Question

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 σ = 0.20 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (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 error of estimate E = 0.07 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

Given that,

= 3.15

= 0.20

n = 17

A ) At 80% confidence level the z is ,

  = 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.281

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

= 1.281 * (0.20 / 17 )

= 0.06

Margin of error = E = 0.06

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

- E < < + E

3.15 - 0.06 < < 3.15 + 0.06

3.09< < 3.21

Lower limit = 3.09

Upper limit = 3.21

B ) margin of error = E = 0.07

At 80% confidence level the z is ,

  = 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.281

Sample size = n = ((Z_/2 * ) / E)²

= ((1.281 * 0.20 ) / 0.07)²

= 13.3956

= 13

Sample size = 13