document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1186384>Question 1186384</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 20 greeting cards indicates a mean value of $3.55 and a standard deviation of $0.54. <BR>\n" );
document.write( "	Construct a 95% confidence interval estimate for the mean value of all greeting cards in the store’s inventory.<BR>\n" );
document.write( "	If the manager wants to estimate, with 95% confidence, the mean value of all greeting cards to within ± $0.03  sample error, what sample size is needed(assuming normal)?<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 #849447 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=849447\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem:\r<BR>\n" );
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document.write( "**1. Constructing the 95% Confidence Interval:**\r<BR>\n" );
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document.write( "*   **Sample mean (x̄):** $3.55<BR>\n" );
document.write( "*   **Sample standard deviation (s):** $0.54<BR>\n" );
document.write( "*   **Sample size (n):** 20<BR>\n" );
document.write( "*   **Confidence level:** 95%\r<BR>\n" );
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document.write( "Since the sample size is small (n < 30) and the population standard deviation is unknown, we use the t-distribution.\r<BR>\n" );
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document.write( "*   **Degrees of freedom (df):** n - 1 = 20 - 1 = 19<BR>\n" );
document.write( "*   **Critical t-value:** For a 95% confidence level and 19 degrees of freedom, the critical t-value is approximately 2.093 (you can find this using a t-table or calculator).\r<BR>\n" );
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document.write( "*   **Standard error:** SE = s / √n = 0.54 / √20 ≈ 0.1208\r<BR>\n" );
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document.write( "*   **Margin of error:** ME = t * SE = 2.093 * 0.1208 ≈ 0.2529\r<BR>\n" );
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document.write( "*   **Confidence interval:**  x̄ ± ME = 3.55 ± 0.2529\r<BR>\n" );
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document.write( "Therefore, the 95% confidence interval is approximately **($3.2971, $3.8029)**.\r<BR>\n" );
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document.write( "**2. Determining the required sample size:**\r<BR>\n" );
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document.write( "*   **Desired margin of error (E):** $0.03<BR>\n" );
document.write( "*   **Confidence level:** 95% (z-score = 1.96, since we are assuming normality for sample size calculation)<BR>\n" );
document.write( "*   **Estimated standard deviation (s):** $0.54 (we use the sample standard deviation as an estimate)\r<BR>\n" );
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document.write( "The formula for sample size is:\r<BR>\n" );
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document.write( "n = (z * s / E)²\r<BR>\n" );
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document.write( "n = (1.96 * 0.54 / 0.03)²<BR>\n" );
document.write( "n = (1.96 * 18)²<BR>\n" );
document.write( "n = (35.28)²<BR>\n" );
document.write( "n ≈ 1244\r<BR>\n" );
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document.write( "Therefore, a sample size of approximately **1244** greeting cards is needed to estimate the mean value to within ±$0.03 with 95% confidence.<BR>\n" );
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