Question

2. Wildlife biologists inspect 153 deer taken by hunters and find 32 of them carrying Lyme disease ticks. a) Calculate the standard error of the sample proportion.

b) Calculate the margin of error with confidence level of 98%.

c) Determine the 98% confidence interval for the proportion of deer that carry Lyme disease ticks. Interpret.

d) If the scientists want to cut the margin of error, E in half, calculate the number of deer that must be inspected? Consider the same 98% C.I. Show all mathematics.

Answer 1

Answer :

Given data is :

n = 156

x = 32

So,

= 32 / 153

= 0.209

then

= 1 - 0.209

= 0.791

a)

Standard error SE =

=

=

=

=

Standard error SE = 0.0332

b)

confidence level of 98%.

i.e Z value at 98% is 2.326

Margin of error ME =

or

ME = Z * SE

= 2.326 * 0.0332

= 0.0772

Margin of at 98% is 0.0772.

c)

Confidence interval CI =

or

CI =

= 0.209 +/- 0.0772

= (0.209 - 0.0772 , 0.209 + 0.0772)

= (0.1318 , 0.2862)

Confidence interval at 98% lies between 0.1318 to 0.2862.

d)

Here the margin of error is cut into half.

i.e ME = 0.04

and

therefore,

sample size

Sample size n = 541