How to determine 95% confidence interval for known population variance?

Question

Consider group 1 only and suppose that the population variance is 12. What is the 95% confidence interval for population 1? Group 1 (Experimental Group) X bar 1 = 33 Sample size = 10 Population variance = 12. I know that standard deviation is square root of variance. So sqrt 12 = 3.464101615 So sqrt 10 = 3.16227766 Standard error = population standard error/sqrt 10. So what I did: (X bar 1) +/- (1.96)((sqrt (12)/sqrt (10)) 33 +/- (1.96)(1.095445115) 33 +/- (2.147072425) = 30.85 < u < 35.15 I have to round it to two significant digits. So lower bound = 31 So upper bound = 35

But I am getting this question wrong somehow. Can somehow help clarify where I am making mistakes. Any help is appreciated it. Thank you

Answer 1

From here: https://www.socscistatistics.com/confidenceinterval/default3.aspx

Yours looks fine.

M = 33
Z = 1.96
sM = √(3.4642/10) = 1.1

μ = M ± Z(sM)
μ = 33 ± 1.96*1.1
μ = 33 ± 2.15 

Result:

M = 33, 95% CI [30.85, 35.15].
You can be 95% confident that the population mean (μ) falls between 30.85 and 35.15.