91影视

The heights of women and men (in cm) are tabulated.

The following formula is used to construct the confidence interval for estimating the difference in the mean heights of women and men:

\(CI = \left( {\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E,\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E} \right)\)

Where,

• \({\bar x_1}\)denotes the sample mean height of women.
• \({\bar x_2}\)denotes the sample mean height of men.
• E is the margin of error.

The sample size for the heights of women is equal to\({n_1} = 10\).

The sample size for the heights of men is equal to\({n_2} = 10\).

The sample mean height of women is computed below:

\(\begin{aligned}{c}{{\bar x}_1} &= \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_{1i}}} }}{{{n_1}}}\\ &= \frac{{160.3 + 167.7 + ...... + 171.1}}{{10}}\\ &= 162.35\end{aligned}\)

The sample mean height of men is computed below:

\(\begin{aligned}{c}{{\bar x}_2} &= \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_{2i}}} }}{{{n_2}}}\\ &= \frac{{190.3 + 169.8 + ...... + 181.3}}{{10}}\\ &= 178.77\end{aligned}\)

The sample variance of heights of women is computed below:

\(\begin{aligned}{c}s_1^2 &= \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_{1i}} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ &= \frac{{{{\left( {160.3 - 162.35} \right)}^2} + {{\left( {167.7 - 162.35} \right)}^2} + ....... + {{\left( {166.9 - 162.35} \right)}^2}}}{{10 - 1}}\\ &= 140.35\end{aligned}\)

The sample variance of heights of men is computed below:

\(\begin{aligned}{c}s_2^2 &= \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_{2i}} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ &= \frac{{{{\left( {190.3 - 178.77} \right)}^2} + {{\left( {169.8 - 178.77} \right)}^2} + ....... + {{\left( {181.3 - 178.77} \right)}^2}}}{{10 - 1}}\\ &= 28.11\end{aligned}\)

The margin of error has the following formula:

\(E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \)

The confidence level is equal to 95%.Thus, the corresponding level of significance is equal to 0.05.

The degrees of freedom are computed below:

\(\begin{aligned}{c}df &= {\rm{smaller}}\;of\;\left( {{n_1} - 1} \right)\;{\rm{and}}\;\left( {{n_2} - 1} \right)\\ &= {\rm{smaller}}\;of\;\left( {10 - 1} \right)\;{\rm{and}}\;\left( {10 - 1} \right)\\ &= 9\end{aligned}\)

The value of \({t_{\frac{\alpha }{2}}}\), when \(\alpha = 0.05\)and degrees of freedom equal to 9,is equal to 2.2622.

Thus, the margin of error is equal to:

\(\begin{aligned}{c}E &= {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ &= \left( {2.2622} \right)\sqrt {\frac{{140.35}}{{10}} + \frac{{28.11}}{{10}}} \\ &= 9.285\end{aligned}\)

Thus, the 95% confidence interval is equal to:

\(\begin{aligned}{c}CI &= \left( {\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E,\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E} \right)\\ &= \left( {\left( {162.35 - 178.77} \right) - 9.285,\left( {162.35 - 178.77} \right) + 9.285} \right)\\ &= \left( { - 25.705, - 7.135} \right)\\ &\approx \left( { - 25.71, - 7.14} \right)\end{aligned}\)

Therefore, the confidence interval to estimate the difference between the mean heights of women and men is equal to (鈥�25.71, 鈥�7.14).

It can be concluded that the mean height of women is less than the mean height of men and the difference between the means is between 7.14 cm and 25.71 cm.