Question

Im having problems with this question, I can't figure out the sample variance, here is the information given:

Answer 1

\[\sum_{i=1}^{1}X^2i = 1,778,400(m^3/ha)^2\]
and
\[Xbar1= 420m^3 /ha\]

Answer 2

\[ \large \sum_{i=1}^1X_i^2 = 1,778,400\]
\[\large \bar{X}=420 \] is that all they gave? is there more data to this?

Answer 3

here is the original question:
The observations (volume in m3/ha ) of ten prism plots from a given forest type are summarized as above

Using the information, test if the standard deviation is significantly different from 50 m3/ha.

Answer 4

would that summation be \[\large \sum_{i=1}^{\color{red}{10}} \] ?

Answer 5

the original was\[\sum_{i=1}^{1}\]

Answer 6

if it was 10 on top, then we could do something, but with just a 1, that is just information from one data point out of 10 given

Answer 7

okay, perhaps its a typo by my prof.
I'm guessing I have to use a test to discover if the two variances are statistically differenct, but I need to know both sample variances first (s^2)

is that right?

Answer 8

I think they just want you to test if the variance of the prisms (which you have to calculate) differs statistically from the assumed value of 50.

Answer 9

well standard deviation actually.. not variance

Answer 10

\[\large \begin{align} s^2&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2 \\
&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2)\\
&=\frac{1}{n-1}\left(\sum_{i=1}^{n}x_i^2-2\bar{x}\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}\bar{x}^2 \right)\\
&=\frac{1}{n-1}\left(\sum_{i=1}^{n}x_i^2-2\bar{x}[n\bar{x}]+n\bar{x}^2\right) \\
&=\frac{1}{n-1}\left(\sum_{i=1}^{n}x_i^2-n\bar{x}^2 \right)
\end{align}\]
Now here \(n=10\), so
\[\large s^2=\frac{1}{10-1}\left(\sum_{i=1}^{n}x_i^2-10\bar{x}^2 \right)=\frac{1}{9}\left(1,778,400-10(420)^2 \right)\]

Then the standard deviation, just take the square root of that result.

Answer 11

Oh I don't think I have ever seen that formula before with the 1/n-1 in front of it.

Answer 12

the population variance, \(\sigma^2\) has 1/n in front. The sample variance, \(s^2\), has 1/(n-1) in front.