# Confidence Interval and Chi-square

1. Jun 18, 2008

### iza-bella

A container of oil is supposed to contain 1000 ml of oil. We want to be sure that the standard deviation of the oil container is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Using these sample construct a 95% confidence interval for the true value of sigma. Does the confidence interval suggest that the variation in oil containers is at an acceptable level?

x-bar=997 n=10 d.f=10-1=9 s=32

s(sqrt)=Σ(x-xbar)2/n-1
Left and Right End points:
(n-1)2/xsqrtR (n-1)2/xsqrtL

Square root of Left and Right endpoints to get confidence interval for the population standard deviation

Unfortunately, my book is used and has a few pages ripped out from this section and I'm having a really hard time figuring how to put this all together. Any help would be greatly appreciated

2. Jun 19, 2008

### konthelion

I believe this is what you're trying to say?

$$P\left( \chi_{1-\alpha/2,n-1}^2 < \frac{(n-1)s^2}{\sigma^2} < \chi_{1-\alpha/2,n-1}^2 \right) = 1 - \alpha$$
then your 100(1-alpha)% confidence interval for $$\delta^2$$ would be:

$$\frac{(n-1)s^2}{\chi_{\alpha/2,n-1}}$$ for the lower limit (1)
$$\frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}}$$ for the upper limit. (2)

Since you are trying to find the 95%(which is equal to 100(1-alpha)%, so you can find alpha) confidence interval for the true value of sigma. The lower and upper limit would just be the square roots of (1) and (2)

Last edited: Jun 19, 2008