Question

Wildlife biologists inspect 132 deer taken by hunters and find 25 of them carrying ticks that test positive for Lyme disease. Create a​ 90% confidence interval for the percentage of deer that may carry such ticks.

Answer 1

Solution :

n = 132

x = 25

= x / n = 25 / 132 = 0.164

1 - = 1 - 0.164 =0.836

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.164 * 0.836) /132 )

= 0.053

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.164 - 0.053 < p < 0.164 + 0.053

0.111 < p < 0.217