Carolyn Beck

2022-07-01

I need to know how precise a measurement system is, expressed as a standard deviation. I don't really have much of a background in math so please excuse me if I'm a bit unclear or inaccurate about something.

The simplest way I know of doing this is by taking e.g. 100 measurements of the same object and then using the usual sample standard deviation formula.

Unfortunately, I can't use this method. I must measure different objects multiple times. Let's say 10 objects, 10 times each.

The thing that comes to mind is to calculate the mean for each group, subtract the relevant means from each measurement (sort of a normalisation) and then calculate the standard deviation from the resulting 100 numbers. Does this make sense? If not, how do I approach the problem?

The simplest way I know of doing this is by taking e.g. 100 measurements of the same object and then using the usual sample standard deviation formula.

Unfortunately, I can't use this method. I must measure different objects multiple times. Let's say 10 objects, 10 times each.

The thing that comes to mind is to calculate the mean for each group, subtract the relevant means from each measurement (sort of a normalisation) and then calculate the standard deviation from the resulting 100 numbers. Does this make sense? If not, how do I approach the problem?

podesect

Beginner2022-07-02Added 20 answers

Suppose you have ng=100 independent measurements from the distribution $\mathsf{N}\mathsf{o}\mathsf{r}\mathsf{m}(\mu ,\sigma ),$ where there are n=10 measurements on each of g=10 disjoint groups.

Let ${S}_{i}^{2}=\frac{1}{9}\sum _{n=1}^{10}({X}_{ij}-{\overline{X}}_{i}{)}^{2}$ be the sample variance of group i out of g, where ${X}_{ij}$ is the $j$th measurement from group i and ${\overline{X}}_{i}=\frac{1}{10}\sum _{j=1}^{10}{X}_{ij}$ is the sample mean of the $i$th group.

Then the 'pooled' variance for the g groups is

${S}_{p}^{2}=\frac{9{S}_{1}^{2}+9{S}_{2}^{2}+\cdots +9{S}_{10}^{2}}{90}.$

${S}_{p}^{2}$ is an unbiased estimate of the population variance ${\sigma}^{2}.$ For our normal data, its distribution is given by $90{S}_{p}^{2}/{\sigma}^{2}\sim \mathsf{C}\mathsf{h}\mathsf{i}\mathsf{s}\mathsf{q}(df=90).$

If necessary, this relationship would allow you to find a 95% confidence interval for ${\sigma}^{2}$ of the form $(90{S}_{p}^{2}/U,\phantom{\rule{thinmathspace}{0ex}}90{S}_{p}^{2}/L),$ where L and U cut probability 0.025 from the lower and upper tails of Chisq(df=90). For 90 'degrees of freedom' as in your example, L=65.65 and U=118.14 can be found from printed tables of the chi-squared distribution or by using software. (The result below is from R statistical software.)

qchisq(c(.025,.975), 90)## 65.64662 118.13589

Note: This kind of pooled variance is used in a one-factor analysis of variance (ANOVA) with $n$ replications on each of $g$ groups. It is called the 'denominator mean square' or 'mean square for error'. A one-factor ANOVA could be used to check whether it is reasonable to assume that all $g$ groups really do have the same population mean $\mu ,$ as assumed in your example. You can find descriptions of such an ANOVA design in an intermediate level statistics book or online.

Let ${S}_{i}^{2}=\frac{1}{9}\sum _{n=1}^{10}({X}_{ij}-{\overline{X}}_{i}{)}^{2}$ be the sample variance of group i out of g, where ${X}_{ij}$ is the $j$th measurement from group i and ${\overline{X}}_{i}=\frac{1}{10}\sum _{j=1}^{10}{X}_{ij}$ is the sample mean of the $i$th group.

Then the 'pooled' variance for the g groups is

${S}_{p}^{2}=\frac{9{S}_{1}^{2}+9{S}_{2}^{2}+\cdots +9{S}_{10}^{2}}{90}.$

${S}_{p}^{2}$ is an unbiased estimate of the population variance ${\sigma}^{2}.$ For our normal data, its distribution is given by $90{S}_{p}^{2}/{\sigma}^{2}\sim \mathsf{C}\mathsf{h}\mathsf{i}\mathsf{s}\mathsf{q}(df=90).$

If necessary, this relationship would allow you to find a 95% confidence interval for ${\sigma}^{2}$ of the form $(90{S}_{p}^{2}/U,\phantom{\rule{thinmathspace}{0ex}}90{S}_{p}^{2}/L),$ where L and U cut probability 0.025 from the lower and upper tails of Chisq(df=90). For 90 'degrees of freedom' as in your example, L=65.65 and U=118.14 can be found from printed tables of the chi-squared distribution or by using software. (The result below is from R statistical software.)

qchisq(c(.025,.975), 90)## 65.64662 118.13589

Note: This kind of pooled variance is used in a one-factor analysis of variance (ANOVA) with $n$ replications on each of $g$ groups. It is called the 'denominator mean square' or 'mean square for error'. A one-factor ANOVA could be used to check whether it is reasonable to assume that all $g$ groups really do have the same population mean $\mu ,$ as assumed in your example. You can find descriptions of such an ANOVA design in an intermediate level statistics book or online.

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