Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2 and chi Subscript Upper R Superscript 2​, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 98​% ​confidence; nequals28​, sequals0.22 mg.

Accepted Solution

Answer:0.16 <\sigma <0.13Step-by-step explanation:givenS=0.28n=22d_f=n-1 =22-1 =21[tex]\chi_{L}^{2}=\chi _{0.995}^{2}[/tex]=64.278[tex]\chi_{U}^{2}=\chi _{0.005}^{2}[/tex]=96.578now80% confidence interval for Standard deviation is given by[tex]\sqrt{\frac{(n-1)S^{2}}{\chi _{U}^{2}}}<\sigma<\sqrt{\frac{(n-1)S^{2}}{\chi _{L}^{2}}}[/tex]=[tex]\sqrt{\frac{(22-1)(0.28)^{2}}{64.27}}<\sigma<\sqrt{\frac{(22-1)(0.28)^{2}}{96.578}}[/tex]=0.16 <\sigma <0.13