Answer:
Option A) (8.682, 10.12)
Explanation:
We are given the following in the question:
Sample mean,
= 9.4
Sample size, n = 12
Confidence level = 99%
Alpha, α = 0.01
Sample variance,
= 0.64
Sample standard deviation =
= 0.8
Confidence interval:
![\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))](https://img.qammunity.org/2020/formulas/mathematics/high-school/2ndb6n2l5ba2gqql87u24ub8123vsaq1ah.png)
Putting the values, we get,
![t_(critical)\text{ at degree of freedom 11 and }~\alpha_(0.01) = \pm 3.105](https://img.qammunity.org/2020/formulas/mathematics/high-school/zbcc1jszxryucr8nyza5s01s30rbb0l03v.png)
![9.4 \pm 3.105(\displaystyle(0.8)/(√(12)) ) = 9.4 \pm 0.717 = (8.6829,10.1170) \approx (8.682,10.12)](https://img.qammunity.org/2020/formulas/mathematics/high-school/qqrebvml6x1co3b2nsycj61qczxbblt27u.png)