Question

A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the population standard deviation is

Answer 1

The
`100(1-\alpha)\%` confidence interval of a standard deviation is given by:


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

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval,


`\chi^2_{1- (\alpha)/(2) }=43.773 \\ \\ \chi^2_{(\alpha)/(2) }=18.493`

Therefore, the 90% confidence interval for the standard deviation is given by


`\sqrt{ ((31-1)50^2)/(43.773 )} \leq\sigma\leq\sqrt{ ((31-1)50^2)/(18.493 )} \\ \\ \Rightarrow\sqrt{ ((30)2500)/(43.773 )} \leq\sigma\leq\sqrt{ ((30)2500)/(18.493 )} \\ \\ \Rightarrow\sqrt{ (75000)/(43.773 )} \leq\sigma\leq\sqrt{ (75000)/(18.493 )} \\ \\ \Rightarrow √(1713.38) \leq\sigma\leq √(4055.59) \\ \\ \Rightarrow41.4\leq\sigma\leq63.7`