Answer to Question #301838 in Statistics and Probability for secret

Question #301838

A random sample of 14 cigarettes of a certain brand has an average nicotine content of 3.8 milligrams and a standard deviation of 2 milligrams. What is the upper bound of the 99% confidence interval for the average nicotine content of the cigarettes.



1
Expert's answer
2022-02-24T12:36:05-0500

The critical value for α=0.01\alpha = 0.01 and df=n1=13df = n-1 = 13 degrees of freedom is tc=z1α/2;n1=3.012275.t_c = z_{1-\alpha/2; n-1} =3.012275.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉ+tc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})

=(3.83.012275×214,3.8+3.012275×214)=(3.8-3.012275\times\dfrac{2}{\sqrt{14}}, 3.8+3.012275\times\dfrac{2}{\sqrt{14}})

=(2.190,5.410)=(2.190,5.410 )

The upper bound of the 99% confidence interval for the average nicotine content of the cigarettes is 5.410 milligrams.


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