Question

Find the 90% confidence interval for the population standard deviation given the following.

Answer 1

Answer:

b. 2.01 < σ < 2.81

Step-by-step explanation:

The 90% confidence interval for the population standard deviation is calculated as:

$\sqrt{\frac{(n-1)*s^{2} }{X^2_{\alpha/2 } } }$    < σ < $\sqrt{\frac{(n-1)*s^{2} }{X^2_{1-\alpha/2 } } }$

Where n is equal to 51, s is equal to 2.34, $\alpha$ is 10%, $X^{2} _{\alpha /2}$ is the value for the Chi squared distribution with n-1 degrees of freedom that has a probability of $\alpha /2$ in the right tail and $X^{2} _{1-\alpha /2}$ is the value for the Chi squared distribution with n-1 degrees of freedom that has a probability of $1-\alpha /2$ in the right tail.

So, replacing n by 51, s by 2.34, $X^{2} _{\alpha /2}$ by 67.50 and $X^{2} _{1-\alpha /2}$ by 34.76, we get that the The 90% confidence interval is equal to:

$\sqrt{\frac{(51-1)*2.34^{2} }{67.50} } }$ < σ <$\sqrt{\frac{(51-1)*2.34^{2} }{34.76} } }$

2.01 < σ < 2.81