document.write( "Question 861790: in a gallup poll, 1025 randomly selected adults were surveyed and 29% of them said that they used the internet for shopping at least twice a year.
\n" ); document.write( "a)find the point estimate of the percentage of adults who use the internet for shopping?
\n" ); document.write( "b)Find a 99% confidence interval estimate of the percentage of adults who use the internet for shopping?
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Algebra.Com's Answer #519259 by ewatrrr(24785)\"\" \"About 
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\n" ); document.write( "Hi,
\n" ); document.write( "a) for P(x ≥ 2) TI syntax is 1 – binomcdf(n, p, largest x-value).
\n" ); document.write( "P(x ≥ 2) = 1 - binomcdf(1025,.29,1)
\n" ); document.write( "b) \"1025%2A.29%29\" = 297.25 is the mean
\n" ); document.write( "ME = \"2.576sqrt%28.29%2A.61%29%2F1025%29%29\"
\n" ); document.write( "CI: 297.25-ME < u < 297.25 + ME\r
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\n" ); document.write( "\n" ); document.write( "CI z -value
\n" ); document.write( "90% z =1.645
\n" ); document.write( "92% z = 1.751
\n" ); document.write( "95% z = 1.96
\n" ); document.write( "98% z = 2.326
\n" ); document.write( "99% z = 2.576\r
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