A sample of size 9 from a normal population with sample mean \( \bar{X}=15.8 \) and population estimated standard deviation \( \sigma=3.21 \), then 99% confidence interval for population mean is : \( (Use:\ t_{8}(0.01)=3.335) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [12.23, 19.37]
For a small sample (n = 9, so 8 degrees of freedom) the 99% confidence interval for the population mean is
\( \displaystyle \bar{X}\pm t_{8}(0.01)\frac{\sigma}{\sqrt{n}} \)
\( \displaystyle =15.8\pm3.335\times\frac{3.21}{3} \)
\( \displaystyle =15.8\pm3.335\times1.07 \)
\( \displaystyle =15.8\pm3.57 \)
So the interval is from 15.8 − 3.57 = 12.23 to 15.8 + 3.57 = 19.37, that is [12.23, 19.37].