document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1191336>Question 1191336</A>: <I><SPAN STYLE=\"word-break: break-all;\"> he personnel director of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers reveals the following:<BR>\n" );
document.write( "• Absenteeism:  = 9.7 days, S = 4.0 days.<BR>\n" );
document.write( "• 12 clerical workers were absent more than 10 days.<BR>\n" );
document.write( "a. Construct a 95% confidence interval estimate for the mean number of absences for clerical workers during the year.<BR>\n" );
document.write( "b. Construct a 95% confidence interval estimate for the population proportion of clerical workers absent more than 10 days during the year. Suppose that the personnel director also wishes to take a survey in a branch office. Answer these questions:<BR>\n" );
document.write( "c. What sample size is needed to have 95% confidence in estimating the population mean absenteeism to within ±1.5 days if the population standard deviation is estimated to be 4.5 days?<BR>\n" );
document.write( "d. How many clerical workers need to be selected to have 90% confidence in estimating the population proportion to within ± 0.075 if no previous estimate is available?<BR>\n" );
document.write( "e. Based on (c) and (d), what sample size is needed if a single survey is being conducted? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849224 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=849224\">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( "**a) 95% Confidence Interval for Mean Absences:**\r<BR>\n" );
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document.write( "Since the sample size is small (n=25) and the population standard deviation is unknown, we use the t-distribution.\r<BR>\n" );
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document.write( "1. **Find the critical t-value:** For a 95% confidence level and 24 degrees of freedom (n-1 = 25-1 = 24), the critical t-value is approximately 2.064 (you can find this using a t-table or calculator).\r<BR>\n" );
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document.write( "2. **Calculate the margin of error:**<BR>\n" );
document.write( "   Margin of Error = t * (S / √n) = 2.064 * (4.0 / √25) = 2.064 * 0.8 = 1.6512\r<BR>\n" );
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document.write( "3. **Construct the confidence interval:**<BR>\n" );
document.write( "   Confidence Interval = ȳ ± Margin of Error = 9.7 ± 1.6512\r<BR>\n" );
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document.write( "   The 95% confidence interval is approximately (8.05, 11.35) days.\r<BR>\n" );
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document.write( "**b) 95% Confidence Interval for Population Proportion:**\r<BR>\n" );
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document.write( "1. **Calculate the sample proportion (p̂):**<BR>\n" );
document.write( "   p̂ = (Number of workers absent > 10 days) / (Total number of workers) = 12/25 = 0.48\r<BR>\n" );
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document.write( "2. **Find the critical z-value:** For a 95% confidence level, the critical z-value is 1.96.\r<BR>\n" );
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document.write( "3. **Calculate the margin of error:**<BR>\n" );
document.write( "   Margin of Error = z * √(p̂(1 - p̂) / n) = 1.96 * √(0.48 * 0.52 / 25) ≈ 0.196\r<BR>\n" );
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document.write( "4. **Construct the confidence interval:**<BR>\n" );
document.write( "   Confidence Interval = p̂ ± Margin of Error = 0.48 ± 0.196\r<BR>\n" );
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document.write( "   The 95% confidence interval is approximately (0.284, 0.676).\r<BR>\n" );
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document.write( "**c) Sample Size for Mean Absences:**\r<BR>\n" );
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document.write( "1. **Use the sample size formula:**<BR>\n" );
document.write( "   n = (z * σ / E)²\r<BR>\n" );
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document.write( "   Where:<BR>\n" );
document.write( "   * z is the critical z-value for 95% confidence (1.96)<BR>\n" );
document.write( "   * σ is the estimated population standard deviation (4.5 days)<BR>\n" );
document.write( "   * E is the desired margin of error (1.5 days)\r<BR>\n" );
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document.write( "2. **Calculate:**<BR>\n" );
document.write( "   n = (1.96 * 4.5 / 1.5)² = (5.88)² = 34.57\r<BR>\n" );
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document.write( "   Since you can't have a fraction of a worker, round up to the nearest whole number.  Therefore, a sample size of 35 workers is needed.\r<BR>\n" );
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document.write( "**d) Sample Size for Population Proportion:**\r<BR>\n" );
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document.write( "1. **Use the sample size formula (when no prior estimate is available):**<BR>\n" );
document.write( "   n = (z² * 0.25) / E²\r<BR>\n" );
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document.write( "   Where:<BR>\n" );
document.write( "   * z is the critical z-value for 90% confidence (1.645)<BR>\n" );
document.write( "   * E is the desired margin of error (0.075)\r<BR>\n" );
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document.write( "2. **Calculate:**<BR>\n" );
document.write( "   n = (1.645² * 0.25) / 0.075² ≈ 120.278\r<BR>\n" );
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document.write( "   Round up to the nearest whole number. A sample size of 121 workers is needed.\r<BR>\n" );
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document.write( "**e) Sample Size for Single Survey:**\r<BR>\n" );
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document.write( "Since a single survey is being conducted, you need to choose the larger of the two sample sizes calculated in parts (c) and (d).  Therefore, a sample size of 121 workers is needed to satisfy both requirements.<BR>\n" );
document.write( " </SPAN>\n" );
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