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Mila [183]
2 years ago
5

Pleassssssseeeeee help meeeeeeeeeeeee omggggggg a(m)=207-m

Mathematics
1 answer:
Juliette [100K]2 years ago
7 0

Answer:

0.207

Step-by-step explanation: MARK ME BRAINless

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How large a sample should be selected to provide a 95% confidence interval with a margin of error of 6? Assume that the populati
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Using the z-distribution, it is found that a sample of 171 should be selected.

<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.

For this problem, the parameters are:

z = 1.96, \sigma = 40, M = 6

Hence we solve for n to find the needed sample size.

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

6 = 1.96\frac{40}{\sqrt{n}}

6\sqrt{n} = 40 \times 1.96

\sqrt{n} = \frac{40 \times 1.96}{6}

(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2

n = 170.7.

Rounding up, a sample of 171 should be selected.

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

#SPJ1

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