Question

The article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica \([1991]: 63-67\) ) reported that 36 of 193 female bats in flight spent more than 5 min in the air before locating food. For male bats, 64 of 168 spent more than 5 min in the air. Is there sufficient evidence to conclude that the proportion of flights longer than \(5 \mathrm{~min}\) in length differs for males and females? Test the relevant hypotheses using \(\alpha=.01\).

Short answer

Yes, there is sufficient evidence at the 0.01 significance level to conclude that the proportion of flights longer than 5 minutes differs for males and females.

Step-by-step solution

State the Hypotheses

The null hypothesis \(H_0\) is that the proportion of flights longer than 5 minutes for males and females is the same, and the alternative hypothesis \(H_1\) is that they are different. In mathematical notation, these are expressed as: \(H_0: p_m = p_f\) and \(H_1: p_m \neq p_f\) where \(p_m\) is the proportion for male bats and \(p_f\) is for the female bats.

Calculate Sample Proportions

The sample proportion for female bats is the number of successes (flights longer than 5 minutes) divided by the number of trials, or \( \hat{p}_f = 36/193 = 0.1866 \). Similarly, for male bats, the sample proportion is \( \hat{p}_m = 64/168 = 0.3810 \)

Perform the Test

We will perform the two-proportion z-test. The test statistic is calculated as follows: \(Z = \frac{(\hat{p}_m - \hat{p}_f) - 0} { \sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_m}+\frac{1}{n_f})}}\) where \( \hat{p} = \frac{n_m\hat{p}_m+n_f\hat{p}_f}{n_m + n_f} \). Substituting gives \( Z \approx 3.99 \)

Make a Decision

Compare the test statistic to a critical value from the standard normal table. The critical value for a two-tailed test at the 0.01 significance level is approximately 2.58. Since 3.99 > 2.58, we reject the null hypothesis. There is sufficient evidence at the 0.01 significance level to conclude that the proportion of flights longer than 5 minutes differs for males and females.