Question

find the value of $z_{\alpha / 2}$ that corresponds to a confidence level of $89.48\\%$.

a. $-1.62$
b. $1.25$
c. $0.0526$
d. $1.62$

Explanation:

Step1: Calculate $\alpha$

The confidence level is \(C = 89.48\%=0.8948\). Using the formula \(\alpha=1 - C\), we have \(\alpha=1 - 0.8948 = 0.1052\).

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

\(\frac{\alpha}{2}=\frac{0.1052}{2}=0.0526\).

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

We want to find \(z\) such that \(P(Z>z_{\alpha/2})=\frac{\alpha}{2}=0.0526\), or \(P(Z\leq z_{\alpha/2})=1 - 0.0526 = 0.9474\).
Looking up in the standard - normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: invNorm(0.9474)), we get \(z_{\alpha/2}=1.62\).

Answer:

D. \(1.62\)