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The question:

The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on specimens of amber 75 million years ago give these percents of nitrogen: 63.4, 65.0, 64.4, 63.3, 54.8, 64.5, 60.8, 49.1, 51.0. Construct a 95% confidence interval for the true average % nitrogen in the atmosphere at this time.

My attempt:

I used the following formula for confidence intervals from my textbook:

"A 100(1-$\alpha$)% confidence interval for $\mu$ is $(\bar{X} - t_{n-1, 1-\frac{\alpha}{2}} \frac{S}{\sqrt{n}}, \bar{X} + t_{n-1, 1-\frac{\alpha}{2}} \frac{S}{\sqrt{n}})$."

I know $\bar{X}$ is 59.6, and $n$ is 9. But there are 2 things I'm having trouble with:

Firstly, in the answers they say that $S$ is 6.25 when I am calculating it to be 5.89.

Secondly, I'm having trouble finding $t_{n-1, 1-\frac{\alpha}{2}}$. According to the solutions, it is 2.31. But I'm not sure where they got that number from, or how to find $t_{n-1, 1-\frac{\alpha}{2}}$ in general? I thought it was supposed to be 1.96 but that's incorrect. Any help is appreciated.

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    $\begingroup$ 1.96 is $z_{0.975}$, but you’re apparently using $t$ instead of $z$. A z-score uses the normal distribution, but $t$ comes from Student’s t distribution. $\endgroup$ – Joe Aug 29 at 12:23
  • $\begingroup$ Got it, thanks. I also now understand where I went wrong in calculating S - I accidentally calculated the population standard deviation instead of the sample one. $\endgroup$ – scott Aug 29 at 12:33
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    $\begingroup$ My (unbiased) estimated standard deviation is $6.255286653$. Rounded to two decimal places it is $\hat \sigma=6.26$ $\endgroup$ – callculus Aug 29 at 12:49
  • $\begingroup$ How do we know when to use the sample standard deviation vs the population standard deviation, and the t distribution rather than a normal distribution, when calculating confidence intervals? $\endgroup$ – scott Aug 29 at 13:10
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Values of the $t$-distribution are read from a table, for example here: http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf

Since you have nine samples, the degrees of freedom is equal to $n-1 = 8$.

You are using the two-tails confidence interval with $\alpha=0.05$ and therefore one half of the tail is equal to $1-0.025$.

Therefore, you must read off the value $t_{8,0.025}$, which, according to the table, is approximately equal to $2.306$.

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  • $\begingroup$ Thanks, I understand that now. Just a quick question - how do i know when to use $t_{n-1, 1-\frac{\alpha}{2}}$ and when to use $z_{1 - \frac{\alpha}{2}}$? $\endgroup$ – scott Aug 29 at 13:04
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    $\begingroup$ The $t$-distribution applies to cases where you are trying to find out the true mean of a population for which you don't know the true variance. But if the number of samples is large (let's say, more than 30), the $t$-distribution approximates the $z$-distribution quite well. So in summary, if you don't know the true variance AND the number of samples is small, use $t$; otherwise use $z$. $\endgroup$ – Matti P. Aug 30 at 5:16
  • $\begingroup$ So when I use t, I use the sample standard deviation, and when I use z, I use the population standard deviation? Is this correct? $\endgroup$ – scott Aug 30 at 8:58

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