Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 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 = 0.34 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 answers to two decimal places.)

lower limit
upper limit
margin of error

d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.11 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds

Answer 1

Part a)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.2 /2) = 1.282
3.15 ± Z (0.2/2 ) * 0.34/√(12)
Lower Limit = 3.15 - Z(0.2/2) 0.34/√(12)
Lower Limit = 3.02
Upper Limit = 3.15 + Z(0.2/2) 0.34/√(12)
Upper Limit = 3.28
80% Confidence interval is ( 3.02 , 3.28 )

Margin of Error = Z (0.2/2 ) * 0.34/√(12) = 0.13

Part b)
Sample size can be calculated by below formula
n = (( Z(α/2) * σ) / e )2
n = (( Z(0.2/2) * 0.34 ) / 0.11 )2
Critical value Z(α/2) = Z(0.2/2) = 1.2816
n = (( 1.2816 * 0.34 ) / 0.11 )2
n = 16
Required sample size at 80% confident is 16.