In a three-digit lottery, each of the three digits is supposed to have the same
ID: 3238978 • Letter: I
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. (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 d.f. p-valueExplanation / Answer
X^2 = sum of [ (Oi – Ei)^2 / Ei ]
Oi = Observed Value
Ei = Expected Value
Oi
18
15
26
30
30
21
30
34
35
31
270
Ei
27
27
27
27
27
27
27
27
27
27
Oi - Ei
-9
-12
-1
3
3
-6
3
7
8
4
(Oi - Ei)^2
81
144
1
9
9
36
9
49
64
16
(Oi - Ei)^2/Ei
3.00
5.33
0.04
0.33
0.33
1.33
0.33
1.81
2.37
0.59
15.48
d.f = 10 - 1
= 9
p-value = chidist(15.48,9)
= 0.0786
Answer:
Test Statistics = 15.48
d.f = 9
p-value = 0.0786
Oi
18
15
26
30
30
21
30
34
35
31
270
Ei
27
27
27
27
27
27
27
27
27
27
Oi - Ei
-9
-12
-1
3
3
-6
3
7
8
4
(Oi - Ei)^2
81
144
1
9
9
36
9
49
64
16
(Oi - Ei)^2/Ei
3.00
5.33
0.04
0.33
0.33
1.33
0.33
1.81
2.37
0.59
15.48
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