Family Heights. In Exercises 1–5, use the following heights (in.) The data are matched so that each column consists of heights from the same family.

Father	68.0	68.0	65.5	66.0	67.5	70.0	68.0	71.0
Mother	64.0	60.0	63.0	59.0	62.0	69.0	65.5	66.0
Son	71.0	64.0	71.0	68.0	70.0	71.0	71.7	71.0

Confidence Interval Construct a 95% confidence interval estimate of the mean height of sons. Write a brief statement that interprets the confidence interval.

The heights of three members of families are studied.

The data of son’s height is given as follows:

Thus, the 95% confidence interval is

\(\bar x - E < \mu < \bar x + E\).

Here,\(\bar x\)isthe sample mean, and E is the margin of error.

For 8 (n) samples, the sample mean is calculated below:

\(\begin{aligned}{c}\bar x &= \frac{{\sum {{x_i}} }}{n}\\ &= \frac{{71 + 64 + ... + 71}}{8}\\ &= 69.7\;{\rm{in}}\end{aligned}\)

The sample standard deviation is calculated below:

\(\begin{aligned}{c}s &= \sqrt {\frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {71 - 69.7} \right)}^2} + {{\left( {64 - 69.7} \right)}^2} + ... + {{\left( {71 - 69.7} \right)}^2}}}{{8 - 1}}} \\ &= 2.57\;{\rm{in}}{\rm{.}}\end{aligned}\)

The 95% confidence level implies a 0.05 significance level.

The degree of freedom is

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

The critical value of 2.364 is obtained from the t-distribution table at 7 degrees of freedomand 0.05 significance level.

The margin of error is computed below:

\(\begin{aligned} E &= {t_{\frac{\alpha }{2}}} \times \frac{s}{{\sqrt n }}\\ &= 2.364 \times \frac{{2.57}}{{\sqrt 8 }}\\ &= 2.14\\ &\approx 2.1\end{aligned}\)

Thus, the 95% confidence interval is as calculated below:

\(\begin{aligned}{l}\bar x - E < \mu < \bar x + E &= 69.7 - 2.1 < \mu < 69.7 + 2.1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 67.6 < \mu < 71.9\end{aligned}\)

Thus, the 95% confidence interval for the mean height of sons is (67.6 in, 71.9 in).