Question

(1) Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 11 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 = 0.28 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

(2) Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 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 = 0.36 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

(3) Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 16 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.

When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.)

z_c = ______

(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 weights

- normal distribution of weights

- n is large

- σ is known

- σ is unknown

Answer 1

Question 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 ) * 0.28/√(11)
Lower Limit = 3.15 - Z(0.2/2) 0.28/√(11)
Lower Limit = 3.0418
Upper Limit = 3.15 + Z(0.2/2) 0.28/√(11)
Upper Limit = 3.2582
80% Confidence interval is ( 3.04 , 3.26 )
Margin of Error = Z (0.2/2 ) * 0.28/√(11) = 0.12

Question 2

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.2 /2) = 1.282
3.15 ± Z (0.2/2 ) * 0.36/√(14)
Lower Limit = 3.15 - Z(0.2/2) 0.36/√(14)
Lower Limit = 3.0267
Upper Limit = 3.15 + Z(0.2/2) 0.36/√(14)
Upper Limit = 3.2733
80% Confidence interval is ( 3.03 , 3.27 )

Margin of Error = Z (0.2/2 ) * 0.36/√(14) = 0.12

Question 3

Critical value Z(α/2) = Z (0.2 /2) = 1.28

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/√(16)
Lower Limit = 3.15 - Z(0.2/2) 0.34/√(16)
Lower Limit = 3.041
Upper Limit = 3.15 + Z(0.2/2) 0.34/√(16)
Upper Limit = 3.259
80% Confidence interval is ( 3.041 , 3.26 )
Margin of Error = Z (0.2/2 ) * 0.34/√(16) = 0.11

Part b)

- normal distribution of weights

- σ is known