Question

compute the critical value ( z_{alpha / 2} ) that corresponds to a 98% level of confidence
click here to view the standard normal distribution table (page 1)
click here to view the standard normal distribution table (page 2).
( z_{alpha / 2}= )
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the significance level $\alpha$

The confidence level is $C = 0.98$. Using the formula $\alpha=1 - C$, we have $\alpha=1 - 0.98=0.02$.

Step2: Calculate $\frac{\alpha}{2}$

$\frac{\alpha}{2}=\frac{0.02}{2}=0.01$.

Step3: Find the $z -$score

We want to find $z_{\alpha/2}$ such that $P(Z>z_{\alpha/2})=\frac{\alpha}{2}=0.01$. This is equivalent to $P(Z\leq z_{\alpha/2})=1 - 0.01 = 0.99$.
Looking up in the standard - normal distribution table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: invNorm(0.99,0,1)), we find that $z_{\alpha/2}\approx2.33$.

Answer:

$2.33$