Answer
$\bar{x}$ =$\frac{∑X}{n}$
=$\frac{606.35}{20}$
= 30.3175
s^2 = $\frac{ n(∑X^2) - [(∑X)^2] }{n(n-1)}$
= $\frac{524013.4 - 367660.3}{20*19}$
=411.4553
s = $ \sqrt 411.4553 \ $
s =20.2844
n = 20, s = 20.2844 , df = 20-1=19, α = 1-0.90 = 0.10
To find χ2 right ,
α/2=0.05
From the table,
χ2 right = 30.144
To find χ2 left,
1-0.05=0.95
From the table,
χ2 left =10.117
The Confidence Interval for a Variance:
$ \frac{(n-1)s^2}{χ2 right}$ < $σ^{2}$ < $ \frac{(n-1)s^2}{ χ2 left}$
$ \frac{19*20.2844^2}{30.144}$ < $σ^{2}$ < $ \frac{19*20.2844^2}{10.117}$
= 259.3435 dollars < $σ^{2}$< 772.7243 dollars
The Confidence Interval for a Standard Deviation:
$\sqrt 603.6793 $ < σ< $\sqrt 5836.972$
16.1042 dollars < σ < 27.7979 dollars
Hence, we can be 90% confident that the true population variance of stock prices per share (in dollars) is between 259.3435 dollars to 772.7243 dollars, while the true population standard deviation is between 16.1042 dollars to 27.7979 dollars based on a sample of 20 random stock prices.
