Question

One difficulty in measuring the nesting success of birds is that the researchers must count the number of eggs in the nest, which is disturbing to the parents. Even though the researcher does not harm the birds, the flight of the bird might alert predators to the presence of a nest. To see if researcher activity might degrade nesting success, the nest survival of 102 nests that had their eggs counted, was recorded. Sixty-four of the nests failed (i.e. the parent abandoned the nest.) Assuming that it is reasonable to regard the 102 nests in the sample as representative of the population of nests for which the eggs have been counted, find a 95% confidence interval for the proportion of nests that have eggs counted that are then abandoned. Round your answer to four decimal places. (0.6230, 0.6319) (0.5336, 0.7213) (0.5039, 0.7500) (0.5487, 0.7062)

Answer 1

Solution :

Given that,

n = 102

x = 64

Point estimate = sample proportion = = x / n = 64/102 =0.627

1 - = 1-0.627 =0.373

Z_/2 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * ((0.627*(0.373) /102 )

= 0.0934

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.627-0.0934 < p < 0.627 +0.0934

0.5336< p < 0.7213

(0.5336,0.7213)