Question

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 p 1 and p 2 represent the proportion of cyanobacteria in riparian (i.e., river) and coastal populations, respectively. Is there evidence that the proportion of cyanobacteria is higher in coastal samples than in riparian samples? To determine this, you test the given hypotheses. H 0 : p 1 = p 2 , H a : p 1 < p 2

Using technology, what is the P ‑value of your test?

Answer 1

H 0 : p 1 = p 2 , H a : p 1 < p 2

p^1 = 22/50 = 0.44

p^2 = 66/75 = 0.88

pooled proportion = p= (22 + 66)/(50 + 75) = 0.704

standard error of proportion = sqrt [p * (1/n1 + 1/n2)] = sqrt [0.704 * (1/50 + 1/75)] = 0.1532

Z statistc

z = (0.88 - 0.44)/0.1532 = 2.8722

p -value = P(Z > 2.8722) = 1- P(Z < 2.8722) = 1- 0.9980 = 0.0020

so p - value is equal to 0.0020