Wildlife biologists inspect 158 deer taken by hunters and find 38 of them carrying ticks that...

Wildlife biologists inspect 158 deer taken by hunters and find 38 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.

b) If the scientists want to cut the margin of error in half, how many deer must they inspect?

Expert Solution

Solution :

Given that,

n = 158

x = 38

Point estimate = sample proportion = = x / n = 0.241

1 - = 0.759

a)

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 * ( $\sqrt$ ((0.241 * 0.759) / 158)

= 0.056

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.241 - 0.056 < p < 0.241 + 0.056

0.185 < p < 0.297

( 0.185 , 0.297 )

b)

margin of error = E = 0.028

sample size = n = (Z_{/ 2} / E )² * * (1 - )

= (1.645 / 0.028)² * 0.241 * 0.759

= 631.36

sample size = 632

milcah answered 10 months ago

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