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Stels [109]
1 year ago
5

Please help! I don't understand how to solve this problem

Mathematics
1 answer:
Ilia_Sergeevich [38]1 year ago
8 0

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

The margin of error is:

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

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

Assuming an uniform distribution, the standard deviation is given by:

S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

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

0.006 = 1.96\frac{1.1547}{\sqrt{n}}

0.006\sqrt{n} = 1.96 \times 1.1547

\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}

(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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