# Statistics and Confidence Intervals

Given the following set of values:

10,11,14,95,73,30,29,9,97,94,70

How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10

Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is:

$$x-z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$$ and $$x+z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$$ where x denotes the mean. Here the significance level (a) is 0.01.

• What are your thoughts? Are you familiar with any formulas that concern confidence intervals? What computations can you do? – Brian Mar 21 at 17:56
• Well, I know the formula for the confidence interval, but i'm not sure how to find the z-score, I know that I can use R, but i'm not sure what to write in R. – topologicalmagician Mar 21 at 17:57
• Do you know what the "99%" in the confidence interval means? Can you relate this to the Normal distribution in some way? Do you know how z-scores relate to the Normal distribution? – Brian Mar 21 at 18:00
• @Brian Yes, a CI is the 100(1-a)% interval in which the true parameter lies in, a is the significance level. The formula for the CI is : $x+z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$ and : $x-z_{\frac{a}{2}} \frac{\sigma^2}{sqrt(n)}$ where x denotes the mean – topologicalmagician Mar 21 at 18:04

It is likely that you are asked to find a confidence interval for the (unknown) population mean, not the sample mean. The sample mean is not a parameter of interest, it can be calculated from the sample.

A conventional setup for the problem is that you have a sample $$\mathbf X=(X_1,X_2,\ldots,X_n)$$ of size $$n=11$$ from a $$N(\mu,\sigma_0^2)$$ population (by assumption) with $$\sigma_0^2=10$$. You have to find a confidence interval for the mean $$\mu$$.

A suitable pivotal quantity here is $$Q(\mathbf X,\mu)=\frac{\sqrt{n}(\overline X-\mu)}{\sigma_0}\sim N(0,1)$$

, where $$\overline X=\frac{1}{n}\sum\limits_{i=1}^n X_i$$ is the sample mean.

So if $$z_{\alpha/2}$$ be such that $$P(Z>z_{\alpha/2})=\alpha/2$$ where $$Z\sim N(0,1)$$, you have $$P_{\mu}(-z_{\alpha/2}\le Q\le z_{\alpha/2})=1-\alpha\quad\,\forall\,\mu$$

You have to use the above to arrive at the form $$P_{\mu}(c_1\,\le \mu\le\,\, c_2)=1-\alpha\quad\,\forall\,\mu$$

A $$100(1-\alpha)\%$$ confidence interval for $$\mu$$ is then $$[c_1,c_2]$$.