Question

find the critical value ( z_{alpha / 2} ) that corresponds to the confidence level 82%.

( z_{alpha / 2}=square )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate $\alpha$

The confidence level is \(C = 82\%=0.82\). Using the formula \(\alpha=1 - C\), we have \(\alpha=1 - 0.82 = 0.18\).

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

Divide \(\alpha\) by 2. So, \(\frac{\alpha}{2}=\frac{0.18}{2}=0.09\).

Step3: Find the \(z\) - value

We want to find \(z_{\alpha/2}\) such that \(P(Z>z_{\alpha/2})=\frac{\alpha}{2} = 0.09\), or equivalently \(P(Z\leq z_{\alpha/2})=1 - 0.09=0.91\).
Looking up in the standard normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: invNorm(0.91,0,1)), we find \(z_{\alpha/2}\approx1.34\).

Answer:

\(1.34\)