Mathematics, 08.09.2020 14:01 Frankia

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.)

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As per the given problem randomly selects record by researcher is 60 driver in 2009. He found the sample mean $0.16\:\rm g/dL$ and standard deviation is $0.08\:\rm g/dL$ .

The $90\%$ confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Given:

Sample Mean $\overline{X}=0.16\:\rm g/dL$ .

Sample standard deviation $S= 0.080\:\rm g/dL$ .

Sample of drivers n = 60 .

The expression for pivotal quantity is as follows.

$\begin{aligneed}\\\rm P.Q.=\frac{\overline{X}-\mu}{\frac{s}{\sqrt{n} } } \\\end$

Now, use one sample t-test for constructing a 90% confidence interval.

Find the population mean for 90% confidence.

P(-1.672

Here, as the critical value of t at 59 degree freedom are -1.672 & 1.672 with P = 5%.

P(-1.672

Now for 90% population mean is,

$\begin{aligned}\mu={\overline{X}-1.672\times\frac{s}{\sqrt{n} }, {\overline{X}+1.672\times\frac{s}{\sqrt{n} }\\\mu={0.16-1.672\times\frac{0.08}{\sqrt{60} },0.16+1.672\times\frac{0.08}{\sqrt{60} } \\\\\mu=[0.143, 0.177] \end$

Thus, the $90\%$ 9 confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Learn more about t-test here:

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Answer from: vlactawhalm29

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] .

Answer from: Quest

answer: the answer is c.

step-by-step explanation:

she’s dividing 1/3 of the flour into 2 jars.

hope this : )

Use the expression at the top right to answer the question (1/3 divided by 2)

Answer from: Quest

i would need to see the dot plot : )

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