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how do I go about finding the $95\%$ confidence interval when I have $n=12, s_x=0.66$ and $u=35.72$ and also how many more samples would I need to reduce the "length" of the interval by half?

so I looked in the solution and it says:

$\left(35.72-2.201\cdot 0.66/\sqrt{12},35.72+2.201\cdot 0.66/\sqrt{12}\right)$, so.. I got really confused about where $2.201$ comes from? is it something they calculated or do they look up stuff like that?

Asaf Karagila
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jgh123987
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1 Answers1

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You don´t know the the variance of the population. And addtitonally the sample size is not larger than 30. Thus you have to use the t-statistic. The estimated interval is:

$\Large{\left[ \overline x-t_{\left( 1-\frac{\alpha}{2},n-1 \right)}\cdot \frac{s}{\sqrt n};\overline x+t_{\left( 1-\frac{\alpha}{2},n-1 \right)}\cdot \frac{s}{\sqrt n} \right]}$

$1-\frac{\alpha}{2}=0.975$ and $df=n-1=11$

If you look in a table of a t-distribution, then you will find a t-value of 2,201 for $t_{0.975,11}$

callculus42
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    may I ask, how you managed to get that to be 0.975? ty – gger234 Nov 21 '14 at 18:28
  • @gger234 The significance level is 5%. If you have a two sided confidence interval, you cut on every side 2.5%. Thus you have $95%(=100%-2\cdot 2.5% )$ of the sample means inside the interval. – callculus42 Nov 21 '14 at 18:39