Question

Consider the set {1, 2, 3, 4}.

Answer 1

Answer:

(a)

$List = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),\\(4,1),(4,2),(4,3).(4,4)\}$

(b) Sampling Distribution (Table)

$\begin{array}{cccccccc}{\bar x} & {1} & {1.5} & {2} & {2.5} & {3} & {3.5} & {4} & {Pr}& {\frac{1}{16}} & {\frac{1}{8}} & {\frac{3}{16}} & {\frac{1}{4}} & {\frac{3}{16}} & {\frac{1}{8}} & {\frac{1}{16}} \ \end{array}$

(b) Sampling Distribution (Histogram)

See attachment

Step-by-step explanation:

Given

$Set = \{1,2,3,4\}$

$n =4$

Solving (a): A list of sample size 2

We have:

$n =4$

$r = 2$ --- the sample size

First, we calculate the number of list using permutation (orders matter)

$n(List) = n^r$

So, we have:

$n(List) = 4^2$

$n(List) = 16$

And the list is:

$List = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),\\(4,1),(4,2),(4,3).(4,4)\}$

Solving (b): Sample distribution  of sample means of (a)

First, calculate the mean of each set using:

$Mean = \frac{Sum}{2}$

So, we have:

$(1,1) \to \frac{1+1}{2} \to 1$       $(1,2) \to \frac{1+2}{2} \to 1.5$    $(1,3) \to \frac{1+3}{2} \to 2$    $(1,4) \to \frac{1+4}{2} \to 2.5$

$(2,1) \to \frac{2+1}{2} \to 1.5$    $(2,2) \to \frac{2+2}{2} \to 2$     $(2,3) \to \frac{2+3}{2} \to 2.5$    $(2,4) \to \frac{2+4}{2} \to 3$

$(3,1) \to \frac{3+1}{2} \to 2$       $(3,2) \to \frac{3+2}{2} \to 2.5$    $(3,3) \to \frac{3+3}{2} \to 3$    $(3,4) \to \frac{3+4}{2} \to 3.5$

$(4,1) \to \frac{4+1}{2} \to 2.5$    $(4,2) \to \frac{4+2}{2} \to 3$    $(4,3) \to \frac{4+3}{2} \to 3.5$    $(4,4) \to \frac{4+4}{2} \to 4$

Write out the sample means (sorted)

$\bar x =\{1,1.5,1.5,2,2,2,2.5,2.5,2.5,2.5,3,3,3,3.5,3.5,4\}$

Construct a frequency table

$\begin{array}{cc}{\bar x} & {f} & {1} & {1} & {1.5} & {2} & {2} & {3} & {2.5} & {4} & {3} & {3} & {3.5} &{2} & {4} & {1} & Total & 16\ \end{array}$

Construct the sampling distribution where the probability is calculated using: $\frac{f}{Total}$

So, we have:

$\begin{array}{cccccccc}{\bar x} & {1} & {1.5} & {2} & {2.5} & {3} & {3.5} & {4} & {Pr}& {\frac{1}{16}} & {\frac{1}{8}} & {\frac{3}{16}} & {\frac{1}{4}} & {\frac{3}{16}} & {\frac{1}{8}} & {\frac{1}{16}} \ \end{array}$