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…

Question

asked Jun 22, 2021 127k views

1 Answer

Webpat · Answer 1 · 2021-06-26T14:21:57+0000

Answer:

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