Question

Hypothesis Test Use a 0.05 significance level to test the claim that differences between heights of fathers and their sons have a mean of 0 in.

Short answer

There is insufficient evidence to reject the claim that the difference in the heights of fathers and sons has a 0-inch mean difference at 0.05 level of significance.

Step-by-step solution

Given information

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

Fathers	Sons
68.0	71.0
68.0	64.0
65.5	71.0
66.0	68.0
67.5	70.0
70.0	71.0
68.0	71.7
71.0	71.0

Determine the statistic measures from the sample points

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

Father	Son	Difference (D)
68	71	–3
68	64	4
65.5	71	–5.5
66	68	–2
67.5	70	–2.5
70	71	–1
68	71.7	–3.7
71	71	0

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:

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

Identify the hypotheses

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

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

The hypotheses are statedbelow:

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

Determine the test statistic

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 –1.702.

Determine the decision and conclusion of the test

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.