The following example is about Variance and Standard Deviation copied from Table 6.1 in the book "The Statistical Analysis of Experimental Data" by J Mandel.
The Variance given is 667e-8 and Standard Deviation given is 0.00258, but I didn't get the same result, although it's close. What I got is:
Variance = 6.3964e-06, Standard Deviation = 0.00252911
Is my calculation wrong? Could someone please confirm the result? Thanks in advance!
The 24 measurement values are below:
1.0851 1.0834 1.0782 1.0818 1.0810 1.0837 1.0857 1.0768 1.0842 1.0786 1.0812 1.0784 1.0768 1.0842 1.0811 1.0829 1.0803 1.0811 1.0789 1.0831 1.0829 1.0825 1.0796 1.0841
The solution given implies that it asking for sample variance which is given by $$s^2 = \frac{1}{n-1}\sum_{i = 1}^n(x_i-\bar x)^2 = 6.674493\text{e-}06.$$
You calculated the population variance, $$\sigma^2 = \frac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2 = 6.396389\text{e-}06.$$
Same goes for the standard deviation. So no, technically your calculation is not wrong. But you found the incorrect thing. It wants the sample variance, not the population variance. Same for sd.