document.write( "Question 1175556: A sample of 200 customers at a supermarket showed that 28 used a debit card to pay for their purchases.
\n" ); document.write( "(a) Find the 95 percent confidence interval for the population proportion. [2]
\n" ); document.write( "(b) Why is it OK to assume normality in this case? [2]
\n" ); document.write( "(c) What sample size would be needed to estimate the population proportion with 90 percent confidence and an error of +/- 0.03? [2
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Algebra.Com's Answer #801189 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi\r\n" );
document.write( " n = 200, 28 used debit card\r\n" );
document.write( " sample p = 28/200 = .14\r\n" );
document.write( "ME = \"z%2Asqrt%28%28p%281-p%29%29%2Fn%29\"\r\n" );
document.write( "95% confidence interval\r\n" );
document.write( "ME = \"1.96%2Asqrt%28%28.14%28.86%29%29%2F200%29\" = ± .0481\r\n" );
document.write( "CI = .14 ± .0481\r\n" );
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document.write( "b) Yes, np = 28 > 5\r\n" );
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document.write( "c) n = \"%28z%2FME%29%5E2+%28p%281-p%29%29%29\"\r\n" );
document.write( " 90% confidence interval\r\n" );
document.write( "  n =\"%281.645%2F.03%29%5E2+%28.14%28.86%29%29%29\" = 362.006\r\n" );
document.write( "Sample Size 362 needed.\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
document.write( " = CI	z = value\r\n" );
document.write( "90%	z =1.645\r\n" );
document.write( "92%	z = 1.751\r\n" );
document.write( "95%	z = 1.96\r\n" );
document.write( "98%	z = 2.326\r\n" );
document.write( "99%	z = 2.576\r\n" );
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