Assume that the differences are normally distributed. Complete parts (a) through (d) below.
\begin{tabular}{lcccccccc}
Observation & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
$\mathrm{X}_{\mathrm{i}}$ & 47.2 & 47.8 & 40.8 & 42.9 & 44.4 & 48.4 & 50.1 & 46.5 \\
$\mathrm{Y}_{\mathrm{i}}$ & 50.7 & 47.7 & 45.7 & 48.5 & 46.2 & 51.6 & 54.7 & 49.2
\end{tabular}
(a) Determine $d_{i}=X_{i}-Y_{i}$ for each pair of data.
\begin{tabular}{lcccccccc}
Observation & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline$d_{i}$ & -3.5 & 0.1 & -4.9 & -5.6 & -1.8 & -3.2 & -4.6 & -2.7
\end{tabular}
(Type integers or decimals.)
(b) Compute $\bar{d}$ and $s_{d}$
$\overline{\mathrm{d}}=\square$ (Round to three decimal places as needed.)
Answer
Take the square root of the variance to get the standard deviation, $s_{d} = \sqrt{s_{d}^2}$
Step 1 ：Calculate the difference between each pair of data, $d_{i}=X_{i}-Y_{i}$, to get: $-3.5, 0.1, -4.9, -5.6, -1.8, -3.2, -4.6, -2.7$
Step 2 ：Compute the mean of these differences, $\bar{d} = \frac{-3.5 + 0.1 - 4.9 - 5.6 - 1.8 - 3.2 - 4.6 - 2.7}{8} = -3.275$
Step 3 ：Compute the variance of these differences, $s_{d}^2 = \frac{(-3.5 - (-3.275))^2 + (0.1 - (-3.275))^2 + (-4.9 - (-3.275))^2 + (-5.6 - (-3.275))^2 + (-1.8 - (-3.275))^2 + (-3.2 - (-3.275))^2 + (-4.6 - (-3.275))^2 + (-2.7 - (-3.275))^2}{8-1}$
Step 4 ：Take the square root of the variance to get the standard deviation, $s_{d} = \sqrt{s_{d}^2}$
