Question

(3) A random sample of 51 fatal crashes in 2009 in which the driver had a positive blood alcohol concentration (BAC) from the National Highway Traffic Safety Administration results in a mean BAC of 0.167 grams per deciliter (g/dL) with a standard deviation of 0.010 g/dL.

(a) What is a point estimate for the mean BAC of all fatal crashes with a positive BAC?

(b) In 2009, there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval.

(c) Determine and interpret a 98% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. Do not use the T-Interval or Z-Interval features of your calculator.

Answer 1

3.
a.
TRADITIONAL METHOD
given that,
sample mean, x =0.167
standard deviation, s =0.01
sample size, n =51
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.01/ sqrt ( 51) )
= 0.001
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 50 d.f is 2.403
margin of error = 2.403 * 0.001
= 0.003
III.
CI = x ± margin of error
confidence interval = [ 0.167 ± 0.003 ]
= [ 0.164 , 0.17 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =0.167
standard deviation, s =0.01
sample size, n =51
level of significance, α = 0.02
from standard normal table, two tailed value of |t α/2| with n-1 = 50 d.f is 2.403
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 0.167 ± t a/2 ( 0.01/ Sqrt ( 51) ]
= [ 0.167-(2.403 * 0.001) , 0.167+(2.403 * 0.001) ]
= [ 0.164 , 0.17 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 98% sure that the interval [ 0.164 , 0.17 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean
Answer:
point estimate for the mean BAC of all fatal crashes with a positive BAC =0.003

b.
TRADITIONAL METHOD
given that,
standard deviation, σ =0.01
sample mean, x =0.167
population size (n)=25000
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 0.01/ sqrt ( 25000) )
= 0.00006
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
margin of error = 2.326 * 0.00006
= 0.00015
III.
CI = x ± margin of error
confidence interval = [ 0.167 ± 0.00015 ]
= [ 0.16685,0.16715 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, σ =0.01
sample mean, x =0.167
population size (n)=25000
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
since our test is two-tailed
value of z table is 2.326
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 0.167 ± Z a/2 ( 0.01/ Sqrt ( 25000) ) ]
= [ 0.167 - 2.326 * (0.00006) , 0.167 + 2.326 * (0.00006) ]
= [ 0.16685,0.16715 ]
-----------------------------------------------------------------------------------------------
c.
interpretations:
1. we are 98% sure that the interval [0.16685 , 0.16715 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population mean