document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1184231>Question 1184231</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Using the 95% confidence level and the writing scores 84, 80, 90, 90, 98, 94, 91, 86, 90, 91, 90 you would expect the lower end of the confidence interval to be approximately what value? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849865 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849865\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to calculate the lower end of the 95% confidence interval for the writing scores:\r<BR>\n" );
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document.write( "1. **Calculate the Sample Mean (x̄):**\r<BR>\n" );
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document.write( "Sum the scores: 84 + 80 + 90 + 90 + 98 + 94 + 91 + 86 + 90 + 91 + 90 = 1034<BR>\n" );
document.write( "Number of scores (n) = 11<BR>\n" );
document.write( "x̄ = 1034 / 11 = 94\r<BR>\n" );
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document.write( "2. **Calculate the Sample Standard Deviation (s):**\r<BR>\n" );
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document.write( "* Find the squared difference of each score from the mean:<BR>\n" );
document.write( "    (84-94)²=100, (80-94)²=196, (90-94)²=16, (90-94)²=16, (98-94)²=16, (94-94)²=0, (91-94)²=9, (86-94)²=64, (90-94)²=16, (91-94)²=9, (90-94)²=16<BR>\n" );
document.write( "* Sum of squared differences: 100 + 196 + 16 + 16 + 16 + 0 + 9 + 64 + 16 + 9 + 16 = 478<BR>\n" );
document.write( "* Variance (s²): 478 / (11-1) = 47.8<BR>\n" );
document.write( "* s = √47.8 ≈ 6.91\r<BR>\n" );
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document.write( "3. **Find the t-value:**\r<BR>\n" );
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document.write( "Since the sample size is small (n < 30), we use the t-distribution.  For a 95% confidence level and 10 degrees of freedom (n-1 = 11-1 = 10), the t-value is approximately 2.228.  You can find this using a t-table or a calculator with statistical functions.\r<BR>\n" );
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document.write( "4. **Calculate the Margin of Error:**\r<BR>\n" );
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document.write( "Margin of Error = t * (s / √n)<BR>\n" );
document.write( "Margin of Error = 2.228 * (6.91 / √11)<BR>\n" );
document.write( "Margin of Error = 2.228 * (6.91 / 3.317)<BR>\n" );
document.write( "Margin of Error ≈ 4.63\r<BR>\n" );
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document.write( "5. **Calculate the Confidence Interval:**\r<BR>\n" );
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document.write( "* Upper limit = x̄ + Margin of Error = 94 + 4.63 ≈ 98.63<BR>\n" );
document.write( "* Lower limit = x̄ - Margin of Error = 94 - 4.63 ≈ 89.37\r<BR>\n" );
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document.write( "Therefore, the lower end of the 95% confidence interval is approximately 89.37.<BR>\n" );
document.write( " </SPAN>\n" );
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