evidence to conclude that symptom improvement is not independent of whether patients received a placebo, suggesting that the ADHD medication is effective in treating symptoms.

In summary, the analysis demonstrates a significant relationship between treatment type and symptom improvement, reinforcing the effectiveness of the medication in question.

Here’s what students ask on this topic:

What is the null hypothesis in an independence test?

The null hypothesis in an independence test states that the two categorical variables being analyzed are independent, meaning they do not affect each other. For example, if you are testing whether students' heights and grade levels are related, the null hypothesis would be: 'Students' heights are independent of their grade levels.' This assumption serves as the default position, and the test aims to determine whether there is enough evidence to reject it in favor of the alternative hypothesis, which suggests dependence between the variables.

How is the chi-square test statistic calculated in an independence test?

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

$\left(\frac{{(O_i}^2}{E_i}\right)$

Here, O_i represents the observed frequency for a category, and E_i represents the expected frequency. The expected frequencies are calculated using the formula:

$\frac{(\mathrm{Row\; Total}\times \mathrm{Column\; Total})}{\mathrm{Grand\; Total}}$

After summing the values for all categories, the resulting chi-square statistic is used to determine the p-value.

How do you calculate degrees of freedom in an independence test?

Degrees of freedom in an independence test are calculated using the formula:

$(\mathrm{Rows}-1)\times (\mathrm{Columns}-1)$

For example, if your contingency table has 2 rows and 3 columns, the degrees of freedom would be:

$(2-1)\times (3-1)=2$

Degrees of freedom are essential for determining the critical value or p-value from the chi-square distribution table.

What conditions must be met to perform an independence test?

To perform an independence test, the following conditions must be met:

• Random Sampling: The data should be collected randomly to ensure unbiased results.
• Observed Frequencies: There must be observed frequencies for all categories in the contingency table.
• Expected Frequencies: Each expected frequency should be greater than or equal to 5 to ensure the validity of the chi-square approximation.

These conditions help ensure the reliability and accuracy of the test results.

How do you interpret the p-value in an independence test?

The p-value in an independence test indicates the probability of observing the data if the null hypothesis is true. If the p-value is less than the significance level (α), typically 0.05, you reject the null hypothesis, concluding that the variables are dependent. If the p-value is greater than α, you fail to reject the null hypothesis, meaning there is insufficient evidence to suggest dependence between the variables. For example, if the p-value is 0.19 and α is 0.05, you fail to reject the null hypothesis, implying the variables are likely independent.

What is the difference between a goodness-of-fit test and an independence test?

Both tests use the chi-square statistic, but they differ in purpose and calculation:

• Goodness-of-Fit Test: Assesses whether observed frequencies match a claimed distribution. Expected frequencies are calculated based on the claimed probabilities.
• Independence Test: Evaluates whether two categorical variables are independent. Expected frequencies are calculated using row and column totals divided by the grand total.

While the calculation of the chi-square statistic is similar, the hypotheses and expected frequency formulas differ between the two tests.