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Chapter 18: Problem 27

Deer ticks Wildlife biologists inspect 153 deer taken by hunters and find 32 of them carrying ticks that test positive for Lyme disease. a) Create a \(90 \%\) confidence interval for the percentage of deer that may carry such ticks. b) If the scientists want to cut the margin of error in half, how many deer must they inspect? c) What concerns do you have about this sample?

Short Answer

Expert verified
The 90% confidence interval is (15.18%, 26.66%). To half the margin of error, 611 deer must be inspected. Sample representativity is a concern.

Step by step solution

01

Identify the Proportion and Sample Size

Identify the values needed to construct the confidence interval: the number of deer carrying ticks with Lyme disease is 32, and the total sample size is 153. The sample proportion \( \hat{p} \) is \( \frac{32}{153} \), which simplifies to approximately 0.2092.
02

Determine the Z-score for a 90% Confidence Interval

A 90% confidence level corresponds to a Z-score of approximately 1.645 (you can find this value from a standard normal distribution table or calculator).
03

Calculate the Standard Error

The standard error (SE) is calculated using the formula \( SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} \). Substitute \( \hat{p} \approx 0.2092 \) and \( n = 153 \) into the formula: \( SE = \sqrt{\frac{0.2092 (1 - 0.2092)}{153}} \approx 0.0349 \).
04

Find the Margin of Error

The margin of error (ME) can be found using: \( ME = Z \times SE \). Substituting the values gives \( ME = 1.645 \times 0.0349 \approx 0.0574 \).
05

Create the Confidence Interval

To find the confidence interval, use the formula \( (\hat{p} - ME, \hat{p} + ME) \). Substituting the values gives \( (0.2092 - 0.0574, 0.2092 + 0.0574) \) or approximately (0.1518, 0.2666).
06

Calculate Required Sample Size for Reduced Margin of Error

To cut the margin of error in half, use the formula \( n = \left( \frac{Z}{ME/2} \right)^2 \hat{p}(1 - \hat{p}) \). Substituting \(Z = 1.645\), \( \hat{p} = 0.2092\), and \(ME/2 = 0.0574/2 = 0.0287\) gives \( n = \left( \frac{1.645}{0.0287} \right)^2 \times 0.2092 \times (1 - 0.2092) \approx 610.71 \). Thus, they must inspect at least 611 deer.
07

Discuss Concerns About the Sample

Our sample might be biased if the deer inspected are not representative of the entire deer population. There could be seasonal, geographical, or sampling method biases that affect the generalizability of the results.

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Key Concepts

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

Sample Proportion
When conducting a survey or study, the sample proportion is an essential step. It represents the fraction of the sample that possesses the characteristic of interest. In our example of deer carrying ticks, we calculated this by dividing the number of deer with Lyme-positive ticks (32 deer) by the total deer inspected (153 deer). This gives us the sample proportion, denoted by \( \hat{p} \), which is approximately 0.2092.

The sample proportion is crucial because it provides a point estimate of the population proportion, which is unknown. It serves as the foundation for further calculations, like confidence intervals, as it offers a snapshot of the data at hand.
Margin of Error
The margin of error tells us how much we can expect the sample proportion to differ from the true population proportion. In our simplified example, it is calculated using the Z-score and the standard error.

Specifically, the margin of error (ME) is determined through the formula: \( ME = Z \times SE \), where \( Z \) is the Z-score associated with the confidence level, and \( SE \) is the standard error. In the deer example, our margin of error is roughly 0.0574, meaning we expect the population proportion to vary by this amount in either direction from our sample proportion.
  • The role of the margin of error is vital as it gives the range of values likely to contain the true population proportion.
  • It reflects the precision of our sample estimate; a smaller margin indicates a more precise estimate.
Standard Error
The standard error measures the variability of the sample proportion. It helps us understand how much our sample proportion might fluctuate from the actual population proportion simply due to random sampling error.

We calculate the standard error (SE) using the formula: \( SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. In the deer study, this calculation gave us a standard error of approximately 0.0349.

The standard error is essential because:
  • It highlights the amount of sampling variability present.
  • A larger sample size generally reduces the standard error, leading to more reliable estimates.
Z-score
A Z-score is a statistical measure that helps determine how many standard deviations an element is from the mean. In our context, it is used to establish how confident we are that a sample proportion falls within a certain range of the population proportion.

In confidence interval calculations, the Z-score corresponds to the desired level of certainty we have about our estimate. For example, a Z-score of approximately 1.645 is associated with a 90% confidence interval. This means if we were to take many samples, 90% of them would contain the true population proportion within this interval.

Z-scores are crucial because:
  • They help translate our confidence level into a quantitative measure.
  • They ensure that the calculated intervals reflect the desired certainty level, thus aiding in decision-making.
Understanding Z-scores allows us to precisely determine the range where our estimates are expected to fall.

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Most popular questions from this chapter

Contributions, please The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free mailing labels and greeting cards to potential donors on their list and ask for a voluntary contribution. To test a new campaign, they recently sent letters to a random sample of 100,000 potential donors and received 4781 donations. a) Give a \(95 \%\) confidence interval for the true proportion of their entire mailing list who may donate. b) A staff member thinks that the true rate is \(5 \% .\) Given the confidence interval you found, do you find that percentage plausible?

Teachers A 2011 Gallup poll found that \(76 \%\) of Americans believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1002 Americans. a) Find a \(90 \%\) confidence interval for the proportion of Americans who would agree with this. b) Interpret your interval in this context. c) Explain what "90\% confidence" means. d) Do these data refute a pundit's claim that \(2 / 3\) of Americans believe this statement? Explain.

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?

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.

Another pilot study During routine screening, a doctor notices that \(22 \%\) of her adult patients show higher than normal levels of glucose in their blood-a possible warning signal for diabetes. Hearing this, some medical researchers decide to conduct a large-scale study, hoping to estimate the proportion to within \(4 \%\) with \(98 \%\) confidence. How many randomly selected adults must they test?
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