Question

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.

Short answer

The 95% confidence interval estimate is between 67.6 in and 71.9 in.

There is 95% confidence that the true value of the mean for the population mean of the height of sons would fall between 67.6 in and 71.9 in.

Step-by-step solution

Given information

The heights of three members of families are studied.

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

Son

71.0

64.0

71.0

68.0

70.0

71.0

71.7

71.0

Identify the formula to calculate the 95% confidence interval of thepopulation mean

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.

Determine the statistic measures from the sample points

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:

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

Obtain the confidence interval

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).