Book1 - Ho: P(heads) = 0.5 and HA : P(heads) not equal to...

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Manzur Choudhury Professor Johnston Walter BA 3360 10/6/2010 Problem No. 1 Background: An Article in the Washington Post reported an experiment by a grade school student in which 17,950 penny tosses were recorded. Of these tosses, 9207 outcomes were heads and 8743 were tails. The Student concluded that the coins were made significantly heavier tails than heads. Analysis: The Assumed distribution for the two outcomes is uniform, that is
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Unformatted text preview: Ho: P(heads) = 0.5 and HA : P(heads) not equal to 0.5. This gives E (heads) = E (Tails) = 17950/2 = 8975 under Ho. We will use alpha = 0.05. Using the formula presented above, X^2 heads = (O-E)^2/E 6 X^2 tails = (O-E)^2/E 6 so X^2 = 5.997103 + 5.997103 11.99 the critical value for X^2 df = 1, alpha = .05 = 3.84146 since X^2 > X^2df=1, alpha -= .05, we reject Ho. The p-value for this problem = .00053366 *...
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This note was uploaded on 07/21/2011 for the course BUS 1302 taught by Professor Falcon during the Spring '11 term at University of Texas at Dallas, Richardson.

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