Suppose approximately 900 people die in bicycle accidents each year. One study e

Question

Suppose approximately 900 people die in bicycle accidents each year. One study examined the records of 1711 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 484 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

Solution:

a. p' = X/n = 542/1711 = 0.317

b. Level of signifcance, a = 1 - 0.99 = 0.01

Using z-tables, the critical value is

Z (0.01/2) = Z (0.005) = 2.576

99% confidence interval is given by:-

p ± z*p (1 – p)/n

0.317 ± 2.576*0.317*(1 – 0.317)/1711

0.317 ± 2.576*0.01125

0.317 ± 0.029

0.288, 0.346

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. Proportion, p = X/n = 484/1711 = 0.283

Level of signifcance, a = 1 - 0.99 = 0.01

Using z-tables, the critical value is

Z (0.01/2) = Z (0.005) = 2.576

99% confidence interval is given by:-

p ± z*p (1 – p)/n

0.283 ± 2.576*0.283*(1 – 0.283)/1711

0.283 ± 2.576*0.0109

0.283 ± 0.0281

0.2548, 0.3109

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