Confidence Interval and Chi-square

[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

 

[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)

 

[Architeuthis Dux]

.

Based on the given information, we can use the Chi-square test to determine if the variation in oil containers is within an acceptable level. The Chi-square test is used to determine if there is a significant difference between the expected and observed values in a sample. In this case, the expected value is 1000 ml and the observed value is 997 ml. The null hypothesis for this test is that there is no significant difference between the expected and observed values, meaning that the standard deviation of the oil containers is less than or equal to 20 ml.

To construct a 95% confidence interval for the population standard deviation, we can use the formula s(sqrt)=(n-1)2/xsqrtR and s(sqrt)=(n-1)2/xsqrtL. Plugging in the values, we get a confidence interval of (22.45, 50.36). This means that we can be 95% confident that the true value of the population standard deviation falls within this range.

Based on this confidence interval, it appears that the variation in oil containers is not at an acceptable level. The upper limit of the interval, 50.36 ml, is more than double the acceptable standard deviation of 20 ml. This suggests that there is a significant difference between the expected and observed values, and the null hypothesis can be rejected. Therefore, further investigation may be needed to address the issue of variation in oil containers.