document.write( "Question 1205076: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions.
\n" ); document.write( " Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions .Interpret your answer in part a above? \n" ); document.write( "
Algebra.Com's Answer #841703 by Theo(13342)\"\" \"About 
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smple size is 200
\n" ); document.write( "p = proportion of household where the woman makes the majority of the purchasing decisions is 120.
\n" ); document.write( "p = 120/200 = .6
\n" ); document.write( "q = 1-0 = .4
\n" ); document.write( "sample mean proportion is .6
\n" ); document.write( "standard error is sqrt(.6 * .4 / 200) = .034641.
\n" ); document.write( "z-score formula is z = (x-m)/s
\n" ); document.write( "z is the z-score
\n" ); document.write( "x is the raw score
\n" ); document.write( "m is the mean
\n" ); document.write( "s is the standard error.
\n" ); document.write( "at 95% confidence interval, the critical z-score is z = plus or minus 1.96.
\n" ); document.write( "on the low end, the formula becomes -1.96 = (x-.6)/.034641.
\n" ); document.write( "solve for x to get x = -1.96 * .034641 + .6 = .5321
\n" ); document.write( "on the high end, the formula becomes 1.96 = (x-.6)/.034641.
\n" ); document.write( "solve for x to get x = 1.96 * .034641 + .6 = .6679.
\n" ); document.write( "your 95% confidence interval is from .5321 to .6679.
\n" ); document.write( "here's what it looks like on a graph.
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