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
thanks
The confidence limits use the mean and the standard deviation (mean ± 1.96 SD for 95% confidence interval)). The less variable the confidence limits (lower SD), the more accurate/precise the estimate is.
I hope this helps.
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 comment on the precision and accuracy of the given measurements, let's start with some definitions:
1. Precision: Precision refers to the level of consistency and reproducibility in obtaining measurements. It indicates how close repeated measurements are to each other.
2. Accuracy: Accuracy refers to how close a measured value is to the true or accepted value.
Based on the given data:
1. Precision: In this case, we have the measurements k = 36.86 +/- 4.78 and v = 113 +/- 9.92, both with 95% confidence limits. The plus/minus values represent the range within which the true values are expected to lie. The larger the range, the less precise the measurement. Therefore, we can say that k has a precision of 4.78, while v has a precision of 9.92. Therefore, k is measured more precisely than v.
2. Accuracy: Unfortunately, without the true values of k and v given, we cannot determine their accuracy. Accuracy can only be assessed by comparing measured values to some known or accepted reference value. Since no true values are provided, we cannot comment on the accuracy of these measurements.
In summary, we can say that k is measured more precisely than v because the range of k is smaller. However, without true values provided, we cannot determine how accurate these measurements are.