Question

It is desired to estimate the average time it takes Statistics students to finish a computer project to within two hours at 90% reliability. It is estimated that the standard deviation of the times is 14 hours. How large a sample should be taken to get the desired interval? n = 231 n = 189 n = 325 n = 133

Answer 1

Answer:

n = 133.

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 Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so 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.

Standard deviation of the times is 14 hours.

This means that $\sigma = 14$

How large a sample should be taken to get the desired interval?

Within 2 hours, which means that we want n for which M = 2. So

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

$2 = 1.645\frac{14}{\sqrt{n}}$

$2\sqrt{n} = 1.645*14$

Dividing both sides by 2

$\sqrt{n} = 1.645*7$

$(\sqrt{n})^2 = (1.645*7)^2$

$n = 132.6$

Rounding up, n = 133.