Question 9.10: An Olympic decathlon athlete is interested in his performanc......
An Olympic decathlon athlete is interested in his performance compared with the performance of other athletes. He is a good runner and interested in his 100 m results compared with those of other athletes.
(a) He uses the decathlon data from this book (Appendix A.2) to come up with \hat{\sigma } = s = 0.233. What sample size does he need to calculate a 95% confidence interval for the mean running time which is precise to ±0.1s?
(b) Calculate a 95% confidence interval for the mean running time ( \bar{x} = 10.93) of the 30 athletes captured in the data set in Chap. A.2. Interpret the width of this interval compared with the width determined in a).
(c) The runner’s own best time is 10.86 s. He wants to be among the best 10% of all athletes. Calculate an appropriate confidence interval to compare his time with the 10% best times.
(a) Using z_{{1−α}/{2}} = 1.96 and \hat{\sigma }= 0.233, we obtain an optimal sample size of at least
n_{opt}\geq \left[2\frac{z_{{1-\alpha }/{2}}\sigma _{0}}{\Delta } \right] ^{2}=\left[2\cdot \frac{1.96 \cdot 0.233}{0.2} \right]^{2}= 20.85.
To calculate a confidence interval with a width of not more than 0.2 s, the results of at least 21 athletes are needed.
(b) The sample size is 30. Thus, the confidence interval width should be smaller than 0.2 s. This is indeed true as the calculations show:
\left[10.93\pm \underbrace{t_{0.975;29}}_{2.045}\cdot \frac{0.233}{\sqrt{30} } \right] =\left[10.84; 11.02\right] .
The width is only 0.18 s.
(c) If we calculate a 80 % confidence interval, then the lower confidence limit corresponds to the running time which is achieved by only 10 % of the athletes (of the population). Using t_{0.9;29} ≈ 1.31 we get
\left[10.93\pm 1.31\cdot \frac{0.233}{\sqrt{30} } \right] =\left[10.87; 10.99\right].
The athlete’s best time is thus below the lower confidence limit. He is among the top 10 % of all athletes, using the results of the confidence interval.