Wildlife biologists inspect 128 deer taken by hunters and find 23 of them carrying ticks that...

Wildlife biologists inspect 128 deer taken by hunters and find 23 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.

Expert Solution

Solution :

Given that,

n = 128

x = 23

Point estimate = sample proportion = = x / n = 23/128=0.180

1 - = 1-0.180=0.820

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z / 2 * $\sqrt$ (( * (1 - )) / n)

= 1.645 ( $\sqrt$ ((0.180*0.820) /128 )

= 0.0559

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.180-0.0559 < p <0.180+ 0.0559

0.1241< p < 0.2359

The 90% confidence interval for the population proportion p is : 0.1241,0.2359

orchestra answered 3 years ago

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