Question

Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population in turn rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. They added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area.
Required:
A 90% confidence interval for the difference between the proportions of mice ready to breed in good acorn years and bad acorn years is:_______

Answer 1

Answer: 90% confidence interval is; ( - 0.0516, 0.3752 )

 

Step-by-step explanation:

Given the data in the question;  

n1 = 72, n2 = 17

P1 = 54 / 72 = 0.75

P2 = 10 / 17 = 0.5882

so

P_good = 0.75

P_bad = 0.5882

standard ERRROR will be;

SE = √[(0.75×(1-0.75)/72) + (0.5882×(1-0.5882)/17)]

SE = √( 0.002604 + 0.01424)

SE = 0.12978

given confidence interval = 90%

significance level a = (1 - 90/100) = 0.1, |Z( 0.1/2=0.05)| = 1.645   { from standard normal table}

so

93% CI is;

(0.75 - 0.5882) - 1.645×0.12978 <P_good - P_bad<  (0.75 - 0.5882) + 1.645×0.12978

⇒0.1618 - 0.2134 <P_good - P_bad<  0.1618 + 0.2134

⇒ - 0.0516  <P_good - P_bad< 0.3752

Therefore 90% confidence interval is; ( - 0.0516, 0.3752 )