Question

Suppose that a random sample of size 10 from the N(µ, 9) distribution resulted in a sample mean of x = 2.7. Give an 85% condidence interval for µ. (b) Suppose that a random sample of size 100 from any distribution with mean µ and variance σ 2 resulted in a sample mean of x = 2.7 and a sample variance of s 2 = 1.1. Give a 90% condidence interval for µ. (c) Suppose that a random sample of size 10 from the N(µ, σ2 ) distribution resulted in a sample mean of x = 2.7 and a sample variance of s 2 = 1.1. Give a 95% condidence interval for µ

Answer 1

a)
`CI: 1.3<\mu<4.1`

b)
`CI: 2.5<\mu<2.9`

c)
`CI: 2.0<\mu<3.4`

Explanation:

a) We have a sample of size n=10 and a sample mean of x=2.7.

The σ² of the population is known and is σ²=9 or σ=3.

For a CI of 85%, the z-value is 1.44.

Then the lower and upper limits of the CI are:

`LL=M-z*(\sigma)/(√(n) ) =2.7-1.44*(3)/(√(10) )=2.7-1.44*(3)/(3.162) =2.7-1.4=1.3\\\\UL=M+z*(\sigma)/(√(n) ) =2.7+1.4=4.1`

The 85% confidence interval for the mean is defined by the values LL=1.3 and UL=4.1.

`CI: 1.3<\mu<4.1`

b) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 100

x(mean) = 2.7

s² = 1.1

The CI is defined as

`x-t*(s)/(√(n) ) <\mu<x+t*(s)/(√(n) )`

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=100-1=99 and the CI is of 90%.

We look up in a t-table and the t-value for this conditions is 1.6604.

We can now calculate the CI

`LL: x-t*(s)/(√(n) )=2.7-1.6604*(√(1.1) )/(√(100) ) =2.7-0.2=2.5\\\\UL: x+t*(s)/(√(n) )=2.7+0.2=2.9`

The 90% confidence interval for the mean is defined by the values LL=2.5 and UL=2.9.

`CI: 2.5<\mu<2.9`

c) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 10

x(mean) = 2.7

s² = 1.1

The CI is defined as

`x-t*(s)/(√(n) ) <\mu<x+t*(s)/(√(n) )`

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=10-1=9 and the CI is of 95%.

We look up in a t-table and the t-value for this conditions is 2.2622.

We can now calculate the CI

`LL: x-t*(s)/(√(n) )=2.7-2.2622*(√(1.1) )/(√(10) ) =2.7-0.7=2.0\\\\UL: x+t*(s)/(√(n) )=2.7+0.7=3.4`

The 95% confidence interval for the mean is defined by the values LL=2.0 and UL=3.4.

`CI: 2.0<\mu<3.4`