Statistics and Confidence Intervals. Given the following set of values: 10,11,14,95,73,30,29,9,97,94,70

foyerir 2022-09-11 Answered
Statistics and Confidence Intervals
Given the following set of values:
10,11,14,95,73,30,29,9,97,94,70
How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10
Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:
x z a 2 σ 2 s q r t ( n ) and x + z a 2 σ 2 s q r t ( n ) where x denotes the mean. Here the significance level (a) is 0.01.
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Answers (1)

London Maldonado
Answered 2022-09-12 Author has 13 answers
Step 1
A conventional setup for the problem is that you have a sample X = ( X 1 , X 2 , , X n ) of size n = 11 from a N ( μ , σ 0 2 ) population (by assumption) with σ 0 2 = 10. You have to find a confidence interval for the mean μ.
A suitable pivotal quantity here is
Q ( X , μ ) = n ( X ¯ μ ) σ 0 N ( 0 , 1 )
, where X ¯ = 1 n i = 1 n X i is the sample mean.
So if z α / 2 be such that P ( Z > z α / 2 ) = α / 2 where Z N ( 0 , 1 ), you have
P μ ( z α / 2 Q z α / 2 ) = 1 α μ
You have to use the above to arrive at the form
P μ ( c 1 μ c 2 ) = 1 α μ
100 ( 1 α ) % confidence interval for μ is then [ c 1 , c 2 ].

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