Question

Wildlife biologists inspect 153 deer taken by hunters and found 42 of them carrying Lyme disease ticks. Calculate and interpret a 90% confidence interval for the proportion of deer that carry Lyme disease ticks.

Answer 1

90 Confidence Interval for the proportion = [ 0.216,0.334]

Explanation:

The confidence interval for proportion formula =

p ± z × √p(1 - p)/n

Where p = x/n

x = 42

n = number of samples = 153

p = 42/153

p = 0.2745098039 ≈ 0.275

z = z score of 90% confidence interval = 1.645

Confidence interval =

0.275 ± 1.645 × √0.275 - ( 1 - 0.275)/153

= 0.275 ± 1.645 × √0.0013031046

= 0.275 ± 0.0593820985

Confidence Interval

= 0.275 - 0.0593820985

= 0.2156179015

≈ 0.216

= 0.275 + 0.0593820985

= 0.3343820985

≈ 0.334

90 Confidence Interval for the proportion = [ 0.216,0.334]