Respuesta :
Standard normal distribution is normal distribution with mean 0, and variance 1. The margin of error for given condition is 0.66 (Option D)
How to calculate margin of error for large samples?
Suppose that we have:
- Sample size n > 30
- Sample standard deviation = s
- Population standard deviation = [tex]\sigma[/tex]
- Level of significance = [tex]\alpha[/tex]
Then the margin of error(MOE) is obtained as
Case 1: Population standard deviation is known
Margin of Error = [tex]MOE = Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Case 2: Population standard deviation is unknown
[tex]MOE = Z_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
where [tex]Z_{\alpha/2}[/tex] is critical value of the test statistic at level of significance
For the given case, we've got:
Sample size = n = 35
Sample mean = 50
Sample standard deviation = s = 2 days
[tex]Z_{\alpha/2}[/tex] = 1.96 (at 95% confidence, thus [tex]\alpha[/tex] = 100% - 95% = 5% = 0.05)
Thus, margin of error is calculated for the given case as:
[tex]MOE = Z_{\alpha/2}\dfrac{s}{\sqrt{n}}\\\\\\MOE = 1.96 \times \dfrac{2}{\sqrt{35}} \approx 0.6626 \approx 0.66[/tex]
Thus, the margin of error for given condition is 0.66 (Option D)
Learn more about margin of error here:
https://brainly.com/question/13220147
