Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 18 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.26 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.

lower limit
upper limit
margin of error

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 3.15

Population standard deviation =    = 0.26

Sample size n =18

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z/2 = Z0.10 = 1.28 ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.28 * ( 0.26 / 18 )

E= 0.08
At 95% confidence interval estimate of the population mean
is,

- E < < + E

3.15 - 0.08 <   <3.15 + 0.08

3.07 <   < 3.23

( lower limit= 3.07,upper limit= 3.23)

Margin of error = E =0.08