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

b. 224

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (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*(\sigma)/(√(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*(\sigma)/(√(n))`

`0.5 = 2.58*(2.9)/(√(n))`

`0.5√(n) = 7.482`

`√(n) = 14.964`

`(√(n))^(2) = (14.964)^(2)`

`n = 223.9`

The nearest integer number to 223.9 is 224.

So the correct answer is:

b. 224