Question

find each of the following. enter your answers rounded to at least two decimal places.

part 1 of 5

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

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

part 2 of 5

$z_{\alpha / 2}$ for the $87 \\%$ confidence interval

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

Explanation:

Step1: Calculate \(\alpha\) for 99% confidence interval

The confidence level \(C = 0.99\). Then \(\alpha=1 - C=1 - 0.99 = 0.01\). And \(\frac{\alpha}{2}=\frac{0.01}{2}=0.005\).
We look up the \(z\) - value in the standard normal table (or use a calculator with a normal - distribution function). The \(z\) - value corresponding to a right - tail area of \(0.005\) is \(z_{0.005}\). Using a standard normal table or a calculator (\(z=\text{invNorm}(1 - 0.005)\) on a TI - 84 Plus, for example), we get \(z = 2.5758\approx2.58\).

Step2: Calculate \(\alpha\) for 87% confidence interval

The confidence level \(C = 0.87\). Then \(\alpha=1 - C=1 - 0.87=0.13\). And \(\frac{\alpha}{2}=\frac{0.13}{2}=0.065\).
We look up the \(z\) - value. The \(z\) - value corresponding to a right - tail area of \(0.065\) is \(z_{0.065}\). Using a standard normal table or a calculator (\(z=\text{invNorm}(1 - 0.065)\)), we get \(z = 1.51\) (using a more precise calculation: \(z=\text{invNorm}(0.935)\approx1.51\)).

Answer:

For the 99% confidence interval: \(z_{\alpha/2} = 2.58\)
For the 87% confidence interval: \(z_{\alpha/2} = 1.51\)