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What is the z-score for 95% confidence interval?
The z-score for a two-sided 95% confidence interval is 1.959, which is the 97.5-th quantile of the standard normal distribution N(0,1).
What is the z-score for 99% confidence interval?
The z-score for a two-sided 99% confidence interval is 2.807, which is the 99.5-th quantile of the standard normal distribution N(0,1).
How to calculate confidence interval?
To calculate a confidence interval (two-sided), you need to follow these steps:
- Let's say the sample size is
100. - Find the mean value of your sample. Assume it's
3. - Determine the standard deviation of the sample. Let's say it's
0.5. - Choose the confidence level. The most common confidence level is
95%. - In the statistical table find the Z(0.95)-score, i.e., the 97.5th quantile of N(0,1) – in our case, it's
1.959. - Compute the standard error as
σ/√n = 0.5/√100 = 0.05. - Multiply this value by the z-score to obtain the margin of error:
0.05 × 1.959 = 0.098. - Add and subtract the margin of error from the mean value to obtain the confidence interval. In our case, the confidence interval is between 2.902 and 3.098.
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?
$1.96$ is used because the $95\%$ confidence interval has only $2.5\%$ on each side. The probability for a $z$ score below $-1.96$ is $2.5\%$, and similarly for a $z$ score above $+1.96$; added together this is $5\%$. $1.64$ would be correct for a $90\%$ confidence interval, as the two sides ($5\%$ each) add up to $10\%$.
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