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Comparison of Standard Deviation with F Test:

For the comparison of the mean values of two sets of measurements, we have to decide whether the standarddeviations of the two sets are statistically different. This is done by a Ftest. The Ftest with the quotientis given as:

${\mathbf{F}}_{\mathbf{calculated}}\mathbf{=}\frac{{\mathbf{s}}_{\mathbf{1}}^{\mathbf{2}}}{{\mathbf{s}}_{\mathbf{2}}^{\mathbf{2}}}$

Here,${\mathbf{S}}_{\mathbf{1}}$and${\mathbf{S}}_{\mathbf{2}}$are standard deviations for the set of measurements using the original instrument and the substitute instrument.

If${\mathbf{F}}_{\mathbf{calculated}}\mathbf{<}{\mathbf{F}}_{\mathbf{table}}$, then the difference is significant.

Given data:

The given statements about F-test are as follows:

(i) If${\mathrm{F}}_{\mathrm{calculated}}<{\mathrm{F}}_{\mathrm{table}}$, there is more than a 5% chance that the two sets of data are drawn from populations with the same population standard deviation.

(ii) If ${\mathrm{F}}_{\mathrm{calculated}}<{\mathrm{F}}_{\mathrm{table}}$, there is at least 95% probability that the two sets of data are drawn from populations with the same population standard deviation.

Statement (i) is true because when there is more than a 5% probability of two sets of data drawn from the populations with the same population standard deviation, we accept the null hypothesis, and hence, it is acceptable that the two sets of data are drawn from a population with the same population standard deviation. Also, the results of the two sets of data are not significantly different.

The statement (ii) is false statement.

Statement (ii) is false. We know this because when the${\mathrm{F}}_{\mathrm{calculated}}<{\mathrm{F}}_{\mathrm{table}}$ the difference in the standard deviation of the two sets of data are not significant. Hence, the least probability of finding the two sets of data within the same population standard deviation is not possible.