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 2 33
(round to two decimal places as needed.)
c. the z - score used in a 99.9% confidence interval for a population proportion is
(round to two decimal places as needed.)

Explanation:

Step1: Find the significance level ($\alpha$)

The formula for $\alpha$ is $\alpha = 1 -$ confidence level.
For a $99.9\%$ confidence level, $\alpha=1 - 0.999 = 0.001$.

Step2: Find $\frac{\alpha}{2}$

$\frac{\alpha}{2}=\frac{0.001}{2}=0.0005$.

Step3: Find the $z$-score

We want to find $z$ such that $P(Z < z)=1-\frac{\alpha}{2}=1 - 0.0005 = 0.9995$.
Looking at the standard - normal cumulative probabilities table, the $z$-score corresponding to a cumulative probability of $0.9995$ is $3.29$.

Answer:

$3.29$