Respuesta :
Answer:
The answer is "2653".
Step-by-step explanation:
[tex]Sample \ Size = 99\% \ Cl\\\\ E =60 \\\\\sigma= 1200\\\\[/tex]
Minimum sample size for Cl level [tex]= 99\%[/tex] and Desired Margin of Error, [tex]E = 60[/tex] is:
[tex]\to P(|\bar{x}-\mu| < E) \geq 1-\alpha \\\\\to P(\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} <- \frac{E}{\frac{\sigma}{\sqrt{n}}}) \leq \frac{\alpha}{2} \\\\\to - \frac{E}{\frac{\sigma}{\sqrt{n}}} \leq -z_{\frac{\alpha}{2}} \\\\\to n\geq (\frac{ z_{\frac{\alpha}{2} \times \alpha} }{E})^2 \\\\[/tex]
[tex]Margin \ Error = \frac{\text{length of the CI}}{2}\\\\\sigma=1200\\\\\alpha=1 - Confidence =1-0.99=0.01\\\\Z_{\frac{\alpha}{2}}=z_{0.005}=2.58\\\\n\geq (\frac{z_{0.005} \times \sigma }{E})^2 \\\\= (\frac{2.58 \times 1200 }{60})^2\\\\=2662.560\\[/tex]
The minimum n has to be integer, we take the ceiling Of above number and get n = 2663
The exact z-value.
[tex]n \geq (\frac{2.5758293035489 \times 1200}{60})^2 \\\\= 2653.95864[/tex]
using critical value of 2,575, which gives 2652.25
[tex]\to n =2653[/tex]
