Question

What is the critical value ${{z}_{{}^{\alpha }/{}_{2}}}$ that corresponds to $88\%$ confidence level?

Answer 1

For solving this question you should know about confidence level and the critical value of this. In this question we will find the value of z – score for this. And here it corresponds to $88\%$ confidence level so it will be less than the full confidence. So, we will calculate the z – score for less area.

Complete step by step solution:
In this question it is asked to find the critical value ${{z}_{{}^{\alpha }/{}_{2}}}$ which corresponds to the $88\%$ confidence level.
So, if we read the question carefully then it is asking for the critical value ${{z}_{{}^{\alpha }/{}_{2}}}$ of a confidence level and this ${{z}_{{}^{\alpha }/{}_{2}}}$ value corresponds to the $88\%$ confidence level.
So, if we find the confidence, then:
For confidence $\left( 1-\alpha \right)*100$
$\alpha =0.12$
and at level $\alpha =0.12$, the critical values are given by $+{{z}_{0.12/2}},-{{z}_{0.12/2}}$.
So, if we find ${{z}_{0.12/2}}$
Then: ${{z}_{{}^{\alpha }/{}_{2}}}$ is the z – score such that the area to the right of ${{z}_{{}^{\alpha }/{}_{2}}}$ is ${}^{\alpha }/{}_{2}$ (which is under the standard normal curve)
So, the P $\left( z>{{z}_{{}^{\alpha }/{}_{2}}} \right)=0.06$
Then $\left( z<{{z}_{\alpha }} \right)=0.94$
As we know that 0.9400 is not present in the z- score table containing area to left of z, but we can use here linear interpolation.

x	0.03	0.04	0.05	0.06
1.3	0.9082	0.9099	0.9115	0.9131
1.4	0.9236	0.9251	0.9265	0.9279
1.5	0.9370	0.9382	0.9394	0.9406
1.6	0.9484	0.9495	0.9505	0.9515

We note that nearby 0.9394 has a z – score 1.55 and 0.9406 has a z – score 1.56. So, here we can use the linear interpolation to find out the approximate z – score = 1.55.
0.9400 is $\dfrac{0.9400-0.9406}{0.9394-0.9406}=\dfrac{1}{2}$ way from 0.9406 to 0.9394.
So, the z – score of 0.9400 is approximately $\dfrac{1}{2}$ of the way from 1.56 to 1.55.
Z – score $=1.56+\left( \left( 1.55 \right)-\left( 1.56 \right) \right)*\dfrac{1}{2}=1.5550$

Thus, the z – score of ${{z}_{{}^{\alpha }/{}_{2}}}$ or ${{z}_{0.06}}=1.555$.

Note:
During solving this question you should be careful regarding calculating the values of z – score at any alpha. And if the exact z – score at that $\alpha$ is not available then always take the nearly two z – scores and then apply their linear interpolation for getting the values of z – score.