A of 100 12oz ketchup bottles was taken. 1. If μ=12.02andσ=0.01. With what confidence can we claim that 12.02 oz is within 0.001 of the true mu amount of ketchup in a bottle? 2. If bottles yielded σ=0.1oz., what is the 99% confidence interval for the standard deviation of the amount of ketchup in a bottle?

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A of 100 12oz ketchup bottles was taken.   1. If μ=12.02andσ=0.01. With what confidence can we claim that 12.02 oz is within 0.001 of the true mu amount of ketchup in a bottle?  2. If bottles yielded σ=0.1oz., what is the 99% confidence interval for the standard deviation of the amount of ketchup in a bottle?

Jeffrey Jordon · Accepted Answer

Here we have given that,  sample size (n)=100  mean (X―)=12.02  standard deviation (sd)=0.01  Margin of error (E)=0.001  And we have to find confidence level (C).  C we can find by using formula,  E=Zc⋅sdn  where Zc is the critical value for normal distribution.  Putting all the values in this formula,  0.001=Zc⋅0.01100  0.001=Zc⋅0.0110  0.01=Zc⋅0.01  Zc=0.010.01=1  Zc=1  C we can find by using EXCEL.  syntax :  =NORMSDIST(z)  where Z is 1.  C=0.84  COnfidence level in % =C∗100=0.84∗100=84  With 84% confidence can we assert that 12.02 oz. is within 0.001 of the true mean amount of ketchup in a 12 oz. bottle.  b. If bottles yielded a standard deviation of 0.1 oz ., what is a 99% confidence interval for the standard deviation of the amount of ketchup in a 12 oz bottle  Here we have to find 99% confidence interval for standard deviation.  sd(s)=0.1  99% confidence interval for sigma is,  (n−1)⋅s2X2u&amp;lt;σ&amp;lt;(n−1)⋅s2X2l  where X2u and X2l are critical value for CHi square distribution.  These values we can find by using EXCEL.  syntax :  For X2u :  =CHIINV(probability, deg_freedom)  where probability =1−C2  C is confidence level = 99% = 0.99  deg_freedom = n-1 = 100-1 = 99  For X2l :  =CHIINV(probability, deg_freedom)  where probability =1+C2  X2u=138.99  X2l=66.51  lower limit =(n−1)⋅s2X2u=(100−1)⋅0.12138.99=0.007123=0.084  upper limit =(n−1)⋅s2X2l=(100−1)⋅0.1266.51=0.01488=0.122  99% confidence interval for for the standard deviation of the amount of ketchup in a 12 oz bottle is (0.084, 0.122).

A of 100 12oz ketchup bottles was taken. 1. If mu=12.02 and sigma=0.01ZS

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