Question

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 = 0.32 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 What conditions are necessary for your calculations? (Select all that apply.) σ is unknown uniform distribution of weights normal distribution of weights σ is known n is large (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.)

Answer 1

a)

sample mean, xbar = 3.15
sample standard deviation, σ = 0.32
sample size, n = 16

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.2816

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3.15 - 1.2816 * 0.32/sqrt(16) , 3.15 + 1.2816 * 0.32/sqrt(16))
CI = (3.05 , 3.25)

ME = zc * σ/sqrt(n)
ME = 1.2816 * 0.32/sqrt(16)
ME = 0.10

Lower limit = 3.05
Upper limit = 3.25

normal distribution of weights

σ is known

n is large

d)

The following information is provided,
Significance Level, α = 0.2, Margin or Error, E = 0.13, σ = 0.32

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 * 0.32/0.13)^2
n = 9.93

Therefore, the sample size needed to satisfy the condition n >= 9.93 and it must be an integer number, we conclude that the minimum required sample size is n = 10
Ans : Sample size, n = 10