Timeline for Self-study question on Confidence Interval
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 29, 2014 at 13:55 | vote | accept | user1275515 | ||
| Apr 29, 2014 at 13:55 | |||||
| Apr 28, 2014 at 22:19 | comment | added | whuber♦ | You're right: I meant to write that $\sum\bar{x}(x_i-\bar{x})=0$. With that in mind you need to replace the "$\ne$" in your first comment with an "$=$" (assuming that by "$x$" you meant "$\bar x$"). Nevertheless, thank you for checking the calculation and big +1 for editing your answer to help the O.P.! | |
| Apr 28, 2014 at 22:08 | history | edited | Alexis | CC BY-SA 3.0 |
deleted 61 characters in body
|
| Apr 28, 2014 at 21:04 | comment | added | Alexis | Hmmm you may be correct there, I missed that he didn't have the actual data. While $\sum{\varepsilon_{i}}=0$ I don't think $\sum{\bar{x}x} = 0$, for example: $x \in \{1,2,3\}$, $\bar{x} = 2$, $\bar{x}x \in \{2,4,6\}$, and $\sum{\bar{x}x}=12 \ne 0$ | |
| Apr 28, 2014 at 20:48 | comment | added | whuber♦ | Alexis, I fear you may be leading the OP astray. Notice he has only the sum and sum of squares of the data; therefore an algebraically equivalent expression for the variance estimator is needed--and he is using a correct one and appears to be computing with it correctly. (Your first comment seems to overlook the fact that $\sum_i \bar{x} x_i = 0$ when computing the sum of squares of residuals.) | |
| Apr 28, 2014 at 20:44 | comment | added | Alexis | In addition, there's possibly a learning opportunity for you here: $(x_{i} - \bar{x})$ is the deviation of observation $x_{i}$ from the mean ($\bar{x}$). If you stare at the variance formula, you'll see that the variance is more or less the (artithmetic) mean of squared deviations (i.e. add up the number of squared deviations, and divide by the number of squared deviations, or, for the sample variance, divide by $n-1$, because of sampling uncertainty). | |
| Apr 28, 2014 at 20:37 | comment | added | Alexis | You have not used my formula: $\sum^{n}_{i=1}{(x_{i}−\bar{x})^{2}} \ne n(\sum^{n}_{i=1}{x_{i}})^{2}−x^{2}$. The difference of two squares is not the square of the differences. You may recall: $(x−a)^{2}=x^{2}−2ax+a^{2}$ In this case, you need to manually calculate $x_{i}−\bar{x}$ and square it, do this sixty times, add that up. Then proceed with your CIs. | |
| Apr 28, 2014 at 19:27 | comment | added | user1275515 | many thanks for your rapid reverts. I have attempted to use the above formulae, but I am still getting at the exact same output. I was wondering if there's anything fundamentally with how I did my standard deviation calculations? I've shared my working via picpaste.com/pics/image-12hEwqZw.1398713163.jpg | |
| Apr 28, 2014 at 16:59 | history | answered | Alexis | CC BY-SA 3.0 |
