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

iohanetc 2021-10-11
A of 100 12oz ketchup bottles was taken.
1. If $$\displaystyle\mu={12.02}{\quad\text{and}\quad}\sigma={0.01}$$. 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 $$\displaystyle\sigma={0.1}{o}{z}$$., what is the 99% confidence interval for the standard deviation of the amount of ketchup in a bottle?

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

content_user

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).