Question

In order to set​ rates, an insurance company is trying to estimate the number of sick days that full time workers at an auto repair shop take per year. A previous study indicated that the population standard deviation is 2.8 days. How large a sample must be selected if the company wants to be​ 95% confident that the true mean differs from the sample mean by no more than 1​ day?
A. 512
B. 1024
C. 141
D. 31

Answer 1

Answer:

D. 31

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:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

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.

A previous study indicated that the population standard deviation is 2.8 days.

This means that $\sigma = 2.8$

How large a sample must be selected if the company wants to be​ 95% confident that the true mean differs from the sample mean by no more than 1​ day?

This is n for which M = 1. So

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

$1 = 1.96\frac{2.8}{\sqrt{n}}$

$\sqrt{n} = 1.96*2.8$

$(\sqrt{n})^2 = (1.96*2.8)^2$

$n = 30.12$

Rounding up, at least 31 people are needed, and the correct answer is given by option D.