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 = 0.32 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

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

uniform distribution of weights

normal distribution of weights

σ is known

n is large

σ is unknown

(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.

The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.    

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.

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.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.06 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
____________________ hummingbirds

Answer 1

Solution:-

Given that,

Point estimate = sample mean = =3.15

Population standard deviation = =0.32

Sample size = n =18

a)

At 80% confidence level

= 1-0.80% =1-0.80 =0.20

/2 =0.020/ 2= 0.10

Z/2 = Z_{0.10= 1.28}

Z/2 = 1.28

Margin of error = E = Z/2 * ( /n)

= 1.28 * (0.32 /18 )

= 0.0967

At 80 % confidence interval estimate of the population mean is,

- E < < + E

3.15 - 0.0967 <   < 3.15+ 0.0967

3.05<   < 3.25

( 3.05, 3.25 )

Lower limit =3.05

Upper limit = 3.25

b) σ is known

C) 80% of the intervals created using this method will contain the true average blood plasma volume in male firefighters

D)

Margin of error = E = 0.06

Z_/2 =1.28

sample size = n = [Z_/2* / E] ²

n = ( 1.28*0.32 / 0.06 )²

n = 47

Sample size = n = 47