91影视

Refer to Exercise 1 for the heights of fathers and sons.

Define a random variable Das the difference between the heights of fathers and sons.

The sample mean and standard deviation of differences is calculated below:

\(\begin{aligned}{c}\bar d &= \frac{{\sum {{D_i}} }}{n}\\ &= \frac{{ - 13.7}}{8}\\ &= - 1.713\end{aligned}\)

The standard deviation of differencesiscalculated below:

\(\begin{aligned}{c}{s_D} &= \sqrt {\frac{{\sum {{{\left( {{D_i} - \bar d} \right)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( { - 3 - \left( { - 1.713} \right)} \right)}^2} + {{\left( {4 - \left( { - 1.713} \right)} \right)}^2} + ... + {{\left( {0 - \left( { - 1.713} \right)} \right)}^2}}}{{8 - 1}}} \\ &= 2.847\end{aligned}\)

Let be the true mean heights of the population of fathers and sons.

Define \({\mu _F} - {\mu _S} = {\mu _d}\).

The hypotheses are statedbelow:

\(\begin{array}{l}{H_o}:{\mu _d} = 0\\{H_a}:{\mu _d} \ne 0\end{array}\)

The test statistic is shown below:

\(\begin{aligned}{c}t &= \frac{{\bar d - {\mu _D}}}{{\frac{{{s_D}}}{{\sqrt n }}}}\\ &= \frac{{ - 1.713 - 0}}{{\frac{{2.847}}{{\sqrt 8 }}}}\\ &= - 1.702\end{aligned}\)

Thus, the test statistic is 鈥�1.702.

The degree of freedom is

\(\begin{aligned}{c}df &= n - 1\\ &= 8 - 1\\ &= 7.\end{aligned}\)

The p-value is computed from the t-distribution table.

\(\begin{aligned}{c}p{\rm{ - value}} &= 2P\left( {T < - 1.702} \right)\\ &= 0.1325\end{aligned}\)

Since the p-value is greater than 0.05, the null hypothesis cannot be rejected.

Thus, there is not enough evidence to rejectthe claim that the differences in heights of fathers and sons have a true mean of0.