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2 years ago

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Suppose speeds of vehicles traveling on a highway have an unknown distribution with mean 63 and standard deviation 4 miles per h

our. A sample of size n-44 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

1 answer:

Mashutka [201]2 years ago

6 0

Answer:

The standard deviation for the sample mean distribution=0.603

Step-by-step explanation:

We are given that

Mean, $\mu=63$

Standard deviation, $\sigma=4$

n=44

We have to find the standard deviation for the sample mean distribution using Central Limit Theorem for Means.

Standard deviation for the sample mean distribution

$\sigma_x=\frac{\sigma}{\sqrt{n}}$

Using the formula

$\sigma_x=\frac{4}{\sqrt{44}}$

$\sigma_x=\frac{4}{\sqrt{2\times 2\times 11}}$

$\sigma_x=\frac{4}{2\sqrt{11}}$

$\sigma_x=\frac{2}{\sqrt{11}}$

$\sigma_x=0.603$

Hence, the standard deviation for the sample mean distribution=0.603

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A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

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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

<em>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. </em>

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

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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

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

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

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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] .

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