Question

find ( z_{alpha/2} ) for the 99% confidence interval (draw the graph and show all your work)

Explanation:

Step1: Calculate the significance level

The confidence level is \(C = 0.99\). The significance level \(\alpha=1 - C\). So, \(\alpha=1 - 0.99=0.01\).

Step2: Find \(\frac{\alpha}{2}\)

\(\frac{\alpha}{2}=\frac{0.01}{2}=0.005\).

Step3: Determine \(z_{\alpha/2}\)

We want to find \(z\) - value such that \(P(Z>z_{\alpha/2})=\frac{\alpha}{2} = 0.005\), or \(P(Z\leq z_{\alpha/2})=1-\frac{\alpha}{2}=0.995\).
Looking up in the standard normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: invNorm(0.995,0,1)), we get \(z_{\alpha/2}=2.576\).

Graph Description:

The standard normal distribution \(Z\sim N(0,1)\) is a bell - shaped curve. The total area under the curve is \(1\). The confidence level of \(99\%\) means that the middle area of the curve is \(0.99\). The two - tail areas (left and right) each have an area of \(\frac{1 - 0.99}{2}=0.005\). The \(z\) - values that mark the boundaries of the middle \(99\%\) area are \(-z_{\alpha/2}\) and \(z_{\alpha/2}\), where \(z_{\alpha/2}\) is the positive value we calculated.

Answer:

\(z_{\alpha/2}=2.576\)