Our sample size is 6000, with 10 possible outcomes. \(\frac{6000}{10} = 600\) - we expect roughly 600 of each number to occur, which is our null hypothesis. The equation we will use is: \(\sum_{i = 1}^{10} \frac{(O_{i} - 600)^{2}}{600}\), where \(O_{i}\) is the frequency count of our sequence, in this case \(O_{i} = \langle 601, 583, 589, 623, 572, 606, 582, 630, 618, 596\rangle\) The result of said equation is 5.54, our test statistic. With 9 degrees of freedom, assuming a p-value of 0.05, we get 16.92 as our critical value. as \(5.54 < 16.92\), we do not reject the null hypothesis.