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?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

content_user
Answered 2021-10-26 Author has 11827 answers

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).

Not exactly what you’re looking for?
Ask My Question
0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-05-08
For the poll described in the Chapter problem, assume that the respondents had been asked for their political party affiliation, and the responses were coded as 0 (for Democrat), 1 (for Republican), 2 (for Independent), or 3 (for any other response). If we calculate the average (mean) of the numbers and get 0.95, how can that value be interpreted?
asked 2021-10-08
State whether the investigation in question is an observational study or a designed experiment. Justify your answer in each case. In the article by P. Ridker et al. titled "A Randomized Trial of Low-dose Aspirin in the Primary Prevention of Cardiovascular Disease in Women", the researchers noted that "We randomly assigned 39,876 initially healthy women 45 years of age or older to receive 100 mg of aspirin or placebo on alternate days and then monitored them for 10 years for a first major cardiovascular event (i.e., nonfatal myocardial infarction, nonfatal stroke, or death from cardiovascular causes)."
asked 2021-08-06
Assume that \(\displaystyle\sigma\) is unknown, the lower \(\displaystyle{100}{\left({1}-\alpha\right)}\%\) confidence bound on \(\displaystyle\mu\) is:
a) \(\displaystyle\mu\leq\overline{{{x}}}+{t}_{{\alpha,{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)
b) \(\displaystyle\overline{{{x}}}-{t}_{{\alpha,{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\leq\mu\)
c) \(\displaystyle\mu\leq\overline{{{x}}}+{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\)
d) \(\displaystyle\overline{{{x}}}-{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}\leq\mu\)
asked 2021-12-01
The amount of chlorine in 100 cm3 water taken from a water tank was measured, the average was 2.15 mg was found. The expert claims that the amount of chlorine in the water is greater than that measured. For this reason, 100 households were randomly selected, with an average chlorine content of 2.55mg and a variance of 0.49 mg. According to this data, find the account value z.
asked 2021-07-04
How can you use two way-frequency tables to analyze data?
asked 2021-11-19
1. A population with \(\displaystyle\mu={587}\) and \(\displaystyle\sigma={12}\) is transformed into z-scores. After the transformation, what is the standard deviation for the population of z-scores?
2.A population with \(\displaystyle\mu={58}\) and \(\displaystyle\sigma={12}\) is transformed into z-scores. After the transformation, what is the mean for the population of z-scores?
Please help me answer these two!
asked 2021-09-15
In a factory,that produces glass bottles, 5 number of bottles are defected in 1000 bottles produced.
In a random sample of 100 bottles, calculate the probability that more than 4 are defected?
...