document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197705>Question 1197705</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In a sample of credit card holders the mean monthly value of credit card purchases was $ 350 and the sample variance was 40 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate.\r<BR>\n" );
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document.write( "(a) Suppose the sample results were obtained from a random sample of 15 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.\r<BR>\n" );
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document.write( "(b) Suppose the sample results were obtained from a random sample of 22 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.\r<BR>\n" );
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document.write( "Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). Then check that your answer verifies this. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #848357 by </B><A HREF=/tutors/aboutme.mpl?userid=onyulee><B>onyulee(41)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=onyulee><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=848357\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **a) Sample size of 15**\r<BR>\n" );
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document.write( "* **Given:**<BR>\n" );
document.write( "    * Sample mean (x̄) = $350<BR>\n" );
document.write( "    * Sample variance (s²) = 40 <BR>\n" );
document.write( "    * Sample standard deviation (s) = √40 = $6.32<BR>\n" );
document.write( "    * Sample size (n) = 15<BR>\n" );
document.write( "    * Confidence level: 95% <BR>\n" );
document.write( "    * Degrees of freedom (df) = n - 1 = 15 - 1 = 14\r<BR>\n" );
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document.write( "* **Find the critical t-value:**<BR>\n" );
document.write( "    * For a 95% confidence level and 14 degrees of freedom, the critical t-value (tα/2) from a t-distribution table is approximately 2.145.\r<BR>\n" );
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document.write( "* **Calculate the standard error:** <BR>\n" );
document.write( "    * Standard Error (SE) = s / √n = 6.32 / √15 ≈ 1.63\r<BR>\n" );
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document.write( "* **Calculate the margin of error:**<BR>\n" );
document.write( "    * Margin of Error = tα/2 * SE = 2.145 * 1.63 ≈ 3.50\r<BR>\n" );
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document.write( "* **Calculate the confidence interval:**<BR>\n" );
document.write( "    * Lower limit: x̄ - Margin of Error = 350 - 3.50 = $346.50<BR>\n" );
document.write( "    * Upper limit: x̄ + Margin of Error = 350 + 3.50 = $353.50\r<BR>\n" );
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document.write( "* **95% Confidence Interval:** ($346.50, $353.50)\r<BR>\n" );
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document.write( "**b) Sample size of 22**\r<BR>\n" );
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document.write( "* **Prediction:** <BR>\n" );
document.write( "    * The confidence interval with a larger sample size (n = 22) will be narrower than the interval with a smaller sample size (n = 15). <BR>\n" );
document.write( "    * This is because a larger sample size generally provides a more precise estimate of the population parameter.\r<BR>\n" );
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document.write( "* **Calculations:**<BR>\n" );
document.write( "    * Degrees of freedom (df) = 22 - 1 = 21<BR>\n" );
document.write( "    * Critical t-value (tα/2) for 95% confidence and 21 degrees of freedom is approximately 2.080.<BR>\n" );
document.write( "    * Standard Error (SE) = 6.32 / √22 ≈ 1.35<BR>\n" );
document.write( "    * Margin of Error = 2.080 * 1.35 ≈ 2.81<BR>\n" );
document.write( "    * Lower limit: 350 - 2.81 = $347.19<BR>\n" );
document.write( "    * Upper limit: 350 + 2.81 = $352.81\r<BR>\n" );
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document.write( "* **95% Confidence Interval:** ($347.19, $352.81)\r<BR>\n" );
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document.write( "**Verification:**\r<BR>\n" );
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document.write( "* The confidence interval for the sample size of 22 is indeed narrower than the interval for the sample size of 15, as predicted.\r<BR>\n" );
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document.write( "**In Summary:**\r<BR>\n" );
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document.write( "* The 95% confidence interval for the mean monthly value of credit card purchases for a sample size of 15 is ($346.50, $353.50).<BR>\n" );
document.write( "* The 95% confidence interval for the mean monthly value of credit card purchases for a sample size of 22 is ($347.19, $352.81).<BR>\n" );
document.write( "* As expected, the larger sample size results in a narrower confidence interval, indicating a more precise estimate of the population mean.<BR>\n" );
document.write( " </SPAN>\n" );
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