document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165218>Question 1165218</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The Quality Control Department employs five technicians during the day shift. Listed below is the number of times each technician instructed the production foreman to shut down the manufacturing process last week.\r<BR>\n" );
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document.write( "Technician	Shutdowns<BR>\n" );
document.write( "Taylor	3<BR>\n" );
document.write( "Hurley	1<BR>\n" );
document.write( "Gupta	3<BR>\n" );
document.write( "Rousche	3<BR>\n" );
document.write( "Huang	1<BR>\n" );
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document.write( "How many different samples of two technicians are possible from this population?\r<BR>\n" );
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document.write( "List all possible samples of two observations each and compute the mean of each sample. (Round your answers to 2 decimal places.)\r<BR>\n" );
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document.write( "Compare the mean of the sample means with the population mean. (Round your answers to 2 decimal places.)\r<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 #853108 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.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=853108\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem involves concepts from combinatorics and sampling distributions.\r<BR>\n" );
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document.write( "The population is the set of five technicians with their corresponding number of shutdowns: $\{3, 1, 3, 3, 1\}$.<BR>\n" );
document.write( "The population size is $N=5$.<BR>\n" );
document.write( "The sample size is $n=2$.\r<BR>\n" );
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document.write( "## 1. How many different samples of two technicians are possible?\r<BR>\n" );
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document.write( "Since the order of selection does not matter (a sample of Taylor and Hurley is the same as Hurley and Taylor) and sampling is without replacement, we use the combination formula:\r<BR>\n" );
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document.write( "$$\text{Number of Samples} = \binom{N}{n} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$$\r<BR>\n" );
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document.write( "**There are 10 different samples of two technicians possible.**\r<BR>\n" );
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document.write( "## 2. List all possible samples and compute the mean of each sample.\r<BR>\n" );
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document.write( "Let the technicians be represented by the first letter of their names: T, H, G, R, U.<BR>\n" );
document.write( "The shutdown values are: $T=3, H=1, G=3, R=3, U=1$.\r<BR>\n" );
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document.write( "| Sample (Technicians) | Shutdown Values | Sample Mean ($\bar{x}$) |<BR>\n" );
document.write( "| :---: | :---: | :---: |<BR>\n" );
document.write( "| T, H | $\{3, 1\}$ | $(3+1)/2 = 2.00$ |<BR>\n" );
document.write( "| T, G | $\{3, 3\}$ | $(3+3)/2 = 3.00$ |<BR>\n" );
document.write( "| T, R | $\{3, 3\}$ | $(3+3)/2 = 3.00$ |<BR>\n" );
document.write( "| T, U | $\{3, 1\}$ | $(3+1)/2 = 2.00$ |<BR>\n" );
document.write( "| H, G | $\{1, 3\}$ | $(1+3)/2 = 2.00$ |<BR>\n" );
document.write( "| H, R | $\{1, 3\}$ | $(1+3)/2 = 2.00$ |<BR>\n" );
document.write( "| H, U | $\{1, 1\}$ | $(1+1)/2 = 1.00$ |<BR>\n" );
document.write( "| G, R | $\{3, 3\}$ | $(3+3)/2 = 3.00$ |<BR>\n" );
document.write( "| G, U | $\{3, 1\}$ | $(3+1)/2 = 2.00$ |<BR>\n" );
document.write( "| R, U | $\{3, 1\}$ | $(3+1)/2 = 2.00$ |\r<BR>\n" );
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document.write( "## 3. Compare the mean of the sample means with the population mean.\r<BR>\n" );
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document.write( "### A. Calculate the Population Mean ($\mu$)\r<BR>\n" );
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document.write( "The population consists of all 5 shutdown values: $\{3, 1, 3, 3, 1\}$.<BR>\n" );
document.write( "$$\mu = \frac{\sum x}{N} = \frac{3 + 1 + 3 + 3 + 1}{5} = \frac{11}{5}$$<BR>\n" );
document.write( "$$\mathbf{\mu = 2.20}$$\r<BR>\n" );
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document.write( "### B. Calculate the Mean of the Sample Means ($\mu_{\bar{x}}$)\r<BR>\n" );
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document.write( "The mean of the sample means is the sum of the 10 sample means divided by the number of samples (10).<BR>\n" );
document.write( "$$\mu_{\bar{x}} = \frac{\sum \bar{x}}{10}$$<BR>\n" );
document.write( "$$\mu_{\bar{x}} = \frac{2.00 + 3.00 + 3.00 + 2.00 + 2.00 + 2.00 + 1.00 + 3.00 + 2.00 + 2.00}{10}$$<BR>\n" );
document.write( "$$\mu_{\bar{x}} = \frac{22.00}{10}$$<BR>\n" );
document.write( "$$\mathbf{\mu_{\bar{x}} = 2.20}$$\r<BR>\n" );
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document.write( "### C. Comparison\r<BR>\n" );
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document.write( "The mean of the sample means ($\mu_{\bar{x}}$) is **equal** to the population mean ($\mu$).\r<BR>\n" );
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document.write( "$$\mu_{\bar{x}} = 2.20$$<BR>\n" );
document.write( "$$\mu = 2.20$$\r<BR>\n" );
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document.write( "This demonstrates a key principle of sampling distributions: the mean of the sampling distribution of the sample mean ($\mu_{\bar{x}}$) is always equal to the population mean ($\mu$). </SPAN>\n" );
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