Construct the confidence interval estimate of the mean. An FDA guideline is that the mercury in fish should be below 1 part permillion (ppm). Listed below are the amounts of mercury (ppm) found in tunasushi sampled at different stores in New York City. The study was sponsored bythe New York Times, and the stores (in order) are D'Agostino, Eli's Manhattan,Fairway, Food Emporium, Gourmet Garage, Grace's Marketplace, and Whole Foods.Construct a \(98 \%\) confidence interval estimate of the mean amount of mercuryin the population. Does it appear that there is too much mercury in tunasushi? $$\begin{array}{rrrrrrr} 0.56 & 0.75 & 0.10 & 0.95 & 1.25 & 0.54 & 0.88 \end{array}$$
Short Answer
Expert verified
The 98% confidence interval for the mean mercury level in tuna sushi does not include 1 ppm, indicating it is unlikely the true mean exceeds the FDA guideline.
Step by step solution
01
Find the sample mean (\bar{x})
Add all the mercury amounts together and divide by the number of values: \( \bar{x} = \frac{0.56 + 0.75 + 0.10 + 0.95 + 1.25 + 0.54 + 0.88}{7} \).
02
Calculate the sample standard deviation (s)
Use the formula for the sample standard deviation: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \), where \( x_i \) are the sample values, \( \bar{x} \) is the sample mean, and \( n \) is the sample size.
03
Find the critical value (t*)
For a 98% confidence level with 6 degrees of freedom (n-1), find the t critical value from the t-distribution table.
04
Calculate the margin of error (E)
Use the formula: \( E = t* \times \frac{s}{\sqrt{n}} \), where \( t* \) is the critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size.
05
Construct the confidence interval
Find the lower and upper bounds of the confidence interval using the formulas: \( \bar{x} - E \) and \( \bar{x} + E \).
06
Interpret the confidence interval
Compare the confidence interval to the FDA guideline of 1 ppm to determine if it appears that there is too much mercury in the tuna sushi.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sample Mean
The sample mean, denoted as \(\bar{x}\), is the average of the sampled values. To find the sample mean, sum all the mercury amounts and divide by the number of samples. This provides a central value around which the data points are distributed. For instance, in the given problem, we calculate the sample mean as follows: \(\bar{x} = \frac{0.56 + 0.75 + 0.10 + 0.95 + 1.25 + 0.54 + 0.88}{7} \) The result represents the average mercury level in the sampled tuna sushi. This measure is crucial as it provides an estimate of the central tendency of the data. Testing whether the sample mean is below or above the guideline helps in assessing compliance with the FDA guideline of 1 ppm.
Sample Standard Deviation
The sample standard deviation (\(s\)) quantifies the amount of variation or dispersion in the sample data. It shows how much individual data points differ from the sample mean. Using the formula: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \) where \( x_i \) are the sample values, \( \bar{x} \) is the sample mean, and \( n \) is the sample size. In our example, calculate the deviations of each value from the mean, square these deviations, sum them up, then divide by \( n-1 \) and finally take the square root. This results in a numerical value that indicates the spread of the mercury content around the mean. A smaller \( s \) implies that the mercury levels are closely packed around the mean, whereas a larger \( s \) indicates more variability.
T-distribution
When estimating population parameters from a small sample size, the t-distribution is used. This distribution is similar to the normal distribution but has thicker tails, which account for the increased variability inherent in small samples. The shape of the t-distribution changes based on the sample size, denoted by degrees of freedom (df). To construct a confidence interval for the mean mercury level, you need to determine the t-critical value (\( t* \)). For a 98% confidence level and 6 degrees of freedom (\( n-1 \)), use a t-distribution table or calculator. This critical value helps define the range within which the true population mean is expected to lie. The t-distribution thus corrects for the additional uncertainty in the estimate when the sample size is small (less than 30).
Margin of Error
The margin of error (\( E \)) represents the range above and below the sample mean within which the true population mean is expected to lie with a certain level of confidence. It accounts for the variability in the data and the confidence level chosen. The formula for the margin of error is: \( E = t* \times \frac{s}{\sqrt{n}} \) where \( t* \) is the t-critical value, \( s \) is the sample standard deviation, and \( n \) is the sample size. By calculating the margin of error, we can then construct a confidence interval around the sample mean. For example, for a 98% confidence level, insert the appropriate \( t* \) value, the sample standard deviation, and the sample size into the formula to get \( E \). This margin of error tells us the interval's width, giving us information on how precise our estimate is. A smaller margin of error indicates a more precise estimate.
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Most popular questions from this chapter
Use the data and confidence level to construct a confidence interval estimateof \(p,\) then address the given question. When she was 9 years of age. Emily Rosa did a science fair experiment in whichshe tested professional touch therapists to see if they could sense her energyfield. She flipped a coin to select either her right hand or her left hand,and then she asked the therapists to identify the selected hand by placingtheir hand just under Emily's hand without seeing it and without touching it.Among 280 trials, the touch therapists were correct 123 times (based on data in "A Close Look at Therapeutic Touch," Journal of the AmericanMedical Association, Vol. 279, No. 13). a. Given that Emily used a coin toss to select either her right hand or herleft hand, what proportion of correct responses would be expected if the touchtherapists made random guesses? b. Using Emily's sample results, what is the best point estimate of thetherapists success rate? c. Using Emily's sample results, construct a \(99 \%\) confidence intervalestimate of the proportion of correct responses made by touch therapists.Use the data and confidence level to construct a confidence interval estimateof \(p,\) then address the given question. The drug OxyContin (oxycodone) is used to treat pain, but it is dangerousbecause it is addictive and can be lethal. In clinical trials, 227 subjectswere treated with OxyContin and 52 of them developed nausea (based on datafrom Purdue Pharma L.P.). a. Construct a \(95 \%\) confidence interval estimate of the percentage ofOxyContin users who develop nausea. b. Compare the result from part (a) to this \(95 \%\) confidence interval for 5subjects who developed nausea among the 45 subjects given a placebo instead ofOxyContin: \(1.93 \%Use the data and confidence level to construct a confidence interval estimateof \(p,\) then address the given question. In a survey of 1002 people, \(70 \%\) said that they voted in a recent presidential election (based on data from ICR Research Group). Votingrecords show that \(61 \%\) of eligible voters actually did vote. a. Find a \(98 \%\) confidence interval estimate of the proportion of people whosay that they voted. b. Are the survey results consistent with the actual voter turnout of \(61 \%?\) Why or why not?
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Thus, the crucial value of z for a 97% confidence interval is 2.17, as determined by a z score table, which is as follows: Therefore the obtained probability for the z-score of 2.17 is 0.97.
Strictly speaking a 95% confidence interval means that if we were to take 100 different samples and compute a 95% confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the true mean value (μ).
Since 95% of values fall within two standard deviations of the mean according to the 68-95-99.7 Rule, simply add and subtract two standard deviations from the mean in order to obtain the 95% confidence interval. Notice that with higher confidence levels the confidence interval gets large so there is less precision.
As a result, as the sample size increases, the range of interval values narrows, allowing you to determine the mean with greater precision than with a smaller sample. When we construct a 95% confidence interval, we are 95% sure that the population mean difference lies within the interval.
Finally, calculate the confidence interval using the formula: Confidence Interval = x ± (Z * s / √n), where x is the sample mean, Z is the Z score, s is the sample standard deviation, and n is the sample size.
Step 1: Determine the confidence level, denoted , where is a number (decimal) between 0 and 100. Step 2: Obtain the confidence level, denoted by evaluating α = 1 − C 100 . Step 3: Use the -table (or a calculator) to obtain the -score z α / 2 .
p-values simply provide a cut-off beyond which we assert that the findings are 'statistically significant' (by convention, this is p<0.05). A confidence interval that embraces the value of no difference between treatments indicates that the treatment under investigation is not significantly different from the control.
The confidence level is the percentage of times you expect to get close to the same estimate if you run your experiment again or resample the population in the same way. The confidence interval consists of the upper and lower bounds of the estimate you expect to find at a given level of confidence.
The critical value for a 95% confidence interval is 1.96, where (1-0.95)/2 = 0.025. A 95% confidence interval for the unknown mean is ((101.82 - (1.96*0.49)), (101.82 + (1.96*0.49))) = (101.82 - 0.96, 101.82 + 0.96) = (100.86, 102.78).
For example, the correct interpretation of a 95% confidence interval, [L, U], is that "we are 95% confident that the [population parameter] is between [L] and [U]." Fill in the population parameter with the specific language from the problem.
It's this callous nature that makes 95% confidence intervals so useful. It's a strict gatekeeper that passes statistical signal while filtering a lot of noise out. It dampens false positives in a very measured and unbiased manner.It protects us against experiment owners who are biased judges of their own work.
There is no single accepted name for this number; it is also commonly referred to as the "standard normal deviate", "normal score" or "Z score" for the 97.5 percentile point, the .975 point, or just its approximate value, 1.96.
To find the critical value zα/2 for a 97% confidence level, calculate α from the formula CL = 1 - α, then find α/2. In the standard normal distribution, find the z-score that corresponds to α/2 which will be the critical value. For a 97% confidence level, the critical value zα/2 is approximately 2.17.
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