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3 years ago

5

In a certain study, the chance of encountering a car crash on the roadstudy, the chance of encountering a car crash on the road

is stated as 10%. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.

1 answer:

pshichka [43]3 years ago

8 0

10% as a decimal (a value between 0 and 1) is 0.1

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3 years ago

Read 2 more answers

A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From

lutik1710 [3]

Answer:

a) $z = 1.645$

b) The should sample at least 293 small claims.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ , which means that the answer of question a is z = 1.645.

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

$M = 12, \sigma = 124.88$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$12 = 1.645*\frac{124.88}{\sqrt{n}}$

$12\sqrt{n} = 205.43$

$\sqrt{n} = \frac{205.43}{12}$

$\sqrt{n} = 17.12$

$\sqrt{n}^{2} = (17.12)^{2}$

$n = 293$

The should sample at least 293 small claims.

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3 years ago