Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 15 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.

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 limitupper limitmargin of error

(b)

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

normal distribution of weightsuniform distribution of weightsσ is knownσ is unknownn is large

(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.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 that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

(d)

Which equation is used to find the sample size n for estimating μ when σ is known?

n =

zσ σ

E

	2

n =

zσ E

σ

	2

     n =

zσ E σ

n =

zσ σ E

Find the sample size necessary for an 80% confidence level with a maximal margin of error  E = 0.08 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

hummingbirds

Answer 1

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

a)

sample mean, xbar = 3.15
sample standard deviation, σ = 0.22
sample size, n = 15

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28

ME = zc * σ/sqrt(n)
ME = 1.28 * 0.22/sqrt(15)
ME = 0.07

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3.15 - 1.28 * 0.22/sqrt(15) , 3.15 + 1.28 * 0.22/sqrt(15))
CI = (3.08 , 3.22)
Lower limt = 3.08
Upper limit = 3.22

b)
normal distribution of weight
σ is known

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

d)

The following information is provided,
Significance Level, α = 0.2, Margin or Error, E = 0.08, σ = 0.22

The critical value for significance level, α = 0.2 is 1.28.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.28 * 0.22/0.08)^2
n = 12.39

Therefore, the sample size needed to satisfy the condition n >= 12.39 and it must be an integer number, we conclude that the minimum required sample size is n = 13
Ans : Sample size, n = 13 0r 12