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 \(1-4 .\) $$ \begin{array}{lrlll} \hline & {\text { Category }} & \\ \text { Population } & 1 & 2 & 3 & \text { Total } \\ \hline 1 & 108 & 52 & 40 & 200 \\ 2 & 87 & 51 & 62 & 200 \\ 3 & 112 & 39 & 49 & 200 \\ \hline \end{array} $$ State your conclusions.

Short answer

Why or why not? The proportions of observations in the three categories do not depend on the population they are drawn from. This conclusion is based on the Chi-Square Test for Independence, where the test statistic (5.545) is less than the critical value (9.488) at a 0.05 significance level. Since we fail to reject the null hypothesis, there is insufficient evidence to conclude that the proportions are dependent on the population.

Step-by-step solution

Identify the Expected Frequencies

For each cell in the table, we need to calculate the expected frequency under the assumption that the row and column totals are fixed. The formula for the expected frequency (E) is: E = (row total * column total) / grand total We can use the given data to find the row and column totals: Row Totals: 200, 200, 200 Column Totals: 307, 142, 151 Grand Total: 600

Calculate the Expected Frequencies and Chi-Square Test Statistic

Using the formula from Step 1, we can calculate the expected frequencies for each cell: Expected Frequencies table: $$ \begin{array}{cccc} \hline \text{Population} & \text{Category 1} & \text{Category 2} & \text{Category 3} \\ \hline 1 & \frac{200*307}{600} & \frac{200*142}{600} & \frac{200*151}{600} \\ 2 & \frac{200*307}{600} & \frac{200*142}{600} & \frac{200*151}{600} \\ 3 & \frac{200*307}{600} & \frac{200*142}{600} & \frac{200*151}{600} \\ \hline \end{array} $$ Now, calculate the chi-square test statistic, χ², using the formula: χ² = Σ(O - E)² / E, where O is the observed frequency, E is the expected frequency, and the sum is taken over all cells. χ² = [ (108 - 102.333)² / 102.333 + (52 - 47.333)² / 47.333 + (40 - 50.333)² / 50.333 ] + ...+[ (112 - 102.333)² / 102.333 + (39 - 47.333)² / 47.333 + (49 - 50.333)² / 50.333 ] χ² ≈ 5.545

Determine the Critical Value and Compare it with the Test Statistic

For a Chi-Square Test, the critical value depends on the degrees of freedom and the desired significance level (commonly α = 0.05). The degrees of freedom (df) for the test can be calculated as: df = (number of rows - 1) * (number of columns - 1) In our case, df = (3 - 1) * (3 - 1) = 2 * 2 = 4 Using a chi-square distribution table or calculator, we can find the critical value for df=4 and α=0.05, which is approximately 9.488. Now, compare the test statistic with the critical value: χ² = 5.545 < 9.488 Since the test statistic is less than the critical value, we fail to reject the null hypothesis (H₀).

State the Conclusions

With a test statistic of 5.545 and a critical value of 9.488 at a significance level of 0.05, there is not enough evidence to conclude that the proportions of observations in the three categories depend on the population they are drawn from. Therefore, we fail to reject the null hypothesis, and we can't say that the proportions are dependent on the population.

Key concepts

Expected Frequency

In statistical analysis, Expected Frequency refers to the number of times an outcome is anticipated to occur if the experiment were repeated multiple times under the same conditions. It is crucial in hypothesis testing, especially in the Chi-Square Test which is a common statistical method used to compare observed results with expected ones. To calculate the expected frequency, we apply the formula:
\(E = \frac{{\text{{row total}} \times \text{{column total}}}}{\text{{grand total}}}\).
In the context of our exercise, we're looking at data from three populations and categories. The expected frequencies are uniform across populations because the row and column totals determine them. This statistical presumption helps us to understand what the frequencies would look like if there were no effect of the population on the category distribution. The Chi-Square Test then assesses whether the observed frequencies (actual data) significantly diverge from these expected frequencies, suggesting a possible dependency between the population and category proportions.

Chi-Square Test Statistic

The Chi-Square Test Statistic serves as a measure of how much the observed frequency within a dataset deviates from the expected frequency. The formula to compute the Chi-Square Test Statistic (\(\chi^2\)) is:
\(\chi^2 = \sum\frac{{(O - E)^2}}{E}\),
where \(O\) represents the observed frequency and \(E\) the expected frequency. In our exercise, this statistic is calculated by summing up the squared differences between observed and expected frequencies, divided by the expected frequencies for all categories in each population. This value then allows us to determine if the discrepancies observed are due to random chance or if they are statistically significant. After calculating \(\chi^2\), we compare it with a critical value from the Chi-Square distribution table, which ultimately informs us whether to reject the null hypothesis.

Degrees of Freedom

The term Degrees of Freedom in statistics represents the number of values involved in calculations that are free to vary. It's essential in the context of the Chi-Square Test as it influences the shape of the chi-square distribution and is crucial for determining the critical value against which the test statistic is compared. The degrees of freedom are calculated based on the number of categories minus one, multiplied by the number of populations minus one, which is expressed as:
\(df = (\text{{number of rows}} - 1) \times (\text{{number of columns}} - 1)\).
In our exercise with three populations (rows) and three categories (columns), the degrees of freedom would be \(df = (3-1) \times (3-1) = 4\). This parameter helps to define the particular chi-square distribution we refer to when looking up the critical value, which is pivotal in deciding the significance of the test result.

Null Hypothesis

The Null Hypothesis, symbolized as \(H_0\), is a default hypothesis that there is no effect or no difference. In the context of the Chi-Square Test, it usually states that there is no association between the variables under study. For our example, it posits that the population has no effect on the distribution of categories—essentially, that the proportions of observations in each category are the same for all populations. The entire purpose of conducting the Chi-Square Test is to determine whether the observed data provide enough evidence to reject this null hypothesis. In the step by step solution given, since our calculated \(\chi^2\) value was less than the critical value, we concluded that there is insufficient evidence to reject the null hypothesis, suggesting that the category distributions might not necessarily depend on the population.