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 Mothers and Daughters Listed below are heights (in.) of mothers and their first daughters. The data are from a journal kept by Francis Galton. (See Data Set 5 “FamilyHeights” in Appendix B.) Use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters.

Height of Mother	68	60	61	63.5	69	64	69	64	63.5	66
Height of Daughter	68.5	60	63.5	67.5	68	65.5	69	68	64.5	63

The heights of pairs of mothers and the first daughters are recorded.

It is claimed thatthere is no difference in heights between mothers and their first daughters.

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

Null Hypothesis: The mean of the difference between the heights of the mother and the first daughter is equal to 0.

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

Alternative Hypothesis: The mean of the difference between the heights of the mother and the first daughter is not equal to 0.

\[{H_1}:{\mu _d} \ne 0\]

The test is two-tailed.

The following table shows the differences in the heights of the mother and the first daughter for each matched pair:

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

The mean value of the differences is computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 0.5} \right) + 0 + \ldots + 3}}{{10}}\\ = - 0.95\end{array}\)

The standard deviation of the differences is computed below:

\[\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 0.5} \right) - \left( { - 0.95} \right)} \right)}^2} + {{\left( {0 - \left( { - 0.95} \right)} \right)}^2} + ... + {{\left( {3 - \left( { - 0.95} \right)} \right)}^2}}}{{10 - 1}}} \\ = 2.18\end{array}\]

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

The value of the test statistic is computed as shown:

\(\begin{array}{c}t = \frac{{\bar d - {\mu _d}}}{{\frac{{{s_d}}}{{\sqrt n }}}}\\ = \frac{{ - 0.95 - 0}}{{\frac{{2.18}}{{\sqrt {10} }}}}\\ = - 1.379\end{array}\)

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 values 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 p-value of t equal to -1.379 is equal to 0.2012.

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

There is insufficient evidence to reject the claim that there is no difference in heights between mothers and their first daughters.