Question

Part #1

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 13 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

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

Part #2

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 thirteen blood 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 σ = 1.91 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

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

Answer 1

(1)

(a)

n = Sample Size = 13

= Population SD = 0.32

SE = /

= 0.32/ = 0.0888

= 0.20

From Table, critical values of Z = 1.28

Confidence Interval:

3.15 (1.28 X 0.0888)

= 3.15 0.1136

= ( 3.0364, 3.2636)

So,

Lower limit =3.0364

Upper limit = 3.2636

Margin of error = 0.1136

(d)

So,

Answer is:

35

(2)

(a)

n = Sample Size = 13

= Population SD = 1.91

SE = /

= 1.91 / = 0.5297

= 0.05

From Table, critical values of Z = 1.96

Confidence Interval:

5.35 (1.96 X 0.5297)

= 5.35 1.0382

= (4.3118, 6.3882)

So,

Lower limit =4.3118

Upper limit = 6.3882

Margin of error = 1.0382

(d)

So,

Answer is:

13