Respuesta :
Answer:
A sample size of at least 531 is required.
Step-by-step explanation:
We are lacking the confidence level to solve this question, so i am going to use a 90% confidence level.
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{1-0.9}{2} = 0.05[/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]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]
Now, 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.
Find the required sample size for the new study.
A sample size of at least n is required.
n is found when [tex]M = 0.05[/tex]
We have that [tex]\sigma = 0.7[/tex]
So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.05 = 1.645*\frac{0.7}{\sqrt{n}}[/tex]
[tex]0.05\sqrt{n} = 1.645*0.7[/tex]
[tex]\sqrt{n} = \frac{1.645*0.7}{0.05}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645*0.7}{0.05})^{2}[/tex]
[tex]n = 530.4[/tex]
Rounding up
A sample size of at least 531 is required.
