Question

Find the value of z /2 that corresponds to a confidence level of 89.48%.

Answer 1

Answer:  $1.6202$

Step-by-step explanation:

The given confidence level : c= 89.48%.

⇒ c= 0.8948   [ in decimal]

Then, the significance level = $\alpha= 1-c$

$=\alpha= 1-0.8948=0.1052$

Now, to find the value of $z_{\alpha/2}$ , we need to find the two -tailed z-value using z-table for $\alpha=0.1052$ or we can also find the one-tailed z-value for  for $\alpha/2=0.1052/2=0.0526$

Using z-value table , the value is $1.6202$

Hence , $z_{\alpha/2}=1.6202$ that corresponds to a confidence level of 89.48%.

Answer 2

Using confidence interval concepts, it is found that the critical value is $z_{\frac{\alpha}{2}} = 1.62$

For a confidence level of $\alpha$, the critical value of z is z with a p-value of:

$\frac{1 + \alpha}{2}$

• The p-value is found looking at the z-table.

In this problem, confidence level of 89.48%, thus $\alpha = 0.8948$ the p-value of the z-score is:

$\frac{1 + 0.8948}{2} = 0.9474$

Looking at the z-table, this value is $z_{\frac{\alpha}{2}} = 1.62$.

A similar problem is given at https://brainly.com/question/15354742