Question

1. find the critical value $z_{\alpha/2}$ needed to construct a(n) 93% confidence interval.

Explanation:

Step1: Determine alpha

Confidence level is 93%, so \(\alpha = 1 - 0.93 = 0.07\).

Step2: Find alpha/2

\(\frac{\alpha}{2}=\frac{0.07}{2} = 0.035\).

Step3: Find the z - score

We need to find \(z_{\alpha/2}\), which is the z - score such that the area to the right of it is 0.035. The area to the left of \(z_{\alpha/2}\) is \(1 - 0.035=0.965\). Using the standard normal distribution table or a calculator (e.g., invNorm function on a TI - 84: invNorm(0.965, 0, 1)), we find that \(z_{0.035}\approx1.81\).

Answer:

The critical value \(z_{\alpha/2}\) for a 93% confidence interval is approximately \(\boldsymbol{1.81}\)