Question

In the past two years, rodents (woodrats, squirrels, chipmunks etc.) were tested for Lyme disease (caused by Borrelia) and Anaplasma in Humboldt county. Of 100 rodents, 73 had neither bacterium, 9 had both. 4 had only Anaplasma, 14 only Borrelia. Did the presence of one bacterium influence the prevalence of the other? (Hint: was Anaplasma presence independent of Borrelia presence?) State and test the appropriate null hypothesis.

Answer 1

The hypothesis being tested is:

H₀: p₁ = p₂

H_a: p₁ ≠ p₂

p1	p2	pc
0.04	0.14	0.09	p (as decimal)
4/100	14/100	18/200	p (as fraction)
4.	14.	18.	X
100	100	200	n
	-0.1	difference
	0.	hypothesized difference
	0.0405	std. error
	-2.47	z
	.0135	p-value (two-tailed)

The p-value is 0.0135.

Since the p-value (0.0135) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that Anaplasma presence is independent of Borrelia's presence.