Wildlife biologists inspect 145 deer taken by hunters and find 26 of them carrying ticks that...

Wildlife biologists inspect 145 deer taken by hunters and find 26 of them carrying ticks that test positive for Lyme disease.

a) Create a 90% confidence interval for the percentage of deer that may carry such ticks.

(____%,_____%)

(Round to one decimal place as needed.)

Expert Solution

Solution :

Given that,

n = 145

x = 26

Point estimate = sample proportion = = x / n = 16 / 145 = 0.179

1 - = 1 - 0.179 = 0 .821

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} * $\sqrt$ (( * (1 - )) / n)

= 1.645 * ( $\sqrt$ ((0.179 * 0.8/21) / 145)

= 0.052

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.179 - 0.052< p < 0.179 + 0.52

0.127 < p < 0.231

(12.7% , 23.1%)

orchestra answered 2 years ago

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