The mean of a sample size n = 35 is 1860. The standard deviation of the sample is 102 and the population is normally distributed. Construct a 99% confidence interval estimate of the mean of the population. I know the answer is 1813 < u < 1907, but how do I get that answer? I'm not sure how to solve it.

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First we calculate the test statistic, which we do using a t-table because we don’t know the population standard deviation. This table needs the number of degrees of freedom, n-1=34.

T=(1860-µ)/(102/√35) where µ is the population mean. Look up T for 34 dof and a 2-tail test at a confidence level of 99% (significance level 100-99=1% or 0.01). T=2.73 approx. So µ is defined by a range, T=±2.73, and µ=1860±2.73(102/√35)=1860±47, so 1813<µ<1907.

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