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kondaur [170]
3 years ago
8

A population has a known standard deviation of 1.27 and a sample space contains 85 values find the margin of error needed to cre

ate a 99% confidence interval estimate of the mean of the population
Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

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 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 = 2.575

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.

In this question:

\sigma = 1.27, n = 85

So

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

M = 2.575*\frac{1.27}{\sqrt{85}}

M = 0.3547

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

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