Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 17 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 = 0.40 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.)
normal distribution of weights
is unknown
uniform distribution of weights
is known
n is large
Correct: Your answer is correct.

(c)
Interpret your results in the context of this problem.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval.
We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.
Correct: Your answer is correct.
(d)
Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.13 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Incorrect: Your answer is incorrect.
hummingbirds

c) Interpretation: 80 percent of the time, this range will include the actual average weight of Allen's hummingbirds, hence its reliability may be counted on

d) sample size [tex]n \approx 27[/tex].

What conditions are necessary for your calculations?

Generally,
Here n = 17

T= 3.15

[tex]\sigma=0.40[/tex]

confidence level =c = 0.80

Here we will usethe z critical value.

z critical value for (1+c) /2 = (1+0.80)/2

z critical value for (1+c) /2= 0.9

Zc = 1.28

Critical value = 1.28

a) Margin of error (E) :

[tex]E = Zc*\frac{\sigma}{\sqrt{n}}\\\\\E = 1.28*\frac{0.40}{\sqrt{17}}[/tex]

E =0.124

Margin of error = E =0.124

Lower limit = x-E

L= 3.15 - 0.124

L= 3.026

Lower limit =3.026

Upper limit = x + E

U= 3.15 + 0.124

U= 3.274

Upper limit = 3.274

The following is a confidence interval with a value of 80 percent for the mean weights of Allen's hummingbirds in the area under study: (3.02,3.28)

b) Conditions:[tex]\sigma[/tex] is understood to have a regular distribution of weights

c) Interpretation: 80 percent of the time, this range will include the actual average weight of Allen's hummingbirds, hence its reliability may be counted on.

d) When is already known, the following equation may be used to determine the appropriate number of samples, n:

[tex]n=(\frac{z_c*\sigma }{E})^2[/tex]

Margin of error = 0.13

Sample size (n):

[tex]n= (\frac{z_c*\sigma }{E})^2[/tex]

[tex]n=(\frac{ 1.28*0.36}{0.09})^2[/tex]

n = 26.21

[tex]n \approx 27[/tex]