Question

Find the critical value z Subscript alpha divided by 2 that corresponds to the given confidence level. 93​%

Answer 1

Answer:

$z_{{\frac{\alpha}{2}}}=z_{0.035}=1.81$

Step-by-step explanation:

We are asked to compute $z_{{\frac{\alpha}{2}}$ that corresponds to confidence level of 93%.

$93\%=\frac{93}{100}=0.93$

$\alpha=1-0.93=0.07$

$\frac{\alpha}{2}=\frac{0.07}{2}$

$\frac{\alpha}{2}=0.035$

Now, we will use normal distribution table to find the z-score that corresponds to area of 0.035.

$z_{0.035}=-1.81$

Therefore, the value of $z_{{\frac{\alpha}{2}}$ is $1.81$.