For a 90% confidence interval, a .10 and /2 =.05 so 2/2 Z 05 which is the z-Score leaving .05 in each tail. I could find this scanning my z-table 'backwords' looking for the z-score that has an area of .45000 between 0 and Z. My table doesn't give this to me directly, but I can see that this z-score would be between 1.64 and 1.65. I could also find this z-score using my t-table. Here if I want the area in the tail to be, .05, the area to the left of my mark is .95 (this causes errors for some as they are making a 90% confidence interval, but using the column labeled .95). Looking at the bottom line of the t-table (for df = 00) and the column for .95, we find Z 05 = 1.645. The third way we can get Z 05 is to go to 'distr' on our calculator to find InvNorm(.05) = -1.645. This tells us that -1.645 is the z-score that leaves an area of .05 in the left tail. By symmetry we know that 1.645 will leave .05 in the right tail.

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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For a 90% confidence interval, a = 10% = .10 and a/2 = .05 so Z«, = Z.05 which is the z-score leaving .05 in each tail.
I could find this scanning my z-table 'backwords' looking for the z-score that has an area of .45000 between 0 and Z. My table
doesn't give this to me directly, but I can see that this z-score would be between 1.64 and 1.65.
I could also find this z-score using my t-table. Here if I want the area in the tail to be, .05, the area to the left of my mark is .95
(this causes errors for some as they are making a 90% confidence interval, but using the column labeled .95). Looking at the
bottom line of the t-table (for df = 0) and the column for .95, we find Z 05 = 1. 645.
The third way we can get Z 05 is to go to 'distr' on our calculator to find InvNorm(.05) = -1.645. This tells us that -1.645 is the
z-score that leaves an area of .05 in the left tail. By symmetry we know that 1.645 will leave .05 in the right tail.
Transcribed Image Text:For a 90% confidence interval, a = 10% = .10 and a/2 = .05 so Z«, = Z.05 which is the z-score leaving .05 in each tail. I could find this scanning my z-table 'backwords' looking for the z-score that has an area of .45000 between 0 and Z. My table doesn't give this to me directly, but I can see that this z-score would be between 1.64 and 1.65. I could also find this z-score using my t-table. Here if I want the area in the tail to be, .05, the area to the left of my mark is .95 (this causes errors for some as they are making a 90% confidence interval, but using the column labeled .95). Looking at the bottom line of the t-table (for df = 0) and the column for .95, we find Z 05 = 1. 645. The third way we can get Z 05 is to go to 'distr' on our calculator to find InvNorm(.05) = -1.645. This tells us that -1.645 is the z-score that leaves an area of .05 in the left tail. By symmetry we know that 1.645 will leave .05 in the right tail.
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