Question

The length of a scarlet macaw is normally distributed with a standard deviation of 5.11 cm. An ornithologist is interested in the mean length of scarlet macaws. He randomly selects 77 scarlet macaws to study, and finds their average length to be 89.5 cm.

a) Define the parameter of interest.
b) Define the random variable of interest.
c) Name the distribution used to construct confidence intervals. (Check the relevant criteria.)
d) Construct a 90% confidence interval for the true mean length of scarlet macaws.
e) Interpret your confidence interval.

PLEASE EXPLAIN IN DETAIL

Answer 1

(a)

the parameter of interest. mean length of scarlet macaws.

(b)

the random variable of interest. The length of a scarlet macaw

(c)

the distribution used to construct confidence intervals : Z distribution

the relevant criteria :(i) n = Sample size = 77 > 30 large sample and Population standard deviation = = 5.11 is provided.

(d)

Confidence Interval:

= (88.542,90.458)

So,

Answer is:

(88.542, 90.458)

(e)

The 90% confidence Interval (88.542,90.458) is a range of values we are 90% confident will contain the true unknown population mean length of a scarlet macaw.