Question

A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From past research, it is known that the standard deviation of such claims amounts to $124.88.(a) What is the critical value that corresponds to the given level of confidence? Round to two decimals and remember that critical values are defined to be positive.(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

Answer 1

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.