document.write( "Question 1201155: Hello, please i need assistant in solving this question
\n" ); document.write( "A survey found that 78% of the men questioned preferred computer-assisted instruction to lecture and 68% of the women preferred computer-assisted instruction to lecture. There were 100 randomly selected individuals in each sample. Find the 95% confidence interval for the difference of the two proportions.
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Algebra.Com's Answer #837567 by Theo(13342)\"\" \"About 
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men p = .78
\n" ); document.write( "women p = .68
\n" ); document.write( "sample size for men and women the same at 100.
\n" ); document.write( "pooled proportion is (78 + 68) / (100 + 100) = .73
\n" ); document.write( "pooled standard error = sqrt(.73*.27*(1/100 + 1/100)) = .0627853486.
\n" ); document.write( "z-score = (.78 - .68) / .0627853486 = 1.59 rounded to 2 decimal places.
\n" ); document.write( "critical z-score is plus or minus 1.96.
\n" ); document.write( "test z-score in between those critical z-scores, so the resuls are not significant and you do not have any information to say that the men proportion and the women proportion are not the same.
\n" ); document.write( "this means that the observed difference is most likely due to variations in the mean of samples taken rather than any real difference.\r
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\n" ); document.write( "\n" ); document.write( "i used a calculator and go the same results, so the results are accurate, given the assumption of pooled variance.
\n" ); document.write( "here's the results from the calculator at https://www.socscistatistics.com/tests/ztest/default2.aspx\r
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