Question 130986: Please help me. I am lost with this question. Thank you for your help
In a 3-digit lottery, each of the 3 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 3-digit drawings. Perform the chi-squared test for a uniform distributions. At alpha= .05, can you reject the hypothesis that the digits are from a uniform population?
Digit No. of Occurrences
0 33
1 17
2 25
3 30
4 31
5 28
6 24
7 25
8 32
9 25
Total 270
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In a 3-digit lottery, each of the 3 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 3-digit drawings. Perform the chi-squared test for a uniform distributions. At alpha= .05, can you reject the hypothesis that the digits are from a uniform population?
Digit No. of Occurrences
0 33
1 17
2 25
3 30
4 31
5 28
6 24
7 25
8 32
9 25
Total 270
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You would expect each digit to occur (1/10)*270 = 27 times
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Subtract that "expected" number from each listed No.of Occurrences
to generate a "difference column" : Entries would be 5,-10,-2,3, etc.
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Generate a "difference squared column": entries would be 25,100,4,9, etc
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Divide each of those elements by 27: entries would be 25/27,100/27,4/27,9/27..
----------
Add all those numbers to get a Chi-Sq number for your problem:
Chi-Sq = 7.7037...
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Ho: all the digit proportions are the same
Ha: at least one of the proportions is different
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Critical value for right-tail test with alpha=5% is Chi-sq = 16.919
based on df=9.
-----------
Test statistic: Chi-sq = 7.7037
-----------------
Conclusion: Since the test statistic is less than the critical value
Fail to reject Ho.
The test shows no significant evidence that the proportion of digits
is usual.
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Cheers,
Stan H.
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