Question

1. A calcite sample yields the following data for the determination of calcium as Calcium Oxide (CaO): 55.95, 56.00, 56.04, 56.08, and 56.23. Should we reject 56.23 with 95% confidence?

Answer 1

In this problem, Dixon’s Q Test will be used to find outliers and to decide whether a suspected result should be kept or rejected.

This test is generally used when certain data set include one suspect observation that is much lower or higher than the other values of the data set. We need to remove true outliers as it affects mean and standard deviation calculations.

Data: 55.95, 56.00, 56.04, 56.08, and 56.23

We have check for 56.23 (suspected value)

Q_value = x_suspect - x_nearest / x_max. - x_min.

Q_value = (56.23 - 56.08) / (56.23 - 55.95) = 0.15/0.28 = 0.535

The obtained Q_value value is compared to a critical Q-value, which is found in Q-table.

Test is need to be run at 95% confidence. For a sample size of 5 values, critical Q-value from table was found to be 0.710.

Hence, we have

Q_value = 0.535

Q_critical = 0.710

(0.710 > 0.535)

When, Q_critical > Q_value  the suspect value can be characterized as an outlier and it can be rejected.

Hence 56.23 should be rejected.