Consider an urban ecologist who is interested in whether the presence of high-voltage power lines influences the number of eggs laid by birds nearby. He plans to conduct a two-sample t-test and will assume that the population variances are equal. His data are as follows:
Near (group 1): mean = 6.1, SD = 0.35, n = 9
Far (group 2): mean = 5.8, SD = 0.30, n = 12
What is the standard error of the difference between the two means?

Question

Quizplus · Accepted Answer

The standard error (SE) of the difference between two means when assuming equal variances can be calculated using the formula: $$SE = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}}$$ where $s_p^2$ is the pooled variance, $n_1$ and $n_2$ are the sample sizes. The pooled variance $s_p^2$ is calculated as: $$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$$ Substituting the given values: $$s_p^2 = \frac{(9-1)0.35^2 + (12-1)0.30^2}{9 + 12 - 2} = \frac{8*0.1225 + 11*0.09}{19} = \frac{0.98 + 0.99}{19} = \frac{1.97}{19} = 0.1037$$ Then, the SE is: $$SE = \sqrt{\frac{0.1037}{9} + \frac{0.1037}{12}} = \sqrt{0.01152 + 0.00864} = \sqrt{0.02016} = 0.142$$ Thus, the correct answer is A, not B as initially indicated. The calculation error in the explanation led to the wrong choice. The correct calculation shows that A) 0.142 is the right answer.

Consider an Urban Ecologist Who Is Interested in Whether the Presence