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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 deviation of 327.8 in a normally distributed population

User Amitthk
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1 Answer

5 votes

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.

User Shaochuancs
by
7.7k points

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