91Ó°ÊÓ

Data re paired if a data value in one sample corresponds with a unique data value in the other sample.

In this case, we note that the outbound data values belong to the same days as the return data values which implies that we have two data values per day and thus each of these data values per day forms a pair.

The dots in the dot plot represents the difference between outbound and return in flight time.

We note that dots in the dot plot lie to the right of the zero which means that most of the differences are positive and thus the outbound flight time appears to be higher than the return flight time in general.

It is given that the table of the difference which ids the outbound data less by the return data value.

The mean is:

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i-1}{\overset{n}{∑}}}{x}_i}{n} \\ =\frac{33+5+25+19+12+6+11+6+1+4-5+4}{12} \\ =\frac{121}{12} \\ =10.0833 \end{gathered}$$

The sample variance is then as:

$$\begin{gathered} s^2=\frac{\underset{}{∑}{(x-\stackrel{¯}{x})}^2}{n-1} \\ =\frac{{(33-10.0833)}^2+(5-10.0833{)}^2+(25-10.0833{)}^2+(19-10.0833{)}^2+(12-10.0833{)}^2+(6-10.0833{)}^2+(11-10.0833{)}^2+(6-10.0833{)}^2+(1-10.0833{)}^2+(4-10.0833{)}^2+(-5-10.0833{)}^2+(4-10.0833{)}^2}{12-1} \end{gathered}$$

$=115.9015$

The sample standard deviation is then,

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{115.9015} \\ =10.7658 \end{gathered}$$

The difference are on average$10.0833$ minutes which tend to vary on average by$10.7658$ minutes from the average of$10.0833$minutes.