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

The correct option is 1.148 < σ < 6.015

Step-by-step explanation:

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

`\sqrt{\frac{(n-1)s^(2)}{\chi^(2)_{(0.01)/(2) }} } <\sigma<\sqrt{\frac{(n-1)s^(2)}{\chi^(2)_{1-(0.01)/(2) }} }`

Where:

`\chi^(2)_{(0.01)/(2) }=18.548`

`\chi^(2)_{1-(0.01)/(2)}=0.676`

Therefore, the 99% confidence interval is:

`\sqrt{((7-1)2.019^(2))/(18.548) } <\sigma<\sqrt{((7-1)2.019^(2))/(0.676) }`

`1.148<\sigma<6.015`

Therefore, the option 1.148 < σ < 6.015 is correct