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 = 0.38 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 (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

Answer 1

We have given :

n = sample size = 15

x̄ = sample mean = 3.15

σ = population standard deviation = 0.38

## 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

Answer : μ = [ x̄  ± Margin of error ]

Margin of error = critical value * standard error

critical value = z (α /2 ) = ± 1.28

standard error =  σ / √n = 0.09811

Margin of error = 1.28 * 0.09811 = 0.125587 ie 0.13

lower limit =  x̄​​​​​​​ -  Margin of error = 3.15 - 0.13 = 3.02

upper limit = x̄​​​​​​ ​  +  Margin of error = 3.15 +  0.13 = 3.28

Interpretation : We can say 80 % confident that population mean is lies within 3.02 to 3.28 .

#### (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

Answer : we have to find sample size n

n = [( z (α /2 ) )^2 *  σ^2 ] / ME ^2

z (α /2 ) = 1.28 ( from table )

n = [ ( 1.28)^2 * ( 0.38 ) ^2 ] / ( 0.16 ^2)

n = 9.2416 ie 9

sample size is equal to 9