document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1179607>Question 1179607</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A business consultant wanted to investigate if providing day care facilities on premises<BR>\n" );
document.write( "by companies reduces the absentee rate of working mothers from companies that provide<BR>\n" );
document.write( "day care facilities on premises. Sample of 50 mothers selected from the companies that<BR>\n" );
document.write( "provide day care facilities was taken. These mothers missed an average of 6.4 days from<BR>\n" );
document.write( "work last year with a standard deviation of 1.20 days. Another sample of 50 such mothers<BR>\n" );
document.write( "taken from companies that do not provide day care facilities on premises showed that these<BR>\n" );
document.write( "mothers missed an average of 9.3 days last year with a standard deviation of 1.83 days.<BR>\n" );
document.write( "Construct a 98 % confidence interval for the difference between the two population means. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850206 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=850206\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To construct a 98% confidence interval for the difference between the two population means, we'll use a two-sample t-test. Here's how we can approach this problem:\r<BR>\n" );
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document.write( "**1. Define the given information:**\r<BR>\n" );
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document.write( "* Sample 1 (with daycare):<BR>\n" );
document.write( "    * n1 = 50 (sample size)<BR>\n" );
document.write( "    * x1 = 6.4 (sample mean)<BR>\n" );
document.write( "    * s1 = 1.20 (sample standard deviation)<BR>\n" );
document.write( "* Sample 2 (without daycare):<BR>\n" );
document.write( "    * n2 = 50 (sample size)<BR>\n" );
document.write( "    * x2 = 9.3 (sample mean)<BR>\n" );
document.write( "    * s2 = 1.83 (sample standard deviation)<BR>\n" );
document.write( "* Confidence level = 98% (which means alpha = 1 - 0.98 = 0.02)\r<BR>\n" );
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document.write( "**2. Calculate the degrees of freedom (df):**\r<BR>\n" );
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document.write( "* df = n1 + n2 - 2 = 50 + 50 - 2 = 98\r<BR>\n" );
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document.write( "**3. Calculate the pooled standard deviation (sp):**\r<BR>\n" );
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document.write( "* Since we have two independent samples, we can calculate the pooled standard deviation:<BR>\n" );
document.write( "    * sp = √[((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / df]<BR>\n" );
document.write( "    * sp = √[((49 * 1.20^2) + (49 * 1.83^2)) / 98]<BR>\n" );
document.write( "    * sp ≈ √[(70.56 + 163.6329) / 98]<BR>\n" );
document.write( "    * sp ≈ √(234.1929 / 98)<BR>\n" );
document.write( "    * sp ≈ √2.389723469<BR>\n" );
document.write( "    * sp ≈ 1.545873\r<BR>\n" );
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document.write( "**4. Calculate the standard error (SE):**\r<BR>\n" );
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document.write( "* SE = sp * √((1/n1) + (1/n2))<BR>\n" );
document.write( "* SE = 1.545873 * √((1/50) + (1/50))<BR>\n" );
document.write( "* SE = 1.545873 * √(0.02 + 0.02)<BR>\n" );
document.write( "* SE = 1.545873 * √0.04<BR>\n" );
document.write( "* SE = 1.545873 * 0.2<BR>\n" );
document.write( "* SE ≈ 0.3091746\r<BR>\n" );
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document.write( "**5. Find the t-value:**\r<BR>\n" );
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document.write( "* For a 98% confidence interval and 98 degrees of freedom, we need to find the t-value that corresponds to an alpha of 0.02. This means we are looking for the t-value that leaves 0.01 in each tail.<BR>\n" );
document.write( "* Using a t-table or calculator, the t-value for df = 98 and alpha/2 = 0.01 is approximately 2.365.\r<BR>\n" );
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document.write( "**6. Calculate the margin of error (ME):**\r<BR>\n" );
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document.write( "* ME = t-value * SE<BR>\n" );
document.write( "* ME = 2.365 * 0.3091746<BR>\n" );
document.write( "* ME ≈ 0.7311187\r<BR>\n" );
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document.write( "**7. Calculate the confidence interval:**\r<BR>\n" );
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document.write( "* Confidence interval = (x1 - x2) ± ME<BR>\n" );
document.write( "* Confidence interval = (6.4 - 9.3) ± 0.7311187<BR>\n" );
document.write( "* Confidence interval = -2.9 ± 0.7311187<BR>\n" );
document.write( "* Lower bound = -2.9 - 0.7311187 = -3.6311187<BR>\n" );
document.write( "* Upper bound = -2.9 + 0.7311187 = -2.1688813\r<BR>\n" );
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document.write( "**8. Round the results:**\r<BR>\n" );
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document.write( "* The 98% confidence interval is approximately (-3.63, -2.17).\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "The 98% confidence interval for the difference between the two population means is approximately (-3.63, -2.17). This suggests that mothers working in companies that provide daycare facilities on premises miss significantly fewer days of work compared to those who don't.<BR>\n" );
document.write( " </SPAN>\n" );
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