Answer: Option 'b' is correct.
Step-by-step explanation:
Since we have given that
s = 0.81
n = 31
[tex]s^2=0.6561[/tex]
We need to find the 93% confidence interval to estimate the true standard deviation.
degrees of freedom = df = n-1=30
Here, α = 0.93
1-α=0.07
so, [tex]\dfrac{\alpha }{2}=\dfrac{0.07}{2}=0.035[/tex]
and [tex]1-\dfrac{\alpha }{2}=1-0.035=0.965[/tex]
So, we will use "Chi square distribution"
[tex]\chi^2_{\frac{\alpha }{2},df}=45.455\\\\\chi^2_{1-\frac{\alpha }{2},df}=17.576[/tex]
So, the interval would be
[tex]\sqrt{\dfrac{(n-1)s^2}{\chi^2_{\frac{\alpha }{2},df}}}<\sigma<\sqrt{\dfrac{(n-1)s^2}{\chi^2_{1-\frac{\alpha }{2},df}}}}\\\\=\sqrt{\dfrac{30\times 0.6561}{45.455}}<\sigma<\sqrt{\dfrac{30\times 0.6561}{17.576}}\\\\=0.658<\sigma<1.058[/tex]
Hence, 93% confidence interval would be (0.6581,1.058).
Therefore, Option 'b' is correct.
