Suppose approximately 900 people die in bicycle accidents each year. One study e
ID: 3219505 • Letter: S
Question
Suppose approximately 900 people die in bicycle accidents each year. One study examined the records of 1511 bicyclists aged 15 or older who were fatally injured in bicycle accidents in a five-year period and were tested for alcohol. Of these, 642 tested positive for alcohol (blood alcohol concentration of 0.01% or higher).
(a) Summarize the data with appropriate descriptive statistics. (Round your answer to four decimal places.)
p = _____________
(b) To do statistical inference for these data, we think in terms of a model where p is a parameter that represents the probability that a tested bicycle rider is positive for alcohol. Find a 99% confidence interval for p. (Round your answers to four decimal places.) ,
_____________,_________________
(c) Can you conclude from your analysis of this study that alcohol causes fatal bicycle accidents? Explain.
Yes, we know that of the 900 people that die in bicycle accidents each year, about 642 test positive for alcohol.
No, we do not know, for example, what percentage of cyclists who were not involved in fatal accidents had alcohol in their systems.
(d) In this study 584 bicyclists had blood alcohol levels above 0.10%, a level defining legally drunk at the time. Give a 99% confidence interval for the proportion who were legally drunk according to this criterion. (Round your answers to four decimal places.) ,
_____________, __________________
Explanation / Answer
SoluutionA:
p^=sample proportion of bicyclists aged 15 or older who were fatally injured in bicycle accidents in a five-year period and were tested for alcohol.
p^=x/n=642/1511=0.4249
SolutionB:
99% Confidence inetrval for population proportion of tested bicycle rider is positive for alcohol is
z alpha /2 for 99%=2.576
sample proportion+- margin of error
=0.4249+-2.576 sqrt[(0.4249(1-0.4249)/1511]
=0.4249+0.0328
=0.4249-0.0328<p<0.4249+0.0328
= 0.3921<p< 0.4577
lower limit=0.3921
upper limit=0.4577
Solutionc:
Yes, we know that form 99% CI the population proportion is from 39.21% to 45.77%
0.3921*100=39.21% to 0.4577*100=45.77%
SOlutiond:
p^ for legally drunk=x/n=584/1511=0.3865
99%confidence interval for the proportion who were legally drunk are
0.3865+-2.576sqrt[0.3865(1-.3865)/1511]
0.3865+-0.0323
0.3865-0.0323<p<0.3865+0.0323
0.3542<p<0.4188
lower limit=0.3542
upper limit=0.4188
99% confidence interval for the proportion who were legally drunk according to this criterion are
0.3542,0.4188