Answer: 99% confidence interval would be (10.78,11.82).
Step-by-step explanation:
Since we have given that
n = 12
Average = 11.3
Variance = 0.49
Standard deviation = √0.49=0.7
so, we need to find 99% confidence interval.
At 99% confidence , z = 2.58
So, interval would be
[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=11.3\pm 2.58\times \dfrac{0.7}{\sqrt{12}}\\\\=11.3\pm 0.521\\\\=(11.3-0.521,11.3+0.521)\\\\=(10.78,11.82)[/tex]
Hence, 99% confidence interval would be (10.78,11.82).