Is the order router really uniform? A chi-squared test that rejects

Question

A smart order router claims it splits orders equally across $5$ venues ( $20\%$ each). Over $250$ orders you observe $[38, 45, 52, 40, 75]$ .

Test the claim. Give the statistic, degrees of freedom, and conclusion.

Accepted Answer

Expected counts under equal splitting are \frac{250}{5} = 50 per venue. Statistic. \chi^2 = \frac{(38-50)^2 + (45-50)^2 + (52-50)^2 + (40-50)^2 + (75-50)^2}{50} = \frac{144 + 25 + 4 + 100 + 625}{50} = \frac{898}{50} = 17.96. Degrees of freedom. The claimed proportions are fully specified, so 	ext{df} = k - 1 = 5 - 1 = 4. Decide. The critical values for \chi^24 are 9.488 at 5\% and 13.28 at 1\%. Since 17.96 13.28, we reject even at the 1\% level: the split is not uniform. Scanning the terms points at the culprit, venue 5 alone contributes \frac{625}{50} = 12.5 of the total, it is receiving fa…

Is the order router really uniform? A chi-squared test that rejects

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