An ecologist studied the spatial distribution of tree species in a wooded area. From a total area of 21 acres, he randomly selected 144 quadrats (plots), each 38 feet square, and noted the presence or absence of maples and hickories in each quadrat. The results are shown in the table.
Maples
| | | Present | Absent |
| :-------- | :------ | :------ | :----- |
| Hickories | Present | 26 | 63 |
| | Absent | 29 | 26 |
She wants to test the hypothesis that the two species are distributed independently of each other.
(c) Compute the chi-square test statistic.
(d) Find the [tex]$p$[/tex]-value.
(e) State the conclusion.
Answer (1)
Calculate the expected frequencies for each cell.
Compute the chi-square test statistic: χ 2 = 7.96 .
Determine the degrees of freedom: df = 1 .
Find the p-value: p < 0.05 , so we reject the null hypothesis. The two variables are NOT independent. χ 2 = 7.96
Explanation
Problem Setup We are given a contingency table showing the presence/absence of maples and hickories in 144 quadrats. We want to test the hypothesis that the two species are distributed independently.
Contingency Table The contingency table is:
Maples
Present Absent Hickories Present 26 63 Absent 29 26
Hypotheses The null hypothesis ( H 0 ) is that the presence of maples and hickories are independent. The alternative hypothesis ( H 1 ) is that they are not independent.
Expected Frequencies We calculate the expected frequencies for each cell in the contingency table under the assumption of independence. The expected frequency for each cell is calculated as (row total * column total) / grand total.
Calculating Expected Frequencies The expected frequencies are:
Expected (Present, Present) = (26+63) * (26+29) / 144 = 89 * 55 / 144 = 33.99 Expected (Present, Absent) = (26+63) * (63+26) / 144 = 89 * 89 / 144 = 55.01 Expected (Absent, Present) = (29+26) * (26+29) / 144 = 55 * 55 / 144 = 21.01 Expected (Absent, Absent) = (29+26) * (63+26) / 144 = 55 * 89 / 144 = 33.99
Chi-Square Formula We compute the chi-square test statistic using the formula: χ 2 = ∑ E ij ( O ij − E ij ) 2 where O ij are the observed frequencies and E ij are the expected frequencies.
Calculating Chi-Square Statistic χ 2 = 33.99 ( 26 − 33.99 ) 2 + 55.01 ( 63 − 55.01 ) 2 + 21.01 ( 29 − 21.01 ) 2 + 33.99 ( 26 − 33.99 ) 2 χ 2 = 33.99 ( − 7.99 ) 2 + 55.01 ( 7.99 ) 2 + 21.01 ( 7.99 ) 2 + 33.99 ( − 7.99 ) 2 χ 2 = 33.99 63.84 + 55.01 63.84 + 21.01 63.84 + 33.99 63.84 χ 2 = 1.878 + 1.161 + 3.038 + 1.878 = 7.955
The chi-square statistic is approximately 7.96.
Degrees of Freedom The degrees of freedom for the chi-square test is calculated as df = ( ro w s − 1 ) ( co l u mn s − 1 ) = ( 2 − 1 ) ( 2 − 1 ) = 1 .
P-Value The p-value associated with a chi-square statistic of 7.96 and 1 degree of freedom is less than 0.05. Using a chi-square calculator, the p-value is approximately 0.0048.
Conclusion Since the p-value (0.0048) is less than the significance level (0.05), we reject the null hypothesis.
Final Answer Therefore, we conclude that the presence/absence of Hickories and Maples are NOT independent.
Examples
Chi-square tests are commonly used in ecological studies to determine if different species' distributions are independent. For example, an ecologist might want to know if the presence of a certain plant species is related to the presence of a particular type of soil. By collecting data on the presence/absence of the plant and the soil type in several plots, they can perform a chi-square test to see if the two variables are independent. If they are not independent, it suggests that there is a relationship between the plant's distribution and the soil type, which could be due to the soil providing necessary nutrients or affecting the plant's ability to grow. This information can be crucial for conservation efforts or understanding ecosystem dynamics.
