if i have
k = 36.86 +/- 4.78
and
v = 113 +/- 9.92
both using 95% confidence limits and no true values of k or v given
am i right in saying that accuracy cannot be determined
and that v is a more precise measurement
my question was to comment on what can be said about
the precision
the accuracy
and which is measured more precisely
i know that the standard deviation and confidence limits are lower numerically for K
but proportionally as a percentage of the mean they are lower for V
i think this means that v is more accurate
i want to know if im correct in saying this
To assess the precision and accuracy of measurements, we need to consider the concepts of standard deviation, confidence limits, and the relative values of these measures for the given variables.
Precision refers to the consistency or reproducibility of measurements. It can be assessed through the standard deviation, which measures the spread of data points around the mean. A lower standard deviation implies higher precision because the data points are tightly clustered around the mean.
Accuracy, on the other hand, refers to the closeness of a measurement to the true or target value. However, in your case, you mentioned that no true values for k or v were provided. So, without a known reference value, accuracy cannot be determined or compared between k and v.
Now, let's focus on precision.
For k, you have provided:
k = 36.86 +/- 4.78
The value after "+/-" represents the uncertainty or the confidence limits. In this case, with a 95% confidence level, it means that there is a 95% probability that the true value of k lies within the range of 36.86 +/- 4.78.
For v, you have provided:
v = 113 +/- 9.92
Similarly, with a 95% confidence level, there is a 95% probability that the true value of v lies within the range of 113 +/- 9.92.
To compare the precision between k and v, we can examine the relative values of their standard deviations or confidence limits. You mentioned that although the standard deviation and confidence limits are lower numerically for k, they are lower proportionally as a percentage of the mean for v.
From this analysis, it appears that v has a higher precision because its confidence limits, as a percentage of the mean, are lower than those of k.
In summary, based on the information given, it is correct to say that accuracy cannot be determined without known true values for k and v. However, it can be inferred that v is more precisely measured due to its lower confidence limits relative to the mean.