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Chapter 9: Problem 38

Dr. Paul Oswiecmiski wants to estimate the mean serum HDL cholesterol of all \(20-\) to \(29-\) year-old males. How many subjects are needed to estimate the mean serum HDL cholesterol of all 20 - to 29 -year-old males within 1.5 points with \(90 \%\) confidence, assuming that \(\sigma=12.5 ?\) Suppose Dr. Oswiecmiski would prefer \(98 \%\) confidence. How does the increase in confidence affect the sample size required?

Short Answer

Expert verified
90% confidence level requires 188 subjects, while 98% confidence level requires 377 subjects. Higher confidence level leads to a larger sample size.

Step by step solution

01

Determine the Formula

To find the sample size needed to estimate the mean within a specific margin of error, we use the formula: \[ n = \left( \frac{Z_{\alpha/2} \sigma}{E} \right)^2\] where \(\ n\) is the sample size, \(\ \sigma\) is the standard deviation, \(\ E\) is the margin of error, and \(\ Z_{\alpha/2}\) is the z-value corresponding to the desired confidence level.
02

Find the z-value for 90% Confidence Level

For a 90% confidence level, the z-value \(\ Z_{\alpha/2}\) can be found using Z-tables. \(\ Z_{\alpha/2}\) for 90% confidence is approximately 1.645.
03

Calculate the Sample Size for 90% Confidence Level

Substitute the given values into the formula: \(\ Z_{\alpha/2} = 1.645\), \(\ \sigma = 12.5\), and \(\ E = 1.5\). \[\ n = \left( \frac{1.645 \times 12.5}{1.5} \right)^2 \] \[\ n = \left( \frac{20.5625}{1.5} \right)^2 \] \[\ n = \left( 13.7083 \right)^2 \] \[\ n = 187.88 \] Since the sample size cannot be a fraction, round up to 188.
04

Find the z-value for 98% Confidence Level

For a 98% confidence level, the z-value \( Z_{\alpha/2} \) is approximately 2.33.
05

Calculate the Sample Size for 98% Confidence Level

Substitute the given values into the formula: \( Z_{\alpha/2} = 2.33 \), \( \sigma = 12.5 \), and \( E = 1.5 \). \[ n = \left( \frac{2.33 \times 12.5}{1.5} \right)^2 \] \[ n = \left( \frac{29.125}{1.5} \right)^2 \] \[ n = \left( 19.4167 \right)^2\] \[ n = 376.99 \] Since the sample size cannot be a fraction, round up to 377.
06

Conclusion

At 90% confidence level, 188 subjects are needed, while at 98% confidence level, 377 subjects are needed. Increasing the confidence level increases the required sample size to ensure accuracy.

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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 measure of how sure you are that your sample accurately reflects the population. It is expressed as a percentage. For instance, a 90% confidence level means you're 90% certain your sample mean is close to the true population mean. Higher confidence levels increase certainty but require larger sample sizes.
The **z-value** corresponding to the confidence level is crucial. It is found using Z-tables or standard normal distributions. For example:
- **90% confidence level:** z-value 鈮� 1.645
- **98% confidence level:** z-value 鈮� 2.33
The higher the z-value, the wider the margin of error, making higher confidence levels demand more sample individuals.
Margin of Error
The margin of error (MOE) quantifies the range within which the true population parameter lies, considering the sample estimate. In simpler terms, it tells you how much you expect your survey results to reflect the actual population.
In the context of sample size determination, MOE is designated as 'E' in the formula:
n = ( Z 伪 / 2 蟽 ) 2 E Being precise with the MOE ensures that your estimated parameter stays within an acceptable range of the true population parameter.
In our exercise, Dr. Paul Oswiecmiski wanted the mean serum HDL cholesterol to be estimated within 1.5 points. That exact number (1.5) represents the margin of error and greatly impacts how big the sample size should be.
Standard Deviation
Standard deviation (蟽) is a measure of how spread out the values in a data set are. It tells you how much individual data points tend to differ from the mean.
In sample size determination, standard deviation helps understand the variability within your data. A larger standard deviation indicates more variability, which means you'll need a larger sample size to achieve the same level of precision (i.e., margin of error) and confidence level.
In the given problem, the standard deviation 蟽 was 12.5. This value was crucial in calculating the necessary sample size using the formula:
n = ( Z 伪 / 2 蟽 ) 2 E For example, with a 90% confidence level (z-value = 1.645) and margin of error of 1.5, the sample size was found to be 188.
Increasing standard deviation while keeping other variables constant would increase the required sample size to achieve the same results.

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Most popular questions from this chapter

Construct the appropriate confidence interval. A simple random sample of size \(n=22\) is drawn from a population that is normally distributed with \(\sigma=37 .\) The sample mean is found to be \(\bar{x}=122.5 .\) Construct a \(90 \%\) confidence interval about the population mean.

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