Suppose approximately 900 people die in bicycle accidents each year. One study e
ID: 3201318 • 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, 542 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 542 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 498 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
(a)The appropriate descriptive statistics is sample proportion p^=bicycle rider tested positive/total number of bicycle riders
=542/1511
=0.36
(b)The 99% confidence interval for population proportion p is:
P^+/-z(0.99)*root over p^(1-p^)/n
=0.36+/-2.58*root over 0.36*(1-0.36)/1511
=0.36+/-2.58*0.0123
=(0.328,0.391)
The above confidence interval contains the true population proportion with 99% confidence.
(c) No, we do not know, for example, what percentage of cyclists who were not involved in fatal accidents had alcohol in their systems.
(d)The sample proportion is p^=498/1511=0.329
The 99% confidence interval for population proportion p is:
P^+/-z(0.99)*root over p^(1-p^)/n
=0.329+/-2.58*root over 0.329*(1-0.329)/1511
=0.329+/-2.58*0.012
=(0.298,0.359)
The above confidence interval contains the true population proportion with 99% confidence.