A phytologist is worried that a large proportion of the algae found in nearby water sources is cyanobacteria. In particular, he obtains a simple random sample of 50 algae from a local river and an independent simple random sample of 75 algae from the local coastline. He finds that 22 algae in the river sample and 66 algae in the coastline sample are cyanobacteria. He considers the river to act as a control group. Let p₁ and p₂ represent the proportion of cyanobacteria in riparian (i.e., river) and coastal populations, respectively. Using technology, what is a 99% confidence interval for p₁-p₂?

Question

Quizplus · Accepted Answer

To find the confidence interval for the difference in proportions, we can use the formula:

(𝑝1−𝑝2)±𝑧∗√[(𝑝1(1−𝑝1)/𝑛1)+(𝑝2(1−𝑝2)/𝑛2)]

where 𝑝1 and 𝑝2 are the proportions of cyanobacteria in the river and coastal populations respectively, 𝑛1 and 𝑛2 are the sample sizes, and 𝑧 is the critical value for a 99% confidence interval, which is 2.576.

Plugging in the values:

(22/50 - 66/75) ± 2.576*sqrt[(22/50)*(28/50)/50 + (66/75)*(9/75)/75]

= -0.44 ± 0.2054

Therefore, the answer is C.

A Phytologist Is Worried That a Large Proportion of the Algae