Respuesta :
Answer:
The 99% confidence interval is
7.558 - 9.042
Explanation:
The formula for the confidence interval is:
[tex]Confidence\text{ }interval=\bar{X}\pm\frac{\sigma}{\sqrt{n}}[/tex]Where:
X is the mean
σ is the standard deviation
z is the z-score for the confidence interval
n is the sample size.
Also, the interval has:
[tex]Upper\text{ }limit=\bar{X}+\frac{\sigma}{\sqrt{n}}[/tex][tex]Lower\text{ }limit=\bar{X}-\frac{\sigma}{\sqrt{n}}[/tex]Then, in this case,
The sample size is n = 64
The mean is X = 8.3
The z-score for a 99% confidence interval is z = 2.58
The standard deviation is σ = 2.3
Then:
[tex]Lower\text{ }limit=8.3-2.58\cdot\frac{2.3}{\sqrt{64}}=9.04175[/tex][tex]Upper\text{ }limit=8.3+2.58\cdot\frac{2.3}{\sqrt{64}}=7.55825[/tex]Thus, the confidence interval, rounded to 3 decimals is
7.558 - 9.042
