Question

Determine the minimum sample size required when you want to be 99% confident that the sample mean is within 0.50 units of the population mean. Assume a standard deviation of 2.9 in a normally distributed population.
a. 130
b. 224
c. 223
d. 129

Answer 1

Answer:

b. 224

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

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 $1-0.005 = 0.995$, so $z = 2.58$

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.

In this problem:

We want to find n, when $M = 0.5, \sigma = 2.9$ So

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

$0.5 = 2.58*\frac{2.9}{\sqrt{n}}$

$0.5\sqrt{n} = 7.482$

$\sqrt{n} = 14.964$

$(\sqrt{n})^{2} = (14.964)^{2}$

$n = 223.9$

The nearest integer number to 223.9 is 224.

So the correct answer is:

b. 224