Question

(A) 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 ? = 0.20 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.)

n is large

? is unknown

uniform distribution of weights

? is known

normal distribution of weights

(c) Find the sample size necessary for an 80% confidence level with a maximal margin of error E= 0.07for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
_____ hummingbirds

(B) Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.†Over a period of months, an adult male patient has taken fiveblood tests for uric acid. The mean concentration was x = 5.35 mg/dl.The distribution of uric acid in healthy adult males can be assumed to be normal, with ? = 1.79 mg/dl.

(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. 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.)

? is unknown

uniform distribution of uric acid

?i s known

n is large

normal distribution of uric acid

(c) Find the sample size necessary for a 95% confidence level with maximal margin of error E= 1.06for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)
_____ blood tests

Answer 1

A)a) At 80% confidence interval the critical value is z_0.1 = 1.28

Margin of error = z_0.1 * /

                         = 1.28 * 0.2/

                        = 0.06

The 80% confidence interval is

+/- ME

= 3.15 +/- 0.06

= 3.09, 3.21

Lower limit = 3.09

Upper limit = 3.21

b) is known.

normal distribution of weights.

c) Margin of error = 0.07

or, z_0.1 * / = 0.07

or, 1.28 * 0.2/ = 0.07

or, n = (1.28 * 0.2/0.07)^2

or, n = 14

B) a) At 95% confidence interval the critical value is z_0.025 = 1.96

Margin of error = z_0.025 * /

                         = 1.96 * 1.79/

                         = 1.57

The 95% confidence interval is

+/- ME

= 5.35 +/- 1.57

= 3.78, 6.92

b) is known.

normal distribution of uric acid.

c) Margin of error = 1.06

or, z_0.025 * / = 1.06

or, 1.96 * 1.79/ = 1.06

or, n = (1.96 * 1.79/1.06)^2

or, n = 11