Question

A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. He 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.
Determine and interpret a​ 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC.
​(Use ascending order. Round to three decimal places as​ needed.)

Answer 1

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.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~   $t_n_-_1$

where, $\bar X$ = sample mean BAC = 0.16 g/dL

            s = sample standard deviation = 0.080 ​g/dL

            n = sample of drivers = 60

            $\mu$ = 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, $\mu$ is;

P(-1.672 < $t_5_9$ < 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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

                                       = [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

                                       = [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] .