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asambeis [7]
3 years ago
9

you want to create a 99 confidence interval with a margin of error of 25 assuming that the population standard deviation is equa

l to 1.5 what's the minimum size of the Rams Tampa
Mathematics
1 answer:
Ahat [919]3 years ago
4 0

Answer:

The minimum sample size is 239.

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 Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.005 = 0.995, so Z = 2.575.

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.

Population standard deviation is equal to 1.5

This means that \sigma = 1.5

Margin of error of 0.25

This means that M = 0.25

What's the minimum size of the sample?

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

0.25 = 2.575\frac{1.5}{\sqrt{n}}

0.25\sqrt{n} = 2.575*1.5

\sqrt{n} = \frac{2.575*1.5}{0.25}

(\sqrt{n})^2 = (\frac{2.575*1.5}{0.25})^2

n = 238.7

Rounding up:

The minimum sample size is 239.

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