Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 11 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 = 0.28 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)Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.13 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

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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 seven 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 = 1.89 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 = 1.10 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)
blood tests

Answer 1

Solution :

Given that,

1) Point estimate = sample mean = $\bar x$ = 3.15 grams

Population standard deviation =   $\sigma$ = 0.28 grams

Sample size = n = 11

At 80% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.80 =0.20

$\alpha$/2 = 0.10

Z$\alpha$/2 = Z_0.10 = 1.282

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.282 * ( 0.28 /  $\sqrt$11 )

= 0.11

At 90% confidence interval estimate of the population mean is,

$\bar x$  ± E

3.15 ± 0.11   

( 3.04, 3.26 )  

lower limit = 3.04

upper limit = 3.26

margin of error = 0.11

b) margin of error = E = 0.13

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [ 1.282 * 0.28 / 0.13]²

n = 7.62

Sample size = n = 8 hummingbirds

2) Point estimate = sample mean = $\bar x$ = 5.35

Population standard deviation =   $\sigma$ = 1.89

Sample size = n = 7

At 95% confidence level

$\alpha$ = 1 - 95%  

$\alpha$ = 1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * ( 1.89 /  $\sqrt$7 )

= 1.40

At 90% confidence interval estimate of the population mean is,

$\bar x$  ± E

5.35 ± 1.40

( 3.95, 6.75 )  

lower limit = 3.95

upper limit = 6.75

margin of error = 1.40

b) margin of error = E = 1.10

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [ 1.96 * 1.89 / 1.10]²

n = 11.34

Sample size = n = 12 blood tests