document.write( "Question 964030: a dice is thrown 9000 times and 3 or 4 is observed 3240 times can we say dice is fair. \n" ); document.write( "
Algebra.Com's Answer #588948 by Theo(13342)\"\" \"About 
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probably not, based on the following test.\r
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\n" ); document.write( "\n" ); document.write( "n = 9000
\n" ); document.write( "p (3 or 4) = 1/6 + 1/6 = 2/6 = 1/3
\n" ); document.write( "(1-p) = 2/3\r
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\n" ); document.write( "\n" ); document.write( "mean = n * p = 9000 * 1/3 = 3000.\r
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\n" ); document.write( "\n" ); document.write( "standard error equal sqrt (n * p * (1-p) = sqrt ( 9000 * 1/3 * 2/3) = sqrt(2000) = somewhere between 44.72 and 44.73, the exact value not being important for this problem.\r
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\n" ); document.write( "\n" ); document.write( "a raw score of 3240 will generate a z-score of (x-m) / se which would be equal to:\r
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\n" ); document.write( "\n" ); document.write( "z = (3240 - 3000) / 44.72 or 44.73.\r
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\n" ); document.write( "\n" ); document.write( "therefore z = 5.367 or 5.366.\r
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\n" ); document.write( "\n" ); document.write( "both are more than 3 standard deviations from the mean.\r
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\n" ); document.write( "\n" ); document.write( "the probability of getting a z-score greater than 3 standard deviations from the mean would be equal to .00135.\r
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\n" ); document.write( "\n" ); document.write( "this is less than .05 and also less than .01 which are the probabilities normally used as criteria.\r
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\n" ); document.write( "\n" ); document.write( "the .05 is normally used at the .95 confidence level for a one tailed test.\r
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\n" ); document.write( "\n" ); document.write( "the .01 is normally used as the .99 confidence level for a one tailed test.\r
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\n" ); document.write( "\n" ); document.write( "bottom line is the probability of getting such a score out of 9000 tosses just based on random differences between different samples is very remote and therefore not considered feasible.\r
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\n" ); document.write( "\n" ); document.write( "this indicates that the dice are probably biased in some way.\r
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