91Ó°ÊÓ

To find the number of different samples of size \(n=2\) that can be selected without replacement from the population, we use the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), with \(n=4\) and \(k=2\). Thus, the number of possible samples is \(\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4\times 3}{2\times 1} = 6\). So there are 6 different samples of size \(n=2\) without replacement.

To generate all possible samples of size \(n=2\), we can perform pairings of the population elements: \((6,1)\), \((6,3)\), \((6,2)\), \((1,3)\), \((1,2)\), and \((3,2)\). So the possible samples of size \(n=2\) are: \(S_1=(6,1)\), \(S_2=(6,3)\), \(S_3=(6,2)\), \(S_4=(1,3)\), \(S_5=(1,2)\), and \(S_6=(3,2)\).

To compute the sample mean (\(\bar{x}\)) for each sample, add the elements of each sample and then divide the sum by the number of elements (2) in the sample: \(\bar{x}_{S_1}=\frac{6+1}{2} = 3.5\) \(\bar{x}_{S_2}=\frac{6+3}{2} = 4.5\) \(\bar{x}_{S_3}=\frac{6+2}{2} = 4\) \(\bar{x}_{S_4}=\frac{1+3}{2} = 2\) \(\bar{x}_{S_5}=\frac{1+2}{2} = 1.5\) \(\bar{x}_{S_6}=\frac{3+2}{2} = 2.5\)

The sampling distribution of \(\bar{x}\) lists the possible values of \(\bar{x}\) and their respective probabilities. Since all samples are equally likely, the probability of each value of \(\bar{x}\) is the number of samples for that value divided by the total number of samples: 1. \(\bar{x}=1.5\) : Probability \(P(\bar{x}=1.5) = \frac{1}{6}\) 2. \(\bar{x}=2\) : Probability \(P(\bar{x}=2) = \frac{1}{6}\) 3. \(\bar{x}=2.5\) : Probability \(P(\bar{x}=2.5) = \frac{1}{6}\) 4. \(\bar{x}=3.5\) : Probability \(P(\bar{x}=3.5) = \frac{1}{6}\) 5. \(\bar{x}=4\) : Probability \(P(\bar{x}=4) = \frac{1}{6}\) 6. \(\bar{x}=4.5\) : Probability \(P(\bar{x}=4.5) = \frac{1}{6}\) To graph the sampling distribution of \(\bar{x}\), create a histogram with the values of \(\bar{x}\) on the horizontal axis and the probabilities on the vertical axis. Each bar should have a height equal to the probability of the corresponding value of \(\bar{x}\).

Since all four population values are equally likely, the population mean \(\mu\) is the sum of all population values divided by the total number of population values: \(\mu=\frac{6+1+3+2}{4}=\frac{12}{4}=3\) Now, compare the value of \(\mu\) to the sample means calculated in the previous steps. None of the samples (\(S_1\) to \(S_6\)) have a value of \(\bar{x}\) exactly equal to \(\mu\).