Question

In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Heights of Fathers and Sons Listed below are heights (in.) of fathers and their first sons. The data are from a journal kept by Francis Galton. (See Data Set 5 “Family Heights”in Appendix B.) Use a 0.05 significance level to test the claim that there is no difference in heights between fathers and their first sons.

Height of Father	72	66	69	70	70	70	70	75	68.2	65
Height of Son	73	68	68	71	70	70	71	71	70	63

Short answer

There is insufficient evidence to support the claim that the height of fathers and their first sons have no difference.

Step-by-step solution

Given information

The heights of pairs of fathers and their first sons are recorded.

Hypotheses

It is claimed thatthere is no difference in heights between fathers and their first sons.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis: The mean of the differences between the height of fathers and their first sons is equal to 0.

\({H_0}:{\mu _d} = 0\)

Alternative Hypothesis: The mean of the differences between the height of fathers and their first sons is not equal to 0.

\({H_1}:{\mu _d} \ne 0\)

The test is two-tailed.

Differences in the values of each matched pair

The following table shows the differences in the heights of the fathers and sons for each matched pair:

Father	72	66	69	70	70	70	70	75	68.2	65
Son	73	68	68	71	70	70	71	71	70	63
Differences(d)	-1	-2	1	-1	0	0	-1	4	-1.8	2

Mean and sample standard deviation of the differences

The number of pairs (n) is equal to 10.

The mean value of the differences is computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 1} \right) + \left( { - 2} \right) + \ldots + 2}}{{10}}\\ = 0.02\end{array}\)

The standard deviation of the differences is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 1} \right) - 0.02} \right)}^2} + {{\left( {\left( { - 2} \right) - 0.02} \right)}^2} + ..... + {{\left( {2 - 0.02} \right)}^2}}}{{10 - 1}}} \\ = 1.86\end{array}\)

The mean value of the differences for the population of matched pairs \(\left( {{\mu _d}} \right)\) is considered to be equal to 0.

Test statistic

The value of the test statistic is computed as shown:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{0.02 - 0}}{{\frac{{1.86}}{{\sqrt {10} }}}}\\ = 0.034\end{array}\)

The degrees of freedom are computed below:

\(\begin{array}{c}df = n - 1\\ = 10 - 1\\ = 9\end{array}\)

Referring to the t-distribution table, the critical value of t at\(\alpha = 0.05\)and degrees of freedom equal to 9 for a two-tailed test are -2.2622 and 2.2622.

The right-tailed p-value of t equal to 0.034 is equal to 0.9729.

Conclusion

Since the p-value is greater than 0.05 and the test statistic lies between the two critical values, the null hypothesis is failed to reject.

There is insufficient evidence to support the claim that the height of fathers and their first sons have no difference.