Question

In a three-digit lottery, each of the three digits is supposed to have the same probability of occurrence (counting initial blanks as zeros, e.g., 32 is treated as 032). The table shows the frequency of occurrence of each digit for 90 consecutive daily three-digit drawings. Digit 0 1 2 3 4 5 6 7 8 9 Frequency 34 32 30 25 27 29 24 18 25 26 270 Total picture Click here for the Excel Data File (a) Calculate the chi-square test statistic, degrees of freedom, and the p-value. (Perform a uniform goodness-of-fit test. Round your test statistic value to 2 decimal places and the p-value to 4 decimal places.) Test statistic p-value

Answer 1

We know that if all the digits are equally likely than frequency for each of the digits will be 270/10=27.

Oi	Ei	(oi-Ei)^2
34	27	49
32	27	25
30	27	9
25	27	4
27	27	0
29	27	4
24	27	9
18	27	81
25	27	4
26	27	1

i=1 test statistic is given by k x = (E; - 0;) E; which follows chi square with k-1 df under the null hypothesis. x² = (49 + 25 +9+4+0+4+9+81 +4+1)/27 = 6.89 df = 9 P - value = P(x > 6.89) = 0.64957 if level of significance is 10 percent or 5 percent than we don't have enough evidence to reject the null hypothesis.