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A cannery prints “weight 175 grams” on its label. The quality control supervisor selects 64 cans at random and weighs them. He finds the sample mean to be 173 grams, and the sample standard deviation to be 6 grams.

I do not understand how to calculate CI for standard deviation.

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The lower-bound of the confidence interval is given by

$\overline{x} - z_{\alpha/2}SE$

Here you know $\overline{x} = 173$ and since you're looking for a 95% confidence interval, $\alpha = 5\%$ so that $\alpha/2 = 2.5\%$. Now to find $z_{2.5\%}$ you need to either use a table of $z$-scores or some software. But one way or another, you need to know how to look this information up, and when you do, you'll find that it's close to 1.96.

Lastly you need to calculate $SE$. In this case, you're finding the confidence interval for a mean, which tells you that the formula is

$SE = s/\sqrt{n}$ where in this case is $6/\sqrt{64}$ which is $6/8 = 3/4$.

Thus the lower end of the interval is $173 - (1.96)\cdot (3/4)$, whatever that is.

The upper end of the interval is $173 + (1.96)\cdot (3/4)$.

[Edit: update, since the interval is supposed to be for $\sigma$:

Since you want the CI for $\sigma$, we start with the interval for $\sigma^{2}$. The lower-bound is

$\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, n-1}}$

which in this case is $\frac{63\cdot 36}{\chi^{2}_{0.025, 63}}$. At this point you need to look up $\chi^{2}_{0.025, 63}$. I throw that into Wolfram|Alpha and get 42.9503.

So the lower bound is $\frac{63\cdot 36}{42.9503}$.

The upper bound is $\frac{63\cdot 36}{\chi^{2}_{0.975, 63}}$, whatever that is.

To get the interval for $\sigma$, you square-root each of these bounds for $\sigma^{2}$.

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  • $\begingroup$ i'm actually looking for the CI for variance and standard deviation . $\endgroup$ Feb 13 '14 at 3:50
  • $\begingroup$ OK, updated the response. $\endgroup$
    – Addem
    Feb 13 '14 at 4:03

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