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Chapter 18: Problem 11

Mislabeled seafood In December \(2011,\) Consumer Reports published their study of labeling of seafood sold in New York, New Jersey, and Connecticut. They purchased 190 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces to standard gene fragments that can identify the species. Laboratory results indicated that \(22 \%\) of these packages of seafood were mislabeled, incompletely labeled, or misidentified by store or restaurant employees. a) Construct a \(95 \%\) confidence interval for the proportion of all seafood packages in those three states that are mislabeled or misidentified. b) Explain what your confidence interval says about seafood sold in these three states. c) A 2009 report by the Government Accountability Board says that the Food and Drug Administration has spent very little time recently looking for seafood fraud. Suppose an official said, "That's only 190 packages out of the billions of pieces of seafood sold in a year. With the small number tested, I don't know that one would want to change one's buying habits." (An official was quoted similarly in a different but similar context). Is this argument valid? Explain.

Short Answer

Expert verified
The confidence interval is (0.1602, 0.2798). This suggests 16.02% to 27.98% of seafood may be mislabeled. The official's argument is not entirely valid as statistical methods can still offer insights with a small sample.

Step by step solution

01

Calculate Sample Proportion

We are given that 22% of the 190 tested packages were mislabeled. Calculate the sample proportion \( \hat{p} \).\[ \hat{p} = \frac{22}{100} = 0.22 \]
02

Determine the Standard Error

Next, determine the standard error of the proportion using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. Here, \( n = 190 \).\[ SE = \sqrt{\frac{0.22 \times 0.78}{190}} = \sqrt{\frac{0.1716}{190}} \approx 0.0305 \]
03

Calculate the Margin of Error

For a 95% confidence interval, we use the \( z \)-score corresponding to 95%, which is 1.96. The margin of error \( ME \) is calculated as follows:\[ ME = z \times SE = 1.96 \times 0.0305 \approx 0.0598 \]
04

Construct the Confidence Interval

Now, calculate the confidence interval using \( \hat{p} \pm ME \).\[ CI = 0.22 \pm 0.0598 \]So, the interval is approximately \((0.1602, 0.2798)\).
05

Interpret the Confidence Interval

The 95% confidence interval suggests that we are 95% confident that the true proportion of mislabeled or misidentified seafood in these states lies between 16.02% and 27.98%.
06

Evaluate the Official's Statement

The official's argument questions the reliability of the small sample size. While it is true that the sample consists of only 190 packages, the statistical techniques used are valid for making inferences about the population when the sample is random and unbiased. Therefore, the confidence interval provides valuable information about potential labeling inaccuracies, despite the official's concerns.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion is a crucial concept in statistics, especially when it comes to analyzing survey or experimental data. In this exercise, the sample proportion helps determine how many seafood packages are mislabeled from a small sample.
  • The sample size here is 190 packages, and out of these, 22% were found to be mislabeled.
  • The sample proportion, denoted as \( \hat{p} \), is calculated by dividing the number of mislabeled packages by the total number of samples, which is \( \hat{p} = \frac{22}{100} = 0.22 \).
Understanding the sample proportion allows us to estimate what proportion of the entire population (all seafood packages sold in the given states) might be mislabeled. This estimation is the foundation for constructing a confidence interval.
Standard Error
The standard error provides a measure of the variability or dispersion of the sample proportion. It helps us understand how much the sample proportion is likely to vary from the true population proportion.To calculate the standard error, use the formula:\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]For our exercise:
  • \( \hat{p} = 0.22 \)
  • \( n = 190 \)
The standard error is computed as:\[ SE = \sqrt{\frac{0.22 \times 0.78}{190}} = \approx 0.0305 \]The smaller the standard error, the more the sample proportion is likely to reflect the true population proportion. It gives us a sense of how precise the sample's proportion estimate is.
Margin of Error
The margin of error provides a range above and below the sample proportion, giving us a confidence interval. This range is where we expect the true population proportion to lie.To calculate the margin of error for a 95% confidence interval:
  • Determine the corresponding \( z \)-score, which is 1.96 for 95%
  • Multiply the standard error by the \( z \)-score: \( ME = z \times SE = 1.96 \times 0.0305 \approx 0.0598 \)
Thus, the margin of error is approximately 0.0598, meaning we can expect the true proportion of mislabeled seafood to vary this much from our sample proportion under normal conditions.
Statistical Inference
Statistical inference involves making predictions or generalizations about a population based on a sample. In this context, the confidence interval is a key tool.By constructing a 95% confidence interval, we use the sample data to infer a range for the true proportion of mislabeled seafood:
  • The confidence interval is calculated as \( \hat{p} \pm ME \)
  • So, \( CI = 0.22 \pm 0.0598 = (0.1602, 0.2798) \)
We interpret this interval to mean we are 95% confident that the true proportion of mislabeled seafood packages falls between 16.02% and 27.98%. This inference is crucial for understanding the potential scale of the issue, despite the claims regarding the sample size.

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Most popular questions from this chapter

Death penalty, again In the survey on the death penalty you read about in the chapter, the Gallup Poll actually split the sample at random, asking 510 respondents the question quoted earlier, "Generally speaking, do you believe the death penalty is applied fairly or unfairly in this country today?" The other 510 were asked "Generally speaking, do you believe the death penalty is applied unfairly or fairly in this country today?" Seems like the same question, but sometimes the order of the choices matters. Asked the first question, \(58 \%\) said the death penalty was fairly applied; only \(54 \%\) said so with the second wording. a) What kind of bias may be present here? b) If we combine them, considering the overall group to be one larger random sample of 1020 respondents, what is a \(95 \%\) confidence interval for the proportion of the general public that thinks the death penalty is being fairly applied? c) How does the margin of error based on this pooled sample compare with the margins of error from the separate groups? Why?

Take the offer First USA, a major credit card company, is planning a new offer for their current cardholders. The offer will give double airline miles on purchases for the next 6 months if the cardholder goes online and registers for the offer. To test the effectiveness of the campaign, First USA recently sent out offers to a random sample of 50,000 cardholders. Of those, 1184 registered. a) Give a \(95 \%\) confidence interval for the true proportion of those cardholders who will register for the offer. b) If the acceptance rate is only \(2 \%\) or less, the campaign won't be worth the expense. Given the confidence interval you found, what would you say?

Gambling A city ballot includes a local initiative that would legalize gambling. The issue is hotly contested, and two groups decide to conduct polls to predict the outcome. The local newspaper finds that \(53 \%\) of 1200 randomly selected voters plan to vote "yes," while a college Statistics class finds \(54 \%\) of 450 randomly selected voters in support. Both groups will create \(95 \%\) confidence intervals. a) Without finding the confidence intervals, explain which one will have the larger margin of error. b) Find both confidence intervals. c) Which group concludes that the outcome is too close to call? Why?

Conditions For each situation described below, identify the population and the sample, explain what \(p\) and \(\hat{p}\) represent, and tell whether the methods of this chapter can be used to create a confidence interval. a) Police set up an auto checkpoint at which drivers are stopped and their cars inspected for safety problems. They find that 14 of the 134 cars stopped have at least one safety violation. They want to estimate the percentage of all cars that may be unsafe, b) A TV talk show asks viewers to register their opinions on prayer in schools by logging on to a website. Of the 602 people who voted, 488 favored prayer in schools. We want to estimate the level of support among the general public. c) A school is considering requiring students to wear uniforms. The PTA surveys parent opinion by sending a questionnaire home with all 1245 students; 380 surveys are returned, with 228 families in favor of the change. d) A college admits 1632 freshmen one year, and four years later 1388 of them graduate on time. The college wants to estimate the percentage of all their freshman enrollees who graduate on time.

Confidence intervals, again Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true? a) For a given sample size, reducing the margin of error will mean lower confidence. b) For a certain confidence level, you can get a smaller margin of error by selecting a bigger sample. c) For a fixed margin of error, smaller samples will mean lower confidence. d) For a given confidence level, a sample 9 times as large will make a margin of error one third as big.
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