Question

Random samples of 200 observations were selected from each of three populations, and each observation was classified according to whether it fell into one of three mutually exclusive categories. Is there sufficient evidence to indicate that the proportions of observations in the three categories depend on the population from which they were drawn? Use the information in the table to answer the questions in Exercises \(1-4 .\) $$ \begin{array}{lrlll} \hline & {\text { Category }} & \\ \text { Population } & 1 & 2 & 3 & \text { Total } \\ \hline 1 & 108 & 52 & 40 & 200 \\ 2 & 87 & 51 & 62 & 200 \\ 3 & 112 & 39 & 49 & 200 \\ \hline \end{array} $$ Find the approximate \(p\) -value for the test and interpret its value.

Short answer

Based on the Chi-square test for independence, with a calculated test statistic of approximately 11.16 and a p-value of approximately 0.025, we reject the null hypothesis that the proportions of observations in the three categories are the same for all populations. There is sufficient evidence to suggest that the proportions of observations in the three categories depend on the population from which they were drawn.

Step-by-step solution

Set up the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that the proportions of observations in the three categories are the same for all three populations, while the alternative hypothesis (\(H_a\)) states that the proportions are different for at least one population. $$ H_0: \text{The proportions are the same for all populations.} \\ H_a: \text{The proportions are different for at least one population.} $$

Compute the expected frequencies

To compute the expected frequencies, we first find the row and column totals. Then, for each cell, multiply the row total by the column total and divide by the grand total. It’s essential to do this for each cell in the table. The row and column totals are already given in the table. The grand total is 600 (since there are 200 observations for each of the three populations). Expected frequencies for each cell are as follows: $$ \text{Population 1, Category 1: } \frac{200\times307}{600} = 102.33 \\ \text{Population 1, Category 2: } \frac{200\times142}{600} = 47.33 \\ \text{Population 1, Category 3: } \frac{200\times151}{600} = 50.33 \\ \text{Population 2, Category 1: } \frac{200\times307}{600} = 102.33 \\ \text{Population 2, Category 2: } \frac{200\times142}{600} = 47.33 \\ \text{Population 2, Category 3: } \frac{200\times151}{600} = 50.33 \\ \text{Population 3, Category 1: } \frac{200\times307}{600} = 102.33 \\ \text{Population 3, Category 2: } \frac{200\times142}{600} = 47.33 \\ \text{Population 3, Category 3: } \frac{200\times151}{600} = 50.33 $$

Calculate the Chi-square test statistic

To calculate the Chi-square test statistic, we will use the following formula: $$ \chi^2 = \sum \frac{(O - E)^2}{E} $$ where O is the observed frequency and E is the expected frequency. $$ \chi^2 = \frac{(108-102.33)^2}{102.33} + \frac{(52-47.33)^2}{47.33} + \frac{(40-50.33)^2}{50.33} + \frac{(87-102.33)^2}{102.33} + \frac{(51-47.33)^2}{47.33} + \frac{(62-50.33)^2}{50.33} + \frac{(112-102.33)^2}{102.33} + \frac{(39-47.33)^2}{47.33} + \frac{(49-50.33)^2}{50.33} \\ \chi^2 \approx 11.16 $$ The Chi-square test statistic is approximately \(11.16\).

Calculate the degrees of freedom and find the \(p\)-value

To calculate the degrees of freedom (df), we use the formula: $$ \text{df} = (\text{Number of rows - 1}) \times (\text{Number of columns - 1}) \\ \text{df} = (3 - 1) \times (3 - 1) \\ \text{df} = 4 $$ Now, using the chi-square table or an online calculator, we can determine the \(p\)-value for our Chi-square test statistic with 4 degrees of freedom. The \(p\)-value is approximately \(0.025\).

Interpret the \(p\)-value and draw a conclusion

The \(p\)-value of approximately \(0.025\) indicates the probability of observing a Chi-square test statistic as extreme as \(11.16\) occurring by chance, given that the null hypothesis is true. Typically, we reject the null hypothesis if the \(p\)-value is less than a pre-determined significance level, usually \(0.05\). Since our \(p\)-value is approximately \(0.025\), which is less than \(0.05\), we have sufficient evidence to reject the null hypothesis. Therefore, we conclude that there is sufficient evidence to indicate that the proportions of observations in the three categories depend on the population from which they were drawn.

Key concepts

Expected Frequencies

Expected frequencies are crucial in the Chi-square test, serving as theoretical predictions of the observed data under the assumption that the null hypothesis is correct. To calculate them, you need to consider the totals for each row and column, then find the product of the corresponding row and column totals, which is then divided by the grand total. For example, to calculate the expected frequency for Population 1, Category 1, the formula is:
\[\frac{(\text{row total for Population 1}) \times (\text{column total for Category 1})}{\text{grand total}}\]
In our exercise, that calculation would be: \[\text{Population 1, Category 1: } \frac{200\times307}{600} = 102.33\]
This process is repeated for each cell within the contingency table. The expected frequencies represent the distribution we would see if there indeed was no association between the categories and populations—essentially, if everything was purely random.

Null Hypothesis

In any statistical test, the null hypothesis (\(H_0\)) serves as the default statement or proposition that indicates no effect or no difference. It is the hypothesis that the test aims to examine for possible rejection. In the context of the Chi-square test, our null hypothesis states that there is no difference in the proportions of observations across categories within different populations, suggesting that they are independent of each other. The alternative hypothesis (\(H_a\)), on the contrary, suggests that there is an association—that the proportions are not equal and that the observed distribution of frequencies cannot be attributed to random chance alone. The purpose of calculating the Chi-square test statistic is to determine whether the observed frequencies are significantly different from the expected frequencies under the null hypothesis.

Degrees of Freedom

Degrees of freedom (df) are a critical component in statistics, as they determine the number of independent values or quantities which can freely vary when enforcing certain restrictions. In the case of the Chi-square test, degrees of freedom are calculated by using the formula:
\[\text{df} = (\text{Number of rows - 1}) \times (\text{Number of columns - 1})\]
For our exercise, with 3 rows and 3 columns in our contingency table, the degrees of freedom would be: \[\text{df} = (3 - 1) \times (3 - 1) = 4\]
These degrees of freedom are then used in conjunction with the test statistic to determine the p-value from the Chi-square distribution. This helps us understand the likelihood of observing our calculated Chi-square value, given the df, under the assumption that the null hypothesis is true.

P-value Interpretation

The p-value obtained in a statistical test offers insight into the significance of our test results. It is the probability of observing a test statistic as extreme as the one computed from our sample data, assuming that the null hypothesis is true. A common threshold for significance is 0.05; if the p-value is lower than this, we typically reject the null hypothesis.

In this particular exercise, a p-value of approximately 0.025 suggests there is only a 2.5% chance of observing such substantial differences in the category proportions between populations if the null hypothesis were true. As this p-value is less than the common alpha level of 0.05, we conclude that there is sufficient evidence to reject the null hypothesis, indicating a statistically significant association between the populations and the category proportions. This helps in making inferences or decisions based on our sample data regarding the entire population.