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 = 0.22 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

(b) What conditions are necessary for your calculations? (Select all that apply.)

uniform distribution of weightsnormal distribution of weightsσ is unknownσ is knownn is large

(c) Interpret your results in the context of this problem.

There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.    The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds

(f)Use the Student's t distribution to find t_c for a 0.95 confidence level when the sample is 21. (Round your answer to three decimal places.)

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 ) * 0.22/√(18)
Lower Limit = 3.15 - Z(0.2/2) 0.22/√(18)
Lower Limit = 3.08
Upper Limit = 3.15 + Z(0.2/2) 0.22/√(18)
Upper Limit = 3.22
80% Confidence interval is ( 3.08 , 3.22 )

Margin of Error = Z (0.2/2 ) * 0.22/√(18) = 0.07

Part b)

normal distribution of weights

σ is known

part c)

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

Part d)

Sample size can be calculated by below formula
n = (( Z(α/2) * σ) / e )2
n = (( Z(0.2/2) * 0.22 ) / 0.16 )2
Critical value Z(α/2) = Z(0.2/2) = 1.2816
n = (( 1.2816 * 0.22 ) / 0.16 )2
n = 4
Required sample size at 80% confident is 4.

Part e)

Critical value t(α/2, n-1) = t(0.05 /2, 21-1) = 2.086 ( from t table )