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Chapter 9: Problem 37

Dr. Paul Oswiecmiski wants to estimate the Mean serum HDL cholesterol of all 20 - to 29 -year-old females. How many subjects are needed to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old females within 2 points with \(99 \%\) confidence, assuming that \(\sigma=13.4 ?\) Suppose Dr. Oswiecmiski would be content with \(95 \%\) confidence. How does the decrease in confidence affect the sample size required?

Short Answer

Expert verified
For 99% confidence, 184 subjects are needed. For 95% confidence, 137 subjects are needed.

Step by step solution

01

- Identify the Given Information

Given: The standard deviation (蟽) is 13.4, the margin of error (E) is 2 points, and the confidence levels are 99% and 95%.
02

- Find the Z-Score for the Given Confidence Levels

For a 99% confidence level, the Z-score (Z) is approximately 2.576. For a 95% confidence level, the Z-score (Z) is approximately 1.96.
03

- Use the Sample Size Formula

The sample size is given by the formula: o} is given by the formula:o \( n = (\frac{Z蟽}{E})^2 }({\frac{Z蟽}{E}})^ 2 \) Is Given by the formula:o = (\frac{Z蟽}{E})^2\( n= (\frac{Z蟽}{E})^2 \) Plug in the values for 99% Confidence 鈥� \( n = (\frac{2.576脳13.4}{2})^2o= ( \frac{2.576脳13.4}{2})^2 \) Plug in the values for 99% Confidence Level8.07}= For 99% Confidence Level Plug in the values forn(\frac{2.576脳13.4}{2})^2\( 183.95 = 184 \)
04

- Calculate for 95% Confidence Level

\( n = (\frac{1.96脳13.4}{2} )^2o154= 2562 }\) for 95% Confidence Level.a = N = (\frac{1.96脳13.4}{2})^2$ done.
05

Conclusion

With 99% confidence, the necessary sample size is approximately 184. With 95% confidence, the necessary sample size is approximately 137. The decrease in confidence reduces the sample size needed.

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Key Concepts

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

Confidence Level
The confidence level is a key concept in statistical estimation. It measures how confident we can be that a certain interval contains the true population parameter, such as the mean.
For example, a 99% confidence level means we can be 99% certain the interval contains the true mean. It's like saying, 'If we took 100 different samples and computed the confidence interval each time, about 99 of those intervals would contain the true mean.'
Common confidence levels are 90%, 95%, and 99%, and higher confidence levels require larger sample sizes.
Margin of Error
The margin of error (E) indicates the maximum expected difference between the true population parameter and a sample estimate. A smaller margin of error requires a larger sample size, providing more precision in the estimate.
For instance, in the example, the margin of error is set to 2 points. This means Dr. Oswiecmiski wants the estimated mean serum HDL cholesterol to be within 2 points of the actual mean. Smaller margins of error generally provide more reliable estimates but require more subjects.
Standard Deviation
Standard deviation (蟽) measures the amount of variation or dispersion in a set of values. It helps us understand how spread out the data points are from the mean.
In Dr. Oswiecmiski's study, the standard deviation of 13.4 indicates variability in serum HDL cholesterol levels among 20- to 29-year-old females. A higher standard deviation suggests more spread out data, thus impacting the sample size needed to achieve the desired margin of error and confidence level.
Z-Score
The Z-score is a statistical measure that describes the number of standard deviations a data point is from the mean. It's used in estimating confidence levels and can be found in Z-tables.
For a 99% confidence level, the Z-score is approximately 2.576. For a 95% confidence level, it's around 1.96. Higher confidence levels correspond to higher Z-scores. Z-scores are crucial in determining the sample size required for a study, as seen in the sample size formula.
Statistical Estimation
Statistical estimation involves using sample data to estimate a population parameter, like the mean. It combines the concepts of confidence level, margin of error, standard deviation, and Z-score to determine how reliable the estimate is.
In Dr. Oswiecmiski's exercise, statistical estimation is used to determine how many subjects he needs to survey to estimate the mean serum HDL cholesterol accurately. By plugging values into the sample size formula n = ( Z 蟽 E ) 2 , we balance the need for precision with available resources. This careful balance helps ensure that estimates are as accurate and reliable as possible.

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Most popular questions from this chapter

Severe acute respiratory syndrome (or SARS) is a viral respiratory illness. It has the distinction of being the first new communicable disease of the 21st century. Researchers wanted to estimate the incubation period of patients with SARS. Based on interviews with 81 SARS patients, they found that the mean incubation period was 4.6 days with a standard deviation of 15.9 days. Based on this information, construct a \(95 \%\) confidence interval for the mean incubation period of the SARS virus. Interpret the interval. (Source: Gabriel M. Leung et al., The Epidemiology of Severe Acute Respiratory Syndrome in the 2003 Hong Kong Epidemic: An Analysis of All 1755 Patients, Annals of Internal Medicine, \(2004 ; 141: 662-673 .)\)

(a) Find the \(t\) -value such that the area in the right tail is 0.02 with 19 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.10 with 32 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.05 with 6 degrees of freedom. [Hint: Use symmetry. (d) Find the critical \(t\) -value that corresponds to \(95 \%\) confidence. Assume 16 degrees of freedom.

Suppose the arrival of cars at Burger King's drive-through follows a Poisson process with \(\mu=4\) cars every 10 minutes. (a) Simulate obtaining 30 samples of size \(n=35\) from this population. (b) Obtain the sample mean and standard deviation for each of the 30 samples. (c) Construct \(90 \% ~ t\) -intervals for each of the 30 samples. (d) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?

A researcher wishes to estimate the mean number of miles on 4 -year-old Saturn SCIs. (a) How many cars should be in a sample to estimate the mean number of miles within 1000 miles with \(90 \%\) confidence, assuming that \(\sigma=19,700 ?\) (b) How many cars should be in a sample to estimate the mean number of miles within 500 miles with \(90 \%\) confidence, assuming that \(\sigma=19,700 ?\) (c) What effect does doubling the required accuracy have on the sample size? Why is this the expected result?

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be 35.1 , and the sample standard deviation, \(s,\) is found to be 8.7 (a) Construct a \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 40 (b) Construct a \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(100 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(40 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=18,\) what conditions must be satisfied to compute the confidence interval?
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