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Determine the standard deviation of these results

  • Thread starter Tiberious
  • Start date
  • #1
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I've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?


Homework Statement



The ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.

711 N 〖mm〗^(-2),732 N 〖mm〗^(-2),759 N 〖mm〗^(-2),670 N 〖mm〗^(-2),701 N 〖mm〗^(-2),
765 N 〖mm〗^(-2),743 N 〖mm〗^(-2),755 N 〖mm〗^(-2),715 N 〖mm〗^(-2),713 N 〖mm〗^(-2).

[/B]
(a) Determine the mean standard deviation of these results.
(a) Express the values found in (a) in GPa.

Homework Equations



s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )

The Attempt at a Solution



(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:

s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )

Calculating the mean value x ̅

(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10

x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2

(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 625.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56

Determining the sum of the above

∑▒237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56

= 8,030.29 〖N mm〗^(-2)

Divide by N-1

(1/5)∙8030.29= 1606.06 N 〖mm〗^(-2)


Determining the sample standard deviation σ

σ= √((1606.06)=40.08 N 〖mm〗^(-2)=4.008 MPa

(b)4.008 MPa=0.004008 GPa

Or in standard form, 4.008∙10^(-3) GPa.
 

Answers and Replies

  • #2
Charles Link
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##\sigma=40 ## N/mm^2=40 MPa=.040 GPa if my arithmetic is correct. ## \\ ## And you don't need to keep all the extra sig figs on the standard deviation. ## \\ ## And I think they want the mean and standard deviation. You didn't give the value of the mean in GPa.
 
  • #3
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Determine the mean standard deviation of these results.
There's a mean and there's a standard deviation. I'm not aware of a term called mean standard deviation, but there is a standard error, where you look at the variation of a collection of samples.
 
  • #4
FactChecker
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There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
 
Last edited:
  • #5
Charles Link
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There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
 
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  • #6
FactChecker
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Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
They are all very standard things to ask for. But I agree that the sample mean and sample standard deviation are the first two things to calculate.
 
  • #7
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Yes - it does seem my OP was missing the word 'and' apologies for the confusion. Just to confirm;

The answer for standard deviation is 0.040 Gpa ?

So, would the mean be 0.016Gpa ?
 
  • #8
FactChecker
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In your original post, why are you dividing by 5 instead of 9 for the sample variance?

PS. small error: (701-726.4 )^2=645.16, not 625.16.
 
  • #9
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So, amending (701-726.4 )^2 from 625.16 to 645.16 and amending the following would lead to 805.02N mm^-2

Amendments being: -

(701-726.4 )^2 = 645.16

(1/10)∙8050.29= 805.02 N mm^-2

This leads to: -

σ= √((805.02)=28.37 N mm^(-2)= 0.02837 Mpa

Converting to Gpa yields: -

2.837*10^-5 Gpa

Looking any better ?
 
  • #10
FactChecker
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(1/10)∙8050.29= 805.02 N mm^-2
Why are you dividing by 10 instead of N-1=10-1=9?

If you know the true mean, ##\mu##, you should use that in the formula for the sample standard deviation and divide by N. But if you estimate the mean with ##\bar x ##, your results of ##\sum_{i=1}^{i=N} (x_i-\bar x)^2## will be smaller and you should divide by ##N-1##.
 
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  • #11
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Ah! thanks, brief moment of confusion. I assume this would amend the overall error ?
 

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