Question

which z - score is used in a (a) 90%, (b) 98%, and (c) 99.9% confidence interval for a population proportion?
click here to view the table of standard normal cumulative probabilities (page 1).
click here to view the table of standard normal cumulative probabilities (page 2).
a. the z - score used in a 90% confidence interval for a population proportion is
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the significance level

For a \(90\%\) confidence interval, the significance level \(\alpha = 1 - 0.90=0.10\).

Step2: Find the critical value

The critical value \(z_{\alpha/2}\) is the \(z\) - score such that the area in the two tails is \(\alpha\). So the area in one tail is \(\frac{\alpha}{2}=\frac{0.10}{2} = 0.05\).
We want to find \(z\) such that \(P(Z>z)=0.05\), which is equivalent to \(P(Z\leq z)=1 - 0.05 = 0.95\).
Looking up in the standard normal cumulative probabilities table, the \(z\) - score corresponding to a cumulative probability of \(0.95\) is \(z = 1.645\approx1.65\).

Answer:

\(1.65\)