Question

A research firm wants to compute an interval estimate with 90% confidence for the mean time to complete an employment test. Assuming a population standard deviation of three hours, what is the required sample size if the error should be less than a half hour? Select one: a. 196 b. 98 c. 10 d. 16

Answer 1

Answer:

b. 98

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 $z = 1.645$

Now, find 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.

In this problem, we have that:

$M = 0.5, \sigma = 3$

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

$0.5 = 1.645*\frac{3}{\sqrt{n}}$

$0.5\sqrt{n} = 4.935$

$\sqrt{n} = 9.87$

$n = 97.42$

So a sample of at least 98 is required.

The correct answer is:

b. 98