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A sample has: $$\text{a sample size } n = 70,$$ $$\text{sample standard deviation } s = 184.43,$$ $$\text{and a sample mean } \bar{x} = 564.15.$$

Compute a 95% confidence interval for the population that this sample is derived from.


I know how to compute a confidence interval for a sample. I have to translate that to a population confidence interval. What I'm thinking is to use a $t$ distribution.

The $t$ confidence interval would be defined like this:

$$\bar{x}±t_{α/2,n-1}\frac{s}{\sqrt{n}}$$

So the 95% confidence interval is

$$564.15±t_{0.025,69}\frac{184.43}{\sqrt{70}}$$ This $t$ value is $1.9950$, so the interval will be $$(608.13,520.17)$$

Is this correct?

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  • $\begingroup$ Yes, except the interval should be written $[520.17,608.13]$. $\endgroup$ – Nameless Jun 12 '14 at 19:30

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