Question

Simple stats question:
What is the full width of the 99% confidence interval for the mean of the population from which this sample was drawn? (A random sample of size n = 50 yields a sum of squares of 488.)

Answer 1

The \((1-\alpha)\times100\%\) CI for the mean will have the form
\[\left(\bar{x}-Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}},~\bar{x}+Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}}\right)\]
where \(\bar{x}\) is the mean, \(Z_{\alpha/2}\) is the critical value for the significance \(\alpha\), \(\sigma^2\) is the variance, and \(n\) is the sample size.

You want to find the 99% interval, which makes \(\alpha=0.01\) (since 0.01 + 0.99 = 1). The critical value for this confidence level is about \(2.58\).

The variance is constructed using the sum of squares:
\[\sigma^2=\underbrace{\frac{SS}{n}}_{\text{use this one for populations...}}\quad\text{or}\quad\underbrace{\frac{SS}{n-1}}_{\text{...and this one for samples}}\]
Everything else you need is given to you explicitly (mean and sample size).

The width of the interval is the distance from one endpoint to the other. This is just a matter of finding the absolute difference between the values.

Answer 2

Could you assist me further in how to calculate the interval width? How do we calculate x-bar? Is it needed to calculate the width?

Answer 3

Oh sorry, I thought the mean was given to you... Well, you might be able to extract it from the sum of squares.
\[\begin{align*}SS&=\sum_{k=1}^{50}\left(x_k-\bar{x}\right)^2\\\\
488&=\sum_{k=1}^{50}({x_k}^2-2x_k\bar{x}+\bar{x}^2)\\\\
&=\sum_{k=1}^{50}{x_k}^2-2\bar{x}\sum_{k=1}^{50}x_k+50\bar{x}^2\\\\
0&=\sum_{k=1}^{50}{x_k}^2-488-2\bar{x}\sum_{k=1}^{50}x_k+50\bar{x}^2\\\\
\bar{x}&=\frac{\displaystyle2\sum_{k=1}^{50}x_k\pm\sqrt{4\left(\sum_{k=1}^{50}x_k\right)^2-200\left(\sum_{k=1}^{50}{x_k}^2-488\right)}}{100}\end{align*}\]
Unfortunately, you can't find an explicit value for the mean unless you have all the data points \(x_k\)...

Answer 4

Hmmm...

Okay, well thank you for all your assistance!