Question

Based on repeated measurements of the iodine concentration in a solution, a chemist reports the concentration as \(4.614,\) with an "error margin of .006." a. How would you interpret the chemist's "error margin"? b. If the reported concentration is based on a random sample of \(n=30\) measurements, with a sample standard deviation \(s=.017,\) would you agree that the chemist's "error margin" is .006?

Short answer

Answer: The error margin in the chemist's report is an indication of the uncertainty associated with the measurement of the iodine concentration in the solution. It suggests that the true iodine concentration lies between 4.608 and 4.620. The reported error margin of 0.006 is accurate, as our calculations using the sample size and standard deviation produced a similar margin of error of approximately 0.0064.

Step-by-step solution

a. Interpretation of the error margin

The "error margin" reported by the chemist is a measure of the uncertainty associated with the measurement of the iodine concentration in the solution. In this case, the chemist reports an error margin of 0.006. This indicates that the true iodine concentration in the solution is likely to lie within the range of 4.614 ± 0.006, i.e., between 4.608 and 4.620.

b. Verifying the error margin

To verify if the error margin of 0.006 is accurate, we will use the sample size (n=30) and the sample standard deviation (s=0.017). The margin of error can be calculated using the formula: Margin of Error = \(t \times \frac{s}{\sqrt{n}}\) Here, t is the t-score which corresponds to a certain confidence level. Assuming a 95% confidence level, we need to find the t-score for a two-tailed t-distribution with 29 degrees of freedom (30 - 1), which is approximately 2.045. Margin of Error = \(2.045 \times \frac{0.017}{\sqrt{30}}\) Margin of Error ≈ 0.0064 The calculated margin of error is approximately 0.0064, which is close to the chemist's reported error margin of 0.006. Hence, we can agree that the chemist's error margin of 0.006 is reasonable based on the given sample size and standard deviation.