A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Construct a 95\​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?

UkusakazaL 2021-08-01 Answered

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a \(\displaystyle{95}​\%\) confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?
\(\begin{array}{|c|c|}\hline 0.60 & 0.82 & 0.09 & 0.89 & 1.29 & 0.49 & 0.82 \\ \hline \end{array}\)
What is the confidence interval estimate of the population mean \(\displaystyle\mu?\)
\(\displaystyle?\ {p}\pm{<}\mu{<}\ ?\ {p}\pm\)

Expert Community at Your Service

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

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

ddaeeric
Answered 2021-08-01 Author has 5476 answers

Step 1
\(\begin{array}{|c|c|c|}\hline & x & (x-\bar{x}) & (x-\bar{x})^{2} \\ \hline & 0.6 & -0.11429 & 0.01306122 \\ \hline & 0.82 & 0.10571 & 0.01117551 \\ \hline & 0.09 & -0.62429 & 0.38973265 \\ \hline & 0.89 & 0.17571 & 0.03087551 \\ \hline & 1.29 & 0.57571 & 0.33144694 \\ \hline & 0.49 & -0.22429 & 0.05030408 \\ \hline & 0.82 & 0.10571 & 0.01117551 \\ \hline Total & 5 & & 0.83777143 \\ \hline \end{array}\)
Here \(\displaystyle{n}={7}\)
Sampke mean \(\displaystyle=\overline{{{x}}}={\frac{{\sum{x}}}{{{n}}}}={0.714}\)
Sample \(\displaystyle{S}.{D}.={s}=\sqrt{{{\frac{{{1}}}{{{n}-{1}}}}\sum{\left({x}-\overline{{{x}}}\right)}^{{{2}}}}}={0.3737}\)
Step 2
\(\begin{array}{|c|}\hline \text{Confidence Level}=95 \\ \hline \text{Significance Level}=\alpha=(100-95)\%=0.05 \\ \hline \text{Degrees of freedom}=n-1=7-1=6 \\ \hline \text{Critical value}=t^{*}=2.447 [\text{using Excel}=TINV(0.05,\ 6)]\\ \hline \end{array}\)
Standard Error \(\displaystyle={\frac{{{s}}}{{\sqrt{{{n}}}}}}={0.1412}\)
Margin of Error (M.E) \(\displaystyle={t}^{{\cdot}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}={0.3456}\)
\(\displaystyle\text{Lower Limit}=\overline{{{x}}}-{\left({M}.{E}\right)}={0.3687}\)
\(\displaystyle\text{Upper Limit}=\overline{{{x}}}+{\left({M}.{E}\right)}={1.0599}\)

Have a similar question?
Ask An Expert
12
 

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 2021-09-25

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a \(\displaystyle{95}​\%\) confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?
\(\begin{array}{|c|c|}\hline 0.60 & 0.82 & 0.09 & 0.89 & 1.29 & 0.49 & 0.82 \\ \hline \end{array}\)
What is the confidence interval estimate of the population mean \(\displaystyle\mu?\)
\(\displaystyle?\ {p}\pm{<}\mu{<}\ ?\ {p}\pm\)

asked 2021-07-31

The monthly incomes for 12 randomly selected​ people, each with a​ bachelor's degree in​ economics, are shown on the right. Complete parts​ (a) through​ (c) below.
Assume the population is normally distributed.
a) Find the sample mean.
b) Find the sample standard deviation
c) Construct a \(\displaystyle{95}\%\) confidence interval for the population mean \(\displaystyle\mu\). A \(\displaystyle{95}\%\) confidence interval for the population mean is
\(\begin{array}{|c|c|}\hline 4450.42 & 4596.96 & 4366.46 \\ \hline 4455.62 & 4151.52 & 3727.77 \\ \hline 4283.26 & 4527.94 & 4407.68 \\ \hline 3946.49 & 4023.61 & 4221.73\\ \hline \end{array}\)

asked 2021-08-06

It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of \(\displaystyle{95}\%\) confidence level for \(\displaystyle\pi\) ratio of smokers in the population.
a) \(\displaystyle{P}{\left({0.45}{<}{p}{<}{0.78}\right)}={0.95}\)
b) \(\displaystyle{P}{\left({0.11}{<}{p}{<}{0.28}\right)}={0.95}\)
c) \(\displaystyle{P}{\left({0.51}{<}{p}{<}{0.78}\right)}={0.95}\)
d) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.59}\right)}={0.95}\)
e) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.39}\right)}={0.95}\)

asked 2021-07-31

Two different analytical tests can be used to determine the impurity level in steel alloys. Nine specimens are tested using both procedures, and the results are shown in the following tabulation.
\(\begin{array}{|c|c|}\hline Specimen & Test-1 & Test-2 \\ \hline 1 & 1.2 & 1.4 \\ \hline 2 & 1.6 & 1.7 \\ \hline 3 & 1.5 & 1.5 \\ \hline 4 & 1.4 & 1.3 \\ \hline 5 & 1.8 & 2.0 \\ \hline 6 & 1.8 & 2.1 \\ \hline 7 & 1.4 & 1.7 \\ \hline 8 & 1.5 & 1.6 \\ \hline 9 & 1.4 & 1.5 \\ \hline \end{array}\)
Find the \(\displaystyle{95}\%\) confidence interval for the mean difference of the tests, and explain it.

asked 2021-08-08

Based on a simple random sample of 1300 college students, it is found that 299 students own a car. We wish to construct a \(\displaystyle{90}\%\) confidence interval to estimate the proportions ? of all college students who own a car.
A) Read carefully the text and provide each of the following:
The sample size \(\displaystyle?=\)
from the sample, the number of college students who own a car is \(\displaystyle?=\)
the confidence level is \(\displaystyle{C}{L}=\) \(\displaystyle\%\).
B) Find the sample proportion \(\displaystyle\hat{{?}}=\)
and \(\displaystyle\hat{{?}}={1}−\hat{{?}}=\)

asked 2021-08-02

A study a local high school tried to determine the mean height of females in the US. A study surveyed a random sample of 125 females and found a mean height of 64.5 inches with a standard deviation of 5 inches. Determine a \(\displaystyle{95}\%\) confidence interval for the mean.
\(\begin{array}{|c|c|} \hline \text{Confidence Interval} & z \\ \hline 80\% & 1.282 \\ \hline 85\% & 1.440\\ \hline 90\% & 1.645\\ \hline 95\% & 1.960\\ \hline 99\% & 2.576\\ \hline 99.5\% & 2.807\\ \hline 99.9\% & 3.291\\ \hline \end{array}\)

...