Running heats London In Olympic running events, preliminary heats are determined
ID: 3225122 • Letter: R
Question
Running heats London In Olympic running events, preliminary heats are determined by raw draw, so we should expect the abilities of runners in the various heats to be about the same, on average. Here are the times (in seconds) for the 400-m women's run in the 2012 Olympics in London for preliminary heats 2 and 5. Is there any evidence that the mean time to finish is different for randomized heats? Explain. Be sure to include a discussion of assumptions and conditions for your analysis. (Note: one runner in heat 2 did not finish and one runner in heat 5 did not start. (*Use Tukey's test to compare the mean times in each heat. Do your conclustions change? *Use rank sume test to compare mean times in each heat. Do your conclusions change?)
BOT Time 50.4 Heat 2
JAM Time 51.05 Heat 2
GBR Time 52.01 Heat 2
ANT Time 54.25 Heat 2
NCA Time 59.55 Heat 2
BAH Time DNF Heat 2
RUS Time 50.75 Heat 5
UKR Time 52.08 Heat 5
GBR Time 52.23 Heat 5
SWE Time 52.86 Heat 5
MAW Time 54.2 Heat 5
FIJ Time 56.77 Heat 5
GRN Time DNS Heat 5
Explanation / Answer
Heat2=c(50.4,51.05,52.01,54.25,59.55)
Heat5=c(50.75,52.08,52.23,52.86,54.2,56.77)
1) Parametric test
Comparing 2 samples using t-test:
Assumptions:
1) Samples' population have Normal distribution
2) Homoskedasticity
3) samples are iid'
t.test(Heat2,Heat5)
Result:
Welch Two Sample t-test
data: Heat2 and Heat5
t = 0.16269, df = 6.0765, p-value = 0.876
alternative hypothesis: true difference in means is not equal to 0
Thus, there is no significant differnce
2) Non-Parametric test
wilcox.test(Heat2,Heat5)
Result:
Wilcoxon rank sum test
data: Heat2 and Heat5
W = 13, p-value = 0.7922
alternative hypothesis: true location shift is not equal to 0
Thus, there is no significant differnce
In both the tests we find that there is no significant differnce between 2 samples.
(We ignored missing values)
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