Question

part 3 of 5
$z_{\alpha / 2}$ for the $90 \\%$ confidence interval
$z_{\alpha / 2}=\square$
part 4 of 5
$z_{\alpha / 2}$ for the $95 \\%$ confidence interval
$z_{\alpha / 2}=\square$
part 5 of 5
$z_{\alpha / 2}$ for the $92 \\%$ confidence interval
$z_{\alpha / 2}=\square$

Explanation:

Step1: Calculate \(\alpha\) for each confidence level

• For a 90% confidence interval: \(\alpha = 1 - 0.90=0.10\)
• For a 95% confidence interval: \(\alpha = 1 - 0.95 = 0.05\)
• For a 92% confidence interval: \(\alpha=1 - 0.92=0.08\)

Step2: Calculate \(\alpha/2\)

• For 90% confidence interval: \(\alpha/2=\frac{0.10}{2}=0.05\)
• For 95% confidence interval: \(\alpha/2=\frac{0.05}{2}=0.025\)
• For 92% confidence interval: \(\alpha/2=\frac{0.08}{2}=0.04\)

Step3: Find \(z_{\alpha/2}\) using the standard normal distribution table or a calculator

• For \(\alpha/2 = 0.05\), \(z_{0.05}\approx1.645\) (from the standard - normal table, the \(z\) - value such that the area to the right of \(z\) is \(0.05\))
• For \(\alpha/2=0.025\), \(z_{0.025}\approx1.96\) (the \(z\) - value such that the area to the right of \(z\) is \(0.025\))
• For \(\alpha/2 = 0.04\), using a calculator or more detailed standard - normal tables (or the formula \(z=\Phi^{-1}(1 - \alpha/2)\) where \(\Phi\) is the cumulative distribution function of the standard normal distribution), \(z_{0.04}\approx1.75\)

Answer:

• For the 90% confidence interval: \(z_{\alpha/2} = 1.645\)
• For the 95% confidence interval: \(z_{\alpha/2}= 1.96\)
• For the 92% confidence interval: \(z_{\alpha/2}\approx1.75\)