document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1179421>Question 1179421</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A consumer affairs investigator records the repair cost for 20 randomly selected refrigerators. A sample mean of $57.22 and standard deviation of $25.76 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.\r<BR>\n" );
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document.write( "Step 1 of 2: Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.<BR>\n" );
document.write( "A consumer affairs investigator records the repair cost for 20 randomly selected refrigerators. A sample mean of $57.22 and standard deviation of $25.76 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal.<BR>\n" );
document.write( "Step 2 of 2: Construct the 90% confidence interval. Round your answer to two decimal places.<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 #850238 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</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=850238\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **Step 1 of 2: Find the critical value that should be used in constructing the confidence interval.**\r<BR>\n" );
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document.write( "* **Degrees of freedom (df):** n - 1 = 20 - 1 = 19<BR>\n" );
document.write( "* **Confidence level:** 90% (0.90)<BR>\n" );
document.write( "* Since the population standard deviation is unknown and the sample size is small (n < 30), we will use the t-distribution.<BR>\n" );
document.write( "* We need to find the t-value that corresponds to a 90% confidence level with 19 degrees of freedom.\r<BR>\n" );
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document.write( "Using a t-table or calculator, we find the critical t-value to be approximately 1.729.\r<BR>\n" );
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document.write( "**Step 2 of 2: Construct the 90% confidence interval.**\r<BR>\n" );
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document.write( "* **Sample mean (x̄):** $57.22<BR>\n" );
document.write( "* **Sample standard deviation (s):** $25.76<BR>\n" );
document.write( "* **Sample size (n):** 20<BR>\n" );
document.write( "* **Critical t-value (t*):** 1.729\r<BR>\n" );
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document.write( "1.  **Calculate the standard error (SE):**<BR>\n" );
document.write( "    * SE = s / √n = 25.76 / √20 ≈ 5.76\r<BR>\n" );
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document.write( "2.  **Calculate the margin of error (ME):**<BR>\n" );
document.write( "    * ME = t* * SE = 1.729 * 5.76 ≈ 9.96\r<BR>\n" );
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document.write( "3.  **Construct the confidence interval:**<BR>\n" );
document.write( "    * Lower bound: x̄ - ME = 57.22 - 9.96 ≈ 47.26<BR>\n" );
document.write( "    * Upper bound: x̄ + ME = 57.22 + 9.96 ≈ 67.18\r<BR>\n" );
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document.write( "Therefore, the 90% confidence interval for the mean repair cost is approximately ($47.26, $67.18).<BR>\n" );
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