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Chapter 18: Problem 34

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?

Short Answer

Expert verified
Test 583 randomly selected adults.

Step by step solution

01

Identify the Given Information

We know the initial proportion of adults with high glucose levels is \( p = 0.22 \) or 22%. The desired margin of error is \( E = 0.04 \) or 4%, and the confidence level is \( 98\% \).
02

Determine the Critical Value

For a 98% confidence level, we find the critical value \( z \) using a Z-table or calculator. The Z-value corresponding to 98% confidence level is approximately \( z = 2.33 \).
03

Use the Sample Size Formula

The sample size \( n \) can be found using the formula:\[ n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \]Substituting the known values, we get:\[ n = \left( \frac{(2.33)^2 \cdot 0.22 \cdot (1-0.22)}{0.04^2} \right) \]
04

Calculate the Required Sample Size

First, calculate \( p \cdot (1-p) \): \[ 0.22 \cdot 0.78 = 0.1716 \]Then, calculate the other parts:\[ (2.33)^2 = 5.4289 \]So, the sample size is:\[ n \approx \frac{5.4289 \cdot 0.1716}{0.0016} \]Calculate the numerator:\[ 5.4289 \times 0.1716 = 0.93197264 \]Therefore:\[ n \approx \frac{0.93197264}{0.0016} \approx 582.48 \]
05

Conclusion for Sample Size

Since sample size must be a whole number, we round up 582.48 to the nearest whole number, resulting in 583 individuals.

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Key Concepts

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

Understanding Confidence Intervals
A confidence interval is like a safety net in statistics. It gives us a range within which we are pretty sure the true parameter of a population lies. In our case, it's used to estimate the proportion of adults with high blood glucose levels. The confidence interval is constructed around a sample statistic. For instance, if we took a sample and found that 22% of adults have high blood glucose, the confidence interval would tell us a range around this 22% where the actual population proportion is likely to be.

Confidence, such as 98% in this problem, tells us how certain we are that the interval contains the true proportion. So, if we were to repeat the study over and over, 98% of those intervals would contain the true proportion. This concept is crucial because it helps researchers and statisticians make informed guesses about larger groups based on smaller samples.
Calculating Sample Size
Sample size calculation is like planning a road trip 鈥� you need to know how far you鈥檙e going so you can prepare accordingly. To estimate the number of patients needed in this study, we use a special formula: \[ n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \]Here, \( n \) is the sample size, \( z \) is the critical value for our confidence level, \( p \) is the proportion of the sample showing a particular trait (22% here), and \( E \) is the margin of error we鈥檙e willing to accept (4%).

Having enough samples ensures that results are reliable and aren't just due to random chance. In this problem, the calculation shows we need to test 583 individuals for a good estimate. Always round up since testing partial individuals isn't possible!
Estimating Proportions
Proportion estimation is a method used to make predictions about the characteristics of a population. In this study, researchers want to predict the percentage of adults with high glucose levels without testing everyone. Instead, they test a subset (our sample).

Proportional estimation involves calculating a point estimate, which is the sample's proportion (in this case, 22%). Then, the confidence interval is used to extend this estimate over the entire population. It means we use data from a small, manageable sample size to infer details about a much larger group, saving time and resources but still giving us valuable insights.
Understanding Critical Values
Critical values are special numbers that mark the borders of our confidence intervals. They're like the doormen of a club, deciding who gets in and who doesn't. To determine these values, we look at the Z-distribution (a statistical tool used to understand data variability).

In the problem we tackled, a 98% confidence level corresponds to a critical value of about \( z = 2.33 \). This means for our confidence interval to be accurate 98% of the time, our estimates need to fall within this critical value when normally distributed. It's based on the assumption that if repeated, your results will fall under this critical range 98 out of 100 times. Understanding these values helps in effectively planning studies and making accurate predictions.

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Most popular questions from this chapter

Conclusions A catalog sales company promises to deliver orders placed on the Internet within 3 days. Followup calls to a few randomly selected customers show that a \(95 \%\) confidence interval for the proportion of all orders that arrive on time is \(88 \% \pm 6 \% .\) What does this mean? Are these conclusions correct? Explain. a) Between \(82 \%\) and \(94 \%\) of all orders arrive on time. b) \(95 \%\) of all random samples of customers will show that \(88 \%\) of orders arrive on time. c) \(95 \%\) of all random samples of customers will show that \(82 \%\) to \(94 \%\) of orders arrive on time. d) We are \(95 \%\) sure that between \(82 \%\) and \(94 \%\) of the orders placed by the sampled customers arrived on time. e) On \(95 \%\) of the days, between \(82 \%\) and \(94 \%\) of the orders will arrive on time.

Safe food Some food retailers propose subjecting food to a low level of radiation in order to improve safety, but sale of such "irradiated" food is opposed by many people. Suppose a grocer wants to find out what his customers think. He has cashiers distribute surveys at checkout and ask customers to fill them out and drop them in a box near the front door. He gets responses from 122 customers, of whom 78 oppose the radiation treatments. What can the grocer conclude about the opinions of all his customers?

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.

Back to campus In 2004 ACT, Inc., reported that \(74 \%\) of 1644 randomly selected college freshmen returned to college the next year. The study was stratified by type of college - public or private. The retention rates were \(71.9 \%\) among 505 students enrolled in public colleges and \(74.9 \%\) among 1139 students enrolled in private colleges. a) Will the \(95 \%\) confidence interval for the true national retention rate in private colleges be wider or narrower than the \(95 \%\) confidence interval for the retention rate in public colleges? Explain. b) Do you expect the margin of error for the overall

Teenage drivers An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them. a) Create a \(95 \%\) confidence interval for the percentage of all auto accidents that involve teenage drivers. b) Explain what your interval means. c) Explain what "95\% confidence" means. d) A politician urging tighter restrictions on drivers" licenses issued to teens says, "In one of every five auto accidents, a teenager is behind the wheel." Does your confidence interval support or contradict this statement? Explain.
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