Question

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

Answer 1

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, margin of error 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.

Answer 2

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.