Respuesta :
Answer:
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
Step-by-step explanation:
How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?
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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error(width) 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.
For this item, we have:
[tex]M = 0.4, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 2.575*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.93125[/tex]
[tex]\sqrt{n} = \frac{1.93125}{0.4}[/tex]
[tex]\sqrt{n} = 4.828125[/tex]
[tex]\sqrt{n}^{2} = (4.828125)^{2}[/tex]
[tex]n = 23[/tex]
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?
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, find the margin of error(width) 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.
For this item, we have:
[tex]M = 0.5, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 1.47[/tex]
[tex]\sqrt{n} = \frac{1.47}{0.5}[/tex]
[tex]\sqrt{n} = 2.94[/tex]
[tex]\sqrt{n}^{2} = (2.94)^{2}[/tex]
[tex]n \cong 9[/tex]
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
