Question

Construct a 99% confidence interval for the population standard deviation of white blood cell count (in cells per microliter). Assume 7 random samples were selected from a population that has a normal distribution. The sample has a mean of 7.106 cells and a standard deviation of 2.019 cells.
1.098 < σ < 4.973
4.042 < σ < 21.170
1.148 < σ < 6.015
1.319 < σ < 36.181

Answer 1

Answer:

The correct option is 1.148 < σ < 6.015

Explanation:

The 99% confidence interval for the standard deviation is given below:

$\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\frac{0.01}{2} }} }$

Where:

$\chi^{2}_{\frac{0.01}{2} }=18.548$

$\chi^{2}_{1-\frac{0.01}{2}}=0.676$

Therefore, the 99% confidence interval is:

$\sqrt{\frac{(7-1)2.019^{2}}{18.548} }$

$1.148$

Therefore, the option 1.148 < σ < 6.015 is correct