document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1086252>Question 1086252</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A food snack manufacturer samples 7 bags of pretzels off the assembly line and weighs their contents. If the sample mean is 14.2 oz. and the sample standard deviation is 0.60 oz., find the 95% confidence interval of the true mean.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #700628 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=700628\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> standard error(SE) = 0.60 / square root(7) = 0.2268<BR>\n" );
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document.write( "alpha(a) = 1 - (95/100) = 0.05<BR>\n" );
document.write( "critical probability (p*): p* = 1 - &#945;/2 = 1 - 0.05/2 = 0.975<BR>\n" );
document.write( "degrees of freedom(df) = 7 - 1 = 6<BR>\n" );
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document.write( "the critical value(CV) is the t statistic with 6 degrees of freedom and a cumulative probability of 0.975.  We lookup these values in the t-tables to<BR>\n" );
document.write( "find the critical value is 2.447<BR>\n" );
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document.write( "margin of error(ME) = CV * SE = 2.447 * 0.2268 = 0.555<BR>\n" );
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document.write( "confidence interval(CI) = the sample statistic + or - the margin of error<BR>\n" );
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document.write( "95% CI = 14.2 + or - 0.555, (13.645, 14.755)<BR>\n" );
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