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ollegr [7]
3 years ago
8

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

ed margin of error is 5 or less is
Mathematics
2 answers:
Oksanka [162]3 years ago
8 0

Answer:

A sample size of 75 must be needed.

Step-by-step explanation:

We are given that the population variance equals 484 and we have to find the sample size that needs to be taken if the desired margin of error is 5 or less.

As we know that the Margin of error formula is given by;

  Margin of error = Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }

where, \alpha = significance level = 1 - 0.95 = 0.05 and  (\frac{\alpha}{2}) = 0.025.

           \sigma = standard deviation = \sqrt{Variance} = \sqrt{484} = 22

           n = sample size

Also, at 0.025 significance level the z table gives critical value of 1.96.

So,<u> margin of error</u> is ;

                     5=1.96 \times \frac{22}{\sqrt{n} }

                    \sqrt{n} = \frac{1.96 \times 22}{5}

                     \sqrt{n} = 8.624

Squaring both sides we get,

                      n = 8.264^{2} = 74.5

So, we must need at least a sample size of 75.

Ksju [112]3 years ago
6 0

Answer:

We need a sample size of at least 75.

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{1-0.95}{2} = 0.025

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 1-0.025 = 0.975, so z = 1.96

Now, we 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.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

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

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

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