Question

part 3 of 5

$z_{\alpha/2}$ for the $99\\%$ confidence interval

$z_{\alpha/2}=\square$

Explanation:

Step1: Calculate the significance level ($\alpha$)

The confidence level is $99\% = 0.99$. Using the formula $\alpha=1 -$ confidence level, we get $\alpha = 1- 0.99=0.01$.

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

Divide $\alpha$ by 2. So, $\frac{\alpha}{2}=\frac{0.01}{2}=0.005$.

Step3: Find the $z$-score

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

Answer:

$2.576$