Here we have given that,

sample size \((n) = 100\)

mean \(\displaystyle{\left(\overline{{X}}\right)}={12.02}\)

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,

\(\displaystyle{E}=\frac{{{Z}{c}\cdot{s}{d}}}{\sqrt{{{n}}}}\)

where \(Zc\) is the critical value for normal distribution.

Putting all the values in this formula,

\(\displaystyle{0.001}=\frac{{{Z}{c}\cdot{0.01}}}{\sqrt{{{100}}}}\)

\(\displaystyle{0.001}={Z}{c}\cdot\frac{0.01}{{10}}\)

\(\displaystyle{0.01}={Z}{c}\cdot{0.01}\)

\(\displaystyle{Z}{c}=\frac{0.01}{{0.01}}={1}\)

\(Zc = 1\)

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,

\(\displaystyle\sqrt{{{\left({n}-{1}\right)}\cdot\frac{{{s}{2}}}{{{X}{2}{u}}}}}<\sigma<\sqrt{{{\left({n}-{1}\right)}\cdot\frac{{{s}{2}}}{{{X}{2}{l}}}}}\)

where \(X2 u\) and \(X2 l\) are critical value for CHi square distribution.

These values we can find by using EXCEL.

syntax :

For \(X2 u\) :

=CHIINV(probability, deg_freedom)

where probability \(\displaystyle=\frac{{{1}-{C}}}{{2}}\)

C is confidence level = 99% = 0.99

deg_freedom = n-1 = 100-1 = 99

For \(X2 l\) :

=CHIINV(probability, deg_freedom)

where probability \(\displaystyle=\frac{{{1}+{C}}}{{2}}\)

\(X2 u = 138.99\)

\(X2 l = 66.51\)

lower limit \(\displaystyle=\sqrt{{{\left({n}-{1}\right)}\cdot{s}\frac{2}{{X}}{2}{u}}}=\sqrt{{{\left({100}-{1}\right)}\cdot\frac{0.12}{{138.99}}}}=\sqrt{{{0.007123}}}={0.084}\)

upper limit \(\displaystyle=\sqrt{{{\left({n}-{1}\right)}\cdot{s}\frac{2}{{X}}{2}{l}}}=\sqrt{{{\left({100}-{1}\right)}\cdot\frac{0.12}{{66.51}}}}=\sqrt{{{0.01488}}}={0.122}\)

99% confidence interval for for the standard deviation of the amount of ketchup in a 12 oz bottle is (0.084, 0.122).