Standard Deviation sample help

[Hootenanny]

Just a quick question as I'm writing up my coursework. I'm calculating the standard deviation of a sample, my data is rate of reaction defined as \frac{1}{time} \times 10^{-3}. So my input data is say 0.847, but the absolute value is 0.000847. The standard deviation formula returned a value of about 0.6. Is this my standard deviation or is it actually 0.0006? I'm inclined towards 0.6, but I'm not sure. Any help would be appreciated.

 

[3trQN]

f=1/T - sample frequency is in miliseconds?

You made reading of 0.847, so that's 0.847ms? I think the deviation would be in milliseconds also, so 0.6ms or 0.0006s?

or am i way off :?

 

[Hootenanny]

This is a biology pratical. I was timing how long it took for a reaction to occur. For example, if the reaction took 19 minutes 40 seconds, that is a reaction time of 1180 seconds. As rate = \frac{1}{time}, the rate of reaction would be;
rate = \frac{1}{1180} = 8.47\times 10^{-4} = 0.847\times 10^{-3} = 0.000847

I apologise if I mislead you with the information.

 

[Hootenanny]

'bump' anybody else have an opinion?

 

[Cyrus]

Do you have a list of data points?

 

[Hootenanny]

Yes, do you want me to post them? Or attach them as an Excel spreadsheet? I could post them in latex if you like.

 

[Cyrus]

Please do. Just make them standard text so I can just copy and paste it into excel. No need to tex all that.

 

[Hootenanny]

Okay, I've got a tool for converting excel into Latex though

--------------
data
--------------
Temperature (°C) Time (secs) Rate (1/sec) x10-3

17 1180 0.847
1179 0.848
1169 0.855
1217 0.822
Mean 1186.25 0.843
32 180 5.556
182 5.495
180 5.556
178 5.618
Mean 180 5.556
34.5 175 5.714
179 5.587
181 5.525
210 4.762
Mean 186.25 5.369
39.5 100 10.000
116 8.621
116 8.621
120 8.333
Mean 113 8.850
48 90 11.111
90 11.111
91 10.989
93 10.753
Mean 91 10.989
50 105 9.524
104 9.615
104 9.615
109 9.174
Mean 105.5 9.479
57 154 6.494
143 6.993
133 7.519
125 8.000
Mean 138.75 7.207
64 1977 0.506
2220 0.450
2221 0.450
2210 0.452
Mean 2157 0.464

 

[Hootenanny]

Uhh, that doesn't look right :S

 

[Cyrus]

Ok, I need some help understanding your data here. You seems you have 8 runs each of size 4. What do you want to calculate the standard deviation for? The means? For all the data?

 

[Hootenanny]

No, I calculated the standard deviation of each run. I know its only a small data set, so i used sample SD. Is that valid?

 

[Cyrus]

Use the following:

s = \sqrt { \frac{1}{n-1}( \sum^n_{i=1}X^2_i - n \bar{X}^2} )

Yes, you used the right one, because you are taking data from a sample of the population, your data is not for the entire population. It does not matter how big your data set is, if its not the actual population it is still going to be a sample standard deviation.

 

[Hootenanny]

Yes I have done. Can I just ask you what you thing of my origonal question in post #1?

 

[Cyrus]

I'm sorry but I don't understand your original question. Can you rephrase it? You don't have an 'actual' standard deviation, because you are sampling data. The only way you can have an acutal standard deviation is if you had all the population data.

 

[Hootenanny]

I mean I inputed the data direcly from my table (using excel), but my units for rate is \times 10^{-3}. Say if the formula returned a SD value of 0.6, would it actually be 0.0006 due to the \times 10^{-3}? I'm sorry if I'm not very lucid, but I can't think of another way of explaining it.

 

[Cyrus]

well, then in that case:

\sigma_{aX+b} =|a| \sigma_X

so multiply it by the absolute value of 10^{-3}

 

[Hootenanny]

Ahhh, thank-you very much cyrus.