91影视

Referring to the International Conference on Social Robotics and exercise 2.78 (p.111). A study on the current trend in the design of robots. There is a sample of 28 robots with their number of wheels on the table.

Consider the sample mean,\(\mu = \frac{1}{{28}}\sum\limits_{i = 1}^{28} {{X_i}} = \frac{{\left( {4 + 4 + 3 + \cdots + 4 + 2} \right)}}{{28}} = 3.21\)

Consider the sample standard deviation

\(\begin{aligned}\sigma &= \sqrt {\frac{{\sum\limits_{i = 1}^{28} {{{\left( {{X_i} - \mu } \right)}^2}} }}{{n - 1}}} \\ &= \sqrt {\frac{{{{\left( {4 - 3.21} \right)}^2} + \cdots + {{\left( {4 - 3.21} \right)}^2}}}{{27}}} \\ &= \sqrt {1.87} \\ &= 1.37\end{aligned}\)

The confidence interval is 99%.

Therefore, the level of significance\(\alpha = 1 - 0.99 = 0.01\).

So, from the t-table, the statistic\({t_{\alpha ,\left( {n - 1} \right)}} = {t_{0.01,27}} = 2.77\).

Therefore, the confidence interval is\(\left( {\mu \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{\sigma }{{\sqrt n }}} \right)\).

Thus,

\(\begin{aligned}\left( {\mu \pm {t_{\alpha ,\left( {n - 1} \right)}} \times \frac{\sigma }{{\sqrt n }}} \right) &= \left( {3.21 \pm {t_{0.05,27}} \times \frac{{1.37}}{{\sqrt {28} }}} \right)\\ &= \left( {\left( {3.21 - 2.77 \times 0.25} \right),\left( {3.21 + 2.77 \times 0.25} \right)} \right)\\ &= \left( {2.49,3.92} \right)\end{aligned}\)

Therefore, the 99% confidence interval is (2.49,3.92).

b.

The interval implies that a 99% sample of the true population means would lie between the range of 2.49 and 3.92.

More specifically there can be concluded that, if there are 1000 samples then 99%, that is 990 samples have the true mean value within that interval.

c.

Referring to part 鈥榓鈥�, there can be concluded that the proportion of samples that would contain the same as the true mean \(\mu \) contained in the 99% interval.