Question

It is known that the temperature, X, of Eagle flu patients after 3 days of infection is normally distributed with unknown mean, μ degrees and known standard deviation σ = 1.6 degrees. We wish to create a confidence interval for μ. The known standard deviation comes from an analysis of the properties of Eagle flu. We randomly select n=23 eagle flu patients after 3 days of infection. The sample gave a sample mean of x = 101.3 degrees and a sample standard deviation s = 1.8 degrees.

a)What is the critical value for a 96.1% confidence interval for μ?

b) Create a 96.1% confidence interval for μ

,

.

c) How long is the 96.1% confidence interval for μ?

d) How many observations would we need to guarantee that the 96.1% confidence interval has has a length of 1 degree or less?

e) Create a 96.1% prediction interval for a single future measurement of the temperature of an Eagle flu patient after 3 days of infection

,

f) Assuming σ is not known, create a 96.1% confidence interval for μ using this data.

Answer 1

[Used R-Software]

R-commands and output:

alpha=1-96.1/100
alpha
[1] 0.039

xbar=101.3
n=23
sigma=1.6 # Given, so we would refer Z-test

a)
z=qnorm(1-alpha/2)
z
[1] 2.064187
# Critical value for a 96.1% confidence interval for μ is 2.064187

b)
# 96.1% confidence interval for μ is given as

where,

xbar-z*sigma/sqrt(n)
[1] 100.6113
xbar+z*sigma/sqrt(n)
[1] 101.9887
# 96.1% confidence interval for μ is [100.6113, 101.9887].

c)
# length of 96.1% confidence interval for μ = Upper limit minus lower limit

2*z*sigma/sqrt(n)
[1] 1.377321
# 101.9887-100.6113=1.3774

d)
# To find the number of observations to guarantee that the 96.1% confidence interval has has a length of 1 degree or less
2*z*sigma/sqrt(n)<=1
i.e. 2*z*sigma<=sqrt(n)
Squaring on both sides
i.e. (2*z*sigma)^2 <= n

(2*z*sigma)^2
[1] 43.63128
# We would need 44 observations to guarantee that the 96.1% confidence interval has has a length of 1 degree or less.