Question

A random sample of 120 observations produced a mean of ?⎯⎯⎯ =29.4 from a population with a normal distribution and a standard deviation ?=2.45.

(a) Find a 95% confidence interval for ?
___ ≤ ? ≤ ___

(b) Find a 90% confidence interval for ?
___ ≤ ? ≤ ___

(c) Find a 99% confidence interval for ?μ
___ ≤ ? ≤ ___

Answer 1

Solution :

Given that,

(a)

Sample size = n = 120

Z_/2 = 1.96

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

= 1.96 * (2.45 / 120)

Margin of error = E = 0.44

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

- E + E

29.4 - 0.44 29.4 + 0.44

28.96 29.84

(b)

Sample size = n = 120

Z_/2 = 1.645

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

= 1.645 * (2.45 / 120)

Margin of error = E = 0.37

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

- E + E

29.4 - 0.37 29.4 + 0.37

29.03 29.77

(c)

Sample size = n = 120

Z_/2 = 2.576

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

= 2.576 * (2.45 / 120)

Margin of error = E = 0.58

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

- E + E

29.4 - 0.58 29.4 + 0.58

28.82 29.98