Answer:
[ 21.79, 33.0 ]
Step-by-step explanation:
Given:
Sample size, n = 7
Sample mean, μ = 27.4
Standard deviation, σ = 5.75
Confidence level is 99%
also, population is normal i.e normally distributed
Thus,
Confidence interval = μ ± ![z\frac{\sigma}{\sqrt n}](https://tex.z-dn.net/?f=z%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%20n%7D)
For confidence level of 99%
z-value = 2.58
Therefore,
Lower limit of Confidence interval = μ - ![2.58\times\frac{5.75}{\sqrt{7}}](https://tex.z-dn.net/?f=2.58%5Ctimes%5Cfrac%7B5.75%7D%7B%5Csqrt%7B7%7D%7D)
or
Lower limit of Confidence interval = 27.4 - ![2.58\times\frac{5.75}{\sqrt{7}}](https://tex.z-dn.net/?f=2.58%5Ctimes%5Cfrac%7B5.75%7D%7B%5Csqrt%7B7%7D%7D)
or
Lower limit of Confidence interval = 21.79
Upper limit of Confidence interval = μ + ![2.58\times\frac{5.75}{\sqrt{7}}](https://tex.z-dn.net/?f=2.58%5Ctimes%5Cfrac%7B5.75%7D%7B%5Csqrt%7B7%7D%7D)
or
Upper limit of Confidence interval = 27.4 + ![2.58\times\frac{5.75}{\sqrt{7}}](https://tex.z-dn.net/?f=2.58%5Ctimes%5Cfrac%7B5.75%7D%7B%5Csqrt%7B7%7D%7D)
or
Upper limit of Confidence interval = 33.00
Hence, confidence interval = [ 21.79, 33.0 ]