document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169828>Question 1169828</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A bottled water distributor wants to estimate the amount of water contained in 1​-gallon bottles purchased from a nationally known water bottling company. The water bottling​ company's specifications state that the standard deviation of the amount of water is equal to 0.04 gallon. A random sample of 50 bottles is​ selected, and the sample mean amount of water per 1​-gallon bottle is 0.971 gallon. Complete parts​ (a) through​ (d).<BR>\n" );
document.write( "a. Construct a 99 ​% confidence interval estimate for the population mean amount of water included in a​ 1-gallon bottle.<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 #851285 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=851285\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Alright, let's construct the confidence interval for the population mean amount of water in the 1-gallon bottles.\r<BR>\n" );
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document.write( "**Understanding the Problem**\r<BR>\n" );
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document.write( "We're given the sample mean, standard deviation, and sample size. We need to construct a 99% confidence interval for the population mean.\r<BR>\n" );
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document.write( "**Given Information**\r<BR>\n" );
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document.write( "* Sample size (n) = 50<BR>\n" );
document.write( "* Sample mean (x̄) = 0.971 gallon<BR>\n" );
document.write( "* Population standard deviation (σ) = 0.04 gallon<BR>\n" );
document.write( "* Confidence level = 99%\r<BR>\n" );
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document.write( "**a. Constructing the Confidence Interval**\r<BR>\n" );
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document.write( "Since we know the population standard deviation, we'll use the z-distribution.\r<BR>\n" );
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document.write( "1.  **Find the z-critical value:**<BR>\n" );
document.write( "    * For a 99% confidence interval, the area in the middle is 0.99.<BR>\n" );
document.write( "    * The remaining area in the tails is 1 - 0.99 = 0.01.<BR>\n" );
document.write( "    * The area in each tail is 0.01 / 2 = 0.005.<BR>\n" );
document.write( "    * We need to find the z-value that corresponds to an area of 0.995 (0.99 + 0.005) in the cumulative standard normal distribution.<BR>\n" );
document.write( "    * Using a z-table or calculator, the z-critical value for a 99% confidence interval is approximately 2.576.\r<BR>\n" );
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document.write( "2.  **Calculate the standard error:**<BR>\n" );
document.write( "    * The standard error (SE) is calculated as:<BR>\n" );
document.write( "        * SE = σ / √n = 0.04 / √50 ≈ 0.04 / 7.071 ≈ 0.005657\r<BR>\n" );
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document.write( "3.  **Calculate the margin of error (ME):**<BR>\n" );
document.write( "    * The margin of error is calculated as:<BR>\n" );
document.write( "        * ME = z * SE = 2.576 * 0.005657 ≈ 0.01457\r<BR>\n" );
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document.write( "4.  **Construct the confidence interval:**<BR>\n" );
document.write( "    * The confidence interval is calculated as:<BR>\n" );
document.write( "        * x̄ ± ME = 0.971 ± 0.01457\r<BR>\n" );
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document.write( "5.  **Calculate the interval endpoints:**<BR>\n" );
document.write( "    * Lower bound: 0.971 - 0.01457 ≈ 0.95643<BR>\n" );
document.write( "    * Upper bound: 0.971 + 0.01457 ≈ 0.98557\r<BR>\n" );
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document.write( "Therefore, the 99% confidence interval is approximately (0.95643, 0.98557).\r<BR>\n" );
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document.write( "**Final Answer**\r<BR>\n" );
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document.write( "a. The 99% confidence interval estimate for the population mean amount of water is approximately (0.9564, 0.9856) gallons.<BR>\n" );
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