Question

Let \(\bar{x}\) be the observed mean of a random sample of size \(n\) from a distribution having mean \(\mu\) and known variance \(\sigma^{2}\). Find \(n\) so that \(\bar{x}-\sigma / 4\) to \(\bar{x}+\sigma / 4\) is an approximate \(95 \%\) confidence interval for \(\mu\).

Short answer

The sample size should be \(n = 62\) to obtain the desired confidence interval.

Step-by-step solution

Understand the Basic Concept of Confidence Interval

Normally, a \(95\%\) confidence interval for the mean with known variance \(\sigma^{2}\) is given by \(\bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(Z_{\alpha/2}\) is the z-value for a significance level of \(\alpha/2\), and \(n\) is the sample size. The value of \(Z_{\alpha/2}\) for a \(95\%\) confidence is approximately \(1.96\). However, in this case the interval is given by \(\bar{x} \pm \frac{\sigma}{4}\). The task, then, will be to equate these two statements and solve for \(n\).

Set Up the Equation

The equation comes from equating the two ways of creating the confidence interval: \[\frac{\sigma}{4} = Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\] Becaue we're dealing with a \(95 \%\)-confidence interval, \(Z_{\alpha/2}\) is approximately \(1.96\)

Solve the Equation for n

First, cancel out the \(\sigma\) on both sides of the equation: \[\frac{1}{4} = 1.96 \cdot \frac{1}{\sqrt{n}}\] Then solve for \(n\) by squaring both sides and taking the reciprocal: \[n \approx \left(\frac{1.96}{0.25}\right)^2 = 61.4656\] Because the sample size must be an integer, we round up to \(n = 62\)