document.write( "Question 930271: A random sample of 16 female college-aged dancers showed a sample mean height 64.8 of inches and a sample standard deviation of 1.7 inches. \r
\n" ); document.write( "\n" ); document.write( "1.Find a 95% confidence interval for the population mean height of dancers and interpret the interval. \r
\n" ); document.write( "\n" ); document.write( "2. Find a 99% confidence interval for the population mean height and interpret the interval. \r
\n" ); document.write( "\n" ); document.write( "3. Which interval is wider and why? \r
\n" ); document.write( "\n" ); document.write( "Thank you!!
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Algebra.Com's Answer #564865 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
Sample: n = 16, mean= 64.8 , sd = 1.7
\n" ); document.write( "ME = 1.96(1.7/sqrt(16) = .8 (64, 65.6)
\n" ); document.write( "ME = 2.576(1.7/sqrt(16) = 1 (63. 65.8) ***wider
\n" ); document.write( "Including 99% rather than just 95%
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\n" ); document.write( "for Your reference:
\n" ); document.write( "CI
\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
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