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 1.65
(round to two decimal places as needed.)
b. the z - score used in a 98% confidence interval for a population proportion is
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the significance level $\alpha$

For a \(98\%\) confidence interval, \(\alpha = 1 - 0.98=0.02\).

Step2: Find the value of \(\frac{\alpha}{2}\)

\(\frac{\alpha}{2}=\frac{0.02}{2} = 0.01\).

Step3: Determine the \(z\)-score

We want to find \(z\) such that \(P(Z>z)=0.01\), which is equivalent to \(P(Z\leq z)=1 - 0.01 = 0.99\). Looking up in the standard - normal cumulative probabilities table, the \(z\)-score corresponding to a cumulative probability of \(0.99\) is \(z = 2.33\).

Answer:

\(2.33\)