The 90% CI by using normal dist is "[tex]19.585 < \mu < 23.455[/tex]". A complete solution is below.
According to the table,
The sample mean will be:
- [tex]\bar{x} = \frac{\Sigma (x)}{n}[/tex]
[tex]= \frac{1377.28}{64}[/tex]
[tex]= 21.52[/tex]
Standard deviation,
Sample size,
Significance level,
[tex]= 1-0.99[/tex]
[tex]= 0.01[/tex]
Critical value,
- [tex]z_{\frac{\alpha}{2} } = z_{0.005}[/tex]
[tex]= 2.58[/tex]
Now,
The margin of error will be:
→ [tex]E = z_{\frac{\alpha}{2} }\times \frac{\sigma}{\sqrt{n} }[/tex]
By substituting the values, we get
[tex]= 2.58\times \frac{6}{\sqrt{64} }[/tex]
[tex]= 2.58\times \frac{6}{8}[/tex]
[tex]= 1.935[/tex]
Limits will be:
→ Lower = [tex]\bar{x}-E[/tex]
= [tex]21.52-1.935[/tex]
= [tex]19.585[/tex]
→ Upper = [tex]\bar{x} +E[/tex]
= [tex]21.52+1.935[/tex]
= [tex]23.455[/tex]
hence,
The 99% CI is:
= [tex]\bar{x} \pm E[/tex]
= [tex]21.52 \pm 1.935[/tex]
= [tex](19.585, 23.455)[/tex]
Thus the above answer is correct.
Learn more about margin of error here:
https://brainly.com/question/23897124
