Answer
n = 24, s = 2.3 years, df = 24-1=23, α = 1-0.9 = 0.10
To find χ2 right ,
α/2=0.05
From the table,
χ2 right =35.172
To find χ2 left,
1-0.05=0.95
From the table,
χ2 left =13.091
The Confidence Interval for a Variance:
$ \frac{(n-1)s^2}{χ2 right}$ < $σ^{2}$ < $ \frac{(n-1)s^2}{ χ2 left}$
$ \frac{23*2.3^2}{35.172}$ < $σ^{2}$ < $ \frac{23*2.3^2}{13.091}$
= 3.46 years < $σ^{2}$ < 9.29 years
The Confidence Interval for a Standard Deviation:
$\sqrt 3.46$ < σ< $\sqrt 9.29$
1.86 years < σ < 3.05 years
Hence, we can be 90% confident that the true population variance of the ages of seniors at Oak Park College is between 3.46 years to 9,29 years, while the true population standard deviation is between 1.86 years to 3.05 years based on a sample of 24 students.
