Question

A random sample of 100100 observations produced a mean of x¯¯¯=38.6x¯=38.6 from a population with a normal distribution and a standard deviation σ=4.42σ=4.42.

(a) Find a 9090% confidence interval for μμ
≤μ≤≤μ≤

(b) Find a 9595% confidence interval for μμ
≤μ≤≤μ≤

(c) Find a 9999% confidence interval for μμ
≤μ≤≤μ≤

Answer 1

Solution

Given that,

= 38.6

= 4.42

n = 100

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* (/n)

= 1.645 * (4.42 / 100 )

= 0.7

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

- E < < + E

38.6 - 0.7 < < 38.6 +0.7

37.9 < < 39.3

b ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z_/2* (/n)

= 1.960 * (4.42 / 100 )

= 0.9

At 95% confidence interval estimate of the population mean is,

- E < < + E

38.6 - 0.9 < < 38.6 +0.9

37.7 < < 39.5

c ) At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z_/2* (/n)

= 2.576 * (4.42 / 100 )

= 1.1

At 99% confidence interval estimate of the population mean is,

- E < < + E

38.6 - 1.1 < < 38.6 + 1.1

37.5 < < 39.7