Respuesta :
Answer:
We need a sample size of at least 75.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, we find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] 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:
[tex]\sigma = \sqrt{484} = 22[/tex]
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
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]5 = 1.96*\frac{22}{\sqrt{n}}[/tex]
[tex]5\sqrt{n} = 43.12[/tex]
[tex]\sqrt{n} = \frac{43.12}{5}[/tex]
[tex]\sqrt{n} = 8.624[/tex]
[tex](\sqrt{n})^{2} = (8.624)^{2}[/tex]
[tex]n = 74.4[/tex]
We need a sample size of at least 75.
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 = [tex]Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }[/tex]
where, [tex]\alpha[/tex] = significance level = 1 - 0.95 = 0.05 and [tex](\frac{\alpha}{2})[/tex] = 0.025.
[tex]\sigma[/tex] = standard deviation = [tex]\sqrt{Variance}[/tex] = [tex]\sqrt{484}[/tex] = 22
n = sample size
Also, at 0.025 significance level the z table gives critical value of 1.96.
So, margin of error is ;
[tex]5=1.96 \times \frac{22}{\sqrt{n} }[/tex]
[tex]\sqrt{n} = \frac{1.96 \times 22}{5}[/tex]
[tex]\sqrt{n}[/tex] = 8.624
Squaring both sides we get,
n = [tex]8.264^{2}[/tex] = 74.5
So, we must need at least a sample size of 75.
