You mean, a z-table that doesn't have any negative z-values?

Sorry yes z table with no zero

Confidence interval (C.I) for the mean = x̄ ± Z (s/√n),whereby:

x̄ is the sample mean
Z is the Z-value for the desired confidence level (in your case it is 85% or 0.85)
s is the sample standard deviation
n is the number of boxes


So,

C.I = 12.05 ± Z (0.1/√100)
C.I = 12.05 ± 0.01Z


Now I presume your z-table presents the cumulative probability that starts from Z = 0 and goes to the right

i.e z = 0 to z = 3.49



For a confidence level of 85%, it covers 42.5% to the left of Z = 0 and 42.5% to the right of Z = 0. The distribution is symmetrical about Z = 0 so to find each side, we need to divide by 2.


For 42.5% (or 0.425), the closest Z-value is 1.44, where the probability is 0.4251


So we substitute Z = 1.44 into the C.I


C.I = 12.05 ± 0.01(1.44)
C.I = 12.05 ± 0.0144

However, there are a few variations of the z-table that starts from 0. Some will show the probability starting from the left side of z = 0.

If possible, I would like to see the table you are currently using.

May i know how did u get 42.5%? Sadly the comment section doesn't allow me to take pic of my table and post it here

85 ÷ 2 = 42.5

You can post an answer to your own question, in the form of a picture. Just upload a picture of the z table

Thank you for the replies! I uploaded my z test table. Sorry abit confused, y divide by 2? Possible to do it this way? Alpha/2= (100-85)/2? Then without % will be 0.075

Just saw your z-table. It is the same style as the one I expected.

An 85% confidence interval means 85% of the time, the values will lie with that range from the mean.

The way you suggested is based on the method that Yong Kc has posted.

But it will not work for the z-table you have as that α/2 = 0.075 is for the ends of the curve that falls outside the confidence interval.

i.e you cant look at your z-table and look for area = 0.075 because that would give you the probability that started from the middle of the curve, not the ends.

You would have obtained the incorrect Z-value of 0.19 instead.

So, 85% confidence interval means a probability or area of 0.85.

Since this area is symmetrical about the middle of the curve, and given the way your z-table is presented,

You'll need to divide by 2 , such that you'll have 0.425 on the right side and 0.425 on the left side.

The two areas are mirror images of each other.

Look for the closest area in your table, and you'll see that it is 0.4251 (where the z-value is 1.44)


So we can say for a 85% confidence level, there is 85% chance that the values will lie within 1.44 standard deviations from the mean.


Hence the 12.05 ± 1.44 (0.1/√100)

12.05 ± 0.0144


Now,
12.05 - 0.0144 = 12.0356
12.05 + 0.0144 = 12.0644

We can say that there is a 85% chance that the values lie between 12.0356 and 12.0644.

Oh,i finally understand! Thank you very very much for the patience and explanation!

Welcome