Question

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

Answer 1

Hint: 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 \[\left( 1-\alpha \right)*100%=88%\]
\[\alpha =0.12\]
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.