document.write( "Question 1089689: Please help me solve the following problem: A survey of 1000 NUST students find that 48% own a personal computer. Construct a 95% confidence interval on the true proportion of NUST students who own a personal computer. \n" ); document.write( "
Algebra.Com's Answer #704042 by rothauserc(4718)\"\" \"About 
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Our sample size of 1000 is sufficiently large > 30 that we can use the normal distribution
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\n" ); document.write( "sample proportion(p) = 0.48
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\n" ); document.write( "standard error(SE) = square root(0.48 * (1-0.48) / 1000) = 0.0158
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\n" ); document.write( "alpha(a) = 1 - (95/100) = 0.05
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\n" ); document.write( "critical probability(p*) = 1 - (a/2) = 0.975
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\n" ); document.write( "critical value(CV) is the z-value associated with p* which is 1.96
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\n" ); document.write( "margin of error(ME) = CV * SE = 1.96 * 0.0158 = 0.031
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\n" ); document.write( "95% confidence interval is 0.48 + or - 0.031 or (0.449, 0.511)
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