document.write( "Question 1040401: The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza?
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document.write( "A. 0.115 < p < 0.260
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document.write( "B. 0.086 < p < 0.289
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document.write( "C. 0.075 < p < 0.300
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document.write( "D. 0.102 < p < 0.273 \n" );
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Algebra.Com's Answer #655210 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 1 sample proportion \n" ); document.write( "p=15/80=0.1875 \n" ); document.write( "CI interval +/- is z(0.995) sqrt{p(1-p)/n} \n" ); document.write( "2.576*sqrt{(0.1875)(0.8125)/80}; the second term is 0.0436 \n" ); document.write( "multiplied by 2.576 is 0.1124 \n" ); document.write( "CI is (0.0751,0.2999) \n" ); document.write( "It is C. \n" ); document.write( " |