A sample of 31 boxes of cereal has a sample standard deviation of 0.81 ounces. Construct a 93% confidence interval to estimate the true standard deviation of the fllins process for the boxes of cereal. a (0433, 1.120) b)(0.658,1.058) c (0.525, 1095) d (0.731,0.731) e) None of the above.

Answer:  Option 'b' is correct.Step-by-step explanation:Since we have given that s = 0.81n = 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=30Here, α = 0.931-α=0.07so, [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.