Question

Assume we need to estimate the mean of a normally-distributed population with great accuracy. Specifically, for significance level α = .01, we must place a confidence interval around μ such that the width of the interval is no larger than .1. Assume that the variance of the population is known to be σ2. How large must the sample size be to assure that the width of the confidence interval is no larger than .1?

Answer 1

`n = 2662.56\sigma^2`

Explanation:

We are given the following in the question:

Significance level = 0.01

Width of interval = 0.1

Population variance =
`\sigma^2`

We have to find the sample size so that the width of the confidence interval is no larger than 0.1

`z_(critical)\text{ at}~\alpha_(0.01) = \pm 2.58`

Formula for sample size:

`n = \displaystyle(z^2\sigma^2)/(E^2)`

where E is the margin of error. Since the confidence interval width is 0.1,

`E = 0.05`

Putting these values in the equation:

`n = \displaystyle((2.58)^2\sigma^2)/((0.05)^2) = 2662.56\sigma^2`

So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.