statistics - Given a 95% confidence interval why are we using 1.96 and not 1.64? - Mathematics Stack Exchange
Confusion about confidence intervals using Z scores, and t-tests.
How to Calculate z score of Confidence Interval
[Q] Which confidence level (e.g. 90% vs. 95%) and margin of error (1% vs. 9%) is most appropriate?
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is used because the
confidence interval has only
on each side. The probability for a
score below
is
, and similarly for a
score above
; added together this is
.
would be correct for a
confidence interval, as the two sides (
each) add up to
.
To Find a critical value for a 90% confidence level.
Step 1: Subtract the confidence level from 100% to find the α level: 100% – 90% = 10%.
Step 2: Convert Step 1 to a decimal: 10% = 0.10.
Step 3: Divide Step 2 by 2 (this is called “α/2”). 0.10 = 0.05. This is the area in each tail.
Step 4: Subtract Step 3 from 1 (because we want the area in the middle, not the area in the tail): 1 – 0.05 = .95.
Step 5: Look up the area from Step in the z-table. The area is at z=1.645. This is your critical value for a confidence level of 90%.
http://www.statisticshowto.com/find-a-critical-value/
hope this helps
Hi! I have recently been reading about CI, and hypothesis testing.
Suppose I'm trying to find an estimate for the popluation of a mean, so I take a single sample of size that is large enough for CLT to apply, then I calculate the mean of that sample.
Give CLT, I know that the sampling distribution of the means would be normal, so to calculate the 95% CI for this mean, I can use z-scores, i.e. the CI is mean +- 1.96 * SE.
SE here would be the standard deviation of sample means, but I'm not sure how to proceed further.
In practice, it would be impossible to know this value. (I've seen some sources which use the sample standard deviation to approximate but not sure if that makes sense).
Is this a limitation of Z, and consquently Z tests, and perhaps why for practical hypothesis testing, we use t tests, and not Z?