91影视

The heights of three members of families are studied.

The data of son鈥檚 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).