Respuesta :
Answer:
B. 3.52 to 8.48
Step-by-step explanation:
Finding Confidence Interval by using given Confidence Level:
- Find the Z-Scores of the Confidence Level
- Using the Z-Score Formula to find the values of the Confidence Interval
Given:
- Confidence Level = 95%
- population standard deviation (σ) = 4
- mean (μ) = 6
- numbers of sample (n) = 10
For Confidence Level = 95% → which means there are total 5% of the data that lies outside the level with 2.5% on each side, therefore the Confidence Level falls between 2.5% and 97.5% of the total data.
By using the Z-Score table (Normal Distribution Table), we can find:
- the Z-Score for 2.5% = -1.960
- the Z-Score for 97.5% = 1.960
Now, we need to find the values which have Z-Scores -1.960 and 1.960 by using the Z-Score Formula for Sampling Distribution:
[tex]\boxed{Z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } } }[/tex]
(i) value for Z = -1.960
[tex]\displaystyle Z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
[tex]\displaystyle -1.960=\frac{x-6}{\frac{4}{\sqrt{10} } }[/tex]
[tex]\displaystyle x-6=-1.960\left(\frac{4}{\sqrt{10} } \right)[/tex]
[tex]\displaystyle x=-1.960\left(\frac{4}{\sqrt{10} } \right)+6[/tex]
[tex]\displaystyle x=3.52[/tex]
(i) value for Z = 1.960
[tex]\displaystyle Z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
[tex]\displaystyle 1.960=\frac{x-6}{\frac{4}{\sqrt{10} } }[/tex]
[tex]\displaystyle x-6=1.960\left(\frac{4}{\sqrt{10} } \right)[/tex]
[tex]\displaystyle x=1.960\left(\frac{4}{\sqrt{10} } \right)+6[/tex]
[tex]\displaystyle x=38.48[/tex]
Therefore the 95% Confidence Level lies between 3.52 to 38.48
