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Answer:
A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .
Step-by-step explanation:
We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean BAC = 0.16 g/dL
s = sample standard deviation = 0.080 g/dL
n = sample of drivers = 60
[tex]\mu[/tex] = population mean BAC in fatal crashes
Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, a 90% confidence interval for the population mean, [tex]\mu[/tex] is;
P(-1.672 < [tex]t_5_9[/tex] < 1.672) = 0.90 {As the critical value of t at 59 degrees of
freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.672) = 0.90
P( [tex]-1.672 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.672 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
P( [tex]\bar X-1.672 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.672 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.672 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.672 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }[/tex] , [tex]0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }[/tex] ]
= [0.143, 0.177]
Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .
As per the given problem randomly selects record by researcher is 60 driver in 2009. He found the sample mean [tex]0.16\:\rm g/dL[/tex] and standard deviation is [tex]0.08\:\rm g/dL[/tex].
The [tex]90\%[/tex] confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [tex][0.143, 0.177][/tex].
Given:
Sample Mean [tex]\overline{X}=0.16\:\rm g/dL[/tex].
Sample standard deviation [tex]S= 0.080\:\rm g/dL[/tex].
Sample of drivers [tex]n = 60[/tex].
The expression for pivotal quantity is as follows.
[tex]\begin{aligneed}\\\rm P.Q.=\frac{\overline{X}-\mu}{\frac{s}{\sqrt{n} } } \\\end[/tex]
Now, use one sample t-test for constructing a 90% confidence interval.
Find the population mean for 90% confidence.
[tex]P(-1.672<t_{59}<1.672)=0.90[/tex]
Here, as the critical value of t at 59 degree freedom are -1.672 & 1.672 with P = 5%.
[tex]P(-1.672<\frac{\overline{X}-\mu}{\frac{s}{\sqrt{n} } }<1.672)=0.90\\P(-1.672\times \frac{s}{\sqrt{n} } <\overline{X}-\mu<1.672\times \frac{s}{\sqrt{n} })=0.90\\P(\overline{X}-1.672\times \frac{s}{\sqrt{n} } <\mu<\overline{X}<1.672\times \frac{s}{\sqrt{n} })=0.90\\[/tex]
Now for 90% population mean is,
[tex]\begin{aligned}\mu={\overline{X}-1.672\times\frac{s}{\sqrt{n} }, {\overline{X}+1.672\times\frac{s}{\sqrt{n} }\\\mu={0.16-1.672\times\frac{0.08}{\sqrt{60} },0.16+1.672\times\frac{0.08}{\sqrt{60} } \\\\\mu=[0.143, 0.177] \end[/tex]
Thus, the [tex]90\%[/tex] 9 confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [tex][0.143, 0.177][/tex].
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