$$\begin{array}{|c|c|}
\hline
S& P(S) \\ \hline
\{1,3,5,6\}& \frac{1}{8} \\ \hline
\{2,3,7,8\}&\frac{1}{4} \\ \hline
\{1,4,6,8\}& \frac{1}{8} \\ \hline
\{2,4,6,8\}& \frac{3}{8} \\ \hline
\{4,5,7,8\}& \frac{1}{8} \\ \hline
\end{array}$$
Find the probability of selection $\pi_i$ for each unit $i$.
$\pi_i= \sum_j^ P(S_j)$.
\begin{align} \text{So } \pi_1 & = \frac{1}{8}+\frac{1}{8}=\frac{1}{4} \\[10pt] \pi_2 & = \frac{1}{4}+\frac{3}{8}=\frac{5}{8} \\[10pt] \pi_3 & = \frac{1}{8}+\frac{1}{4}=\frac{3}{8} \\[10pt] \pi_4 & = \frac{1}{8}+\frac{3}{8}+\frac{1}{8}=\frac{5}{8} \\[10pt] \pi_5 & = \frac{1}{8}+\frac{1}{8}=\frac{1}{4} \\[10pt] \pi_6 & = \frac{1}{8}+\frac{1}{8}+\frac{3}{8}=\frac{5}{8} \\[10pt] \pi_7 & = \frac{1}{4}+\frac{1}{8}=\frac{3}{8} \\[10pt] \pi_8 & = \frac{1}{8}+\frac{1}{4}+\frac{1}{8}+\frac{3}{8}=\frac{7}{8} \end{align}
What is the sampling distribution of $\hat{t}=8\bar{y}$ I'm stuck on this part, $\bar{y}$ is the sample mean, so i have to find the mean of $S$, and then multiply it by $8$ to get $\hat{t}$ right? $$\begin{array}{|c|c|} \hline S&P(S)&\bar{y}&\hat{t}=\bar{y}*8 \\ \hline \{1,3,5,6\}& \frac{1}{8}& 3.75&30 \\ \hline \{2,3,7,8\}&\frac{1}{4}& 5&40 \\ \hline \{1,4,6,8\}& \frac{1}{8}& 4.75&38 \\ \hline \{2,4,6,8\}& \frac{3}{8}& 5& 40\\ \hline \{4,5,7,8\}& \frac{1}{8}& 6&48 \\ \hline \end{array}$$
Supposedly solution: $$\begin{array}{|c|c|} \hline \text{Sample number}&\text{Sample} S_i & \text{Sampled Observations}& P(S_i)&\hat{t}=\bar{y}*8 \\ \hline 1&\{1,3,5,6\}&1& 4& 7& 7 & \frac{1}{8}&38 \\ \hline 2&\{2,3,7,8\}&2&4& 7&8&\frac{1}{4}& 42 \\ \hline 3&\{1,4,6,8\}&1 &4& 7&8& \frac{1}{8} & 40 \\ \hline 4&\{2,4,6,8\}&2 &4& 7& 8& \frac{3}{8}& 42\\ \hline 5& \{4,5,7,8\}&4& 7& 7&8& \frac{1}{8}& 52 \\ \hline \end{array}$$
I don't know how they got different $\hat{t}$.