Do social robots walk or roll? Refer to the International Conference on Social Robotics(Vol. 6414, 2010) study onthe current trend in the design of social robots, Exercise2.78 (p. 111). Recall that in a random sample of social

robots obtained through a Web search, 28 were built with wheels. The accompanying table shows the number of wheels on each of the 28 robots.

  1. Estimate \({\bf{\mu }}\), the average number of wheels used on all social robots built with wheels, with 99% confidence.
  2. Practically interpret the interval, part a.
  3. Refer to part a. In repeated sampling, what proportion of all similarly constructed confidence intervals will contain the true mean, \({\bf{\mu }}\)?

Short Answer

Expert verified
  1. The average number of wheels used on all social robots built with wheels is 3.21 and the 99% confidence interval is [2.49,3.92].
  2. The interval implies that a 99% sample of the true population means would lie between the range of 2.49 and 3.92.
  3. Referring to part ‘a’, there can be concluded that the proportion of samples that would contain the same as the true mean \(\mu \) contained in the 99% interval.

Step by step solution

01

Given information

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.

02

Estimate the confidence interval

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).

03

Interpreting the confidence interval

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.

04

Determine the proportion

c.

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

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Most popular questions from this chapter

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