Question

Cosmic radiation levels rise with increasing altitude, prompting researchers to consider how pilots and flight crews are affected by increased exposure to cosmic radiation. The Centers for Disease Control and Prevention (CDC) reports that the National Council on Radiation Protection and Measurements estimates that flight crew members have a mean annual radiation exposure of 3.07 millisievert (mSv) per year.† Suppose that the estimated mean exposure for flight crew members is based on a random sample of 100 flight crew members.

Let μ denote the true mean annual cosmic radiation exposure for all flight crew members. Although σ, the true population standard deviation, is not usually known, suppose for illustrative purposes that

σ = 0.35 mSv

is known. Because the sample size is large and σ is known, a 95% confidence interval for μ is

x ± (z critical value)

σ

n

=

3.07 ± (1.96)

0.35

100

=

3.07 ± 0.069

=

(3.001, 3.139)

Based on this sample, plausible values of μ, the true mean annual cosmic radiation exposure for all flight crew members, are between 3.001 and 3.139 mSv. One mSv is the average annual exposure in the United States due to normal exposure to background radiation. This means that it is estimated that the mean annual exposure of flight crew members is somewhere between about 3 and 3.14 times greater than for the general population. A 95% confidence level is associated with the method used to produce this interval estimate.

Which of the following is not required for using the method described above if sample size n is large?

that the population has normal distribution that the population standard deviation σ is known     that the sample mean x is known that the sample is random

Find a 90% confidence interval for μ. (Round your answers to three decimal places.)

,

Answer 1

Which of the following is not required for using the method described above if sample size n is large?

It is not required that the population has normal distribution that the population standard deviation σ is known   

Find a 90% confidence interval for μ. (Round your answers to three decimal places.)

x=3.07
z critical value=1.64
σ=0.35
n=100

a 90% confidence interval for μ is

=x+/- z critical value*σ/sqrt(n)

=(3.07-1.64*0.35/sqrt(100),3.07+1.64*0.35/sqrt(100))

=(3.0126,3.1274)

90% confidence interval for μ=(3.0126,3.1274)