Determine the minimum sample size required when you want to be 75% confident that the sample mean is within 30 units of the population mean assume the population standard deviat…

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asked Jun 3, 2021 69.0k views

1 Answer

Shaochuancs · Answer 1 · 2021-06-07T16:36:25+0000

Answer:

The minimum sample size is 158.

Explanation:

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

$\alpha = (1-0.75)/(2) = 0.125$

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
1-0.125 = 0.875 , so
z = 1.15

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 question:

The minimum sample size is n.

n is found when M = 30.

We have that
$\sigma = 327.8$

So

$M = z*(\sigma)/(√(n))$

30 = 1.15*(327.8)/(√(n))

30√(n) = 1.15*327.8

√(n) = (1.15*327.8)/(30)

(√(n))^(2) = ((1.15*327.8)/(30))^(2)

n = 157.9

Rounding up

The minimum sample size is 158.