Answer
$X^2\lt X_α^2$: the null hypothesis is not rejected.
There is not enough evidence to conclude that the distribution does not follow Benford’s Law.
Work Step by Step
$H_0:$ the distribution follows Benford’s Law.
$H_1:$ the distribution does not follow Benford’s Law.
Total: 335 digits.
Expected count of 1: $335\times0.301=100.835$
Expected count of 2: $335\times0.176=58.96$
Expected count of 3: $335\times0.125=41.875$
Expected count of 4: $335\times0.097=32.495$
Expected count of 5: $335\times0.079=26.465$
Expected count of 6: $335\times0.067=22.445$
Expected count of 7: $335\times0.058=19.43$
Expected count of 8: $335\times0.051=17.085$
Expected count of 9: $335\times0.046=15.41$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(104-100.835)^2}{100.835}+\frac{(55-58.96)^2}{58.96}+\frac{(36-41.875)^2}{41.875}+\frac{(38-32.495)^2}{32.495}+\frac{(24-26.465)^2}{26.465}+\frac{(29-22.445)^2}{22.445}+\frac{(18-19.43)^2}{19.43}+\frac{(14-17.085)^2}{17.085}+\frac{(17-15.41)^2}{15.41}=5.09$
$k=9$. So, $d.f.=9-1=8$
$X_α^2=X_{0.05}^2=15.51$
(According to Table VII, for d.f. = 8 and area to the right of critical value = 0.05)
Since $X^2\lt X_α^2$, we do not reject the null hypothesis.
