A sample of n10 observations from Normal distribution yields

A sample of n=10 observations from Normal distribution yields a sample mean \\bar{X}=2.4. Assuming a population standard deviation \\sigma=0.8,

(a) Construct a 95\\% confidence interval for the population mean.

(b) What is the minimum sample size required for estimating the population mean with a margin of error not exceeding 0.4?

Solution

Given that,

sample size (n) = 10

These observations taken from Normal distribution with mean = 2.4 and

population standard deviation () = 0.8

Construct a 95% confidence interval for the population mean(µ).

Xbar - E < µ < Xbar + E

where Xbar = mean = 2.4

E is the margin of error.

E = Z_c * /sqrt(n)

Here population standard deviation is known so we use z-interval.

Z_c is the critical value for normal distribution.

c is the confidence level = 0.95

a = 1 - c = 1-0.95 = 0.05

a/2 = 0.05/2 = 0.025

This we can find by using EXCEL.

syntax :

=normsdist(probability)

where probability = 1 - a/2

Zc = 1.96

E = Z_c * /sqrt(n)

= 1.96 * 0.8 / sqrt(10) = 0.4958

lower limit = Xbar - E = 2.4 - 0.4958 = 1.9042

upper limit = Xbar + E = 2.4 + 0.4958 = 2.8958

The 95% confidence interval for population mean is (1.9042 , 2.8958)

What is the minimum sample size required for estimating the population mean with a margin of error not exceeding 0.4?

Given E = 0.4

We have to calculate n here.

n = [ (Zc * ) / E ]²

n = [(1.96*0.8)/0.4]² = 10