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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 \(s=13.4\) based on earlier studies? Suppose that Dr. Oswiecmiski would be content with \(95 \%\) confidence. How does the decrease in confidence affect the sample size required?

Short Answer

Expert verified
298 subjects for 99% confidence, 173 subjects for 95% confidence. Reducing confidence decreases the sample size required.

Step by step solution

01

Understand the Problem

The problem involves finding the number of subjects needed to estimate the mean serum HDL cholesterol within 2 points with two different confidence levels: 99% and 95%.
02

Identify the Given Information

Given: - Desired margin of error (E) = 2 points- Standard deviation (s) = 13.4- Confidence levels: 99% and 95%
03

Use the Formula for Sample Size Estimation

The formula for estimating the sample size (n) when the population standard deviation ()s) is known is : \[n = \left( \frac{Z*s}{E} \right)^2\]
04

Determine the Z-Score for 99% Confidence Level

For a 99% confidence level, the Z-score (Z) corresponds to the 99% critical value, which is approximately 2.576.
05

Calculate the Sample Size for 99% Confidence Level

Substitute the values into the sample size formula: \[n = \left( \frac{2.576*13.4}{2} \right)^2\] Calculate the value: \[n = \left( \frac{34.5264}{2} \right)^2 = 297.992\] Rounding up, n \approx 298\text{ subjects}.
06

Determine the Z-Score for 95% Confidence Level

For a 95% confidence level, the Z-score (Z) corresponds to the 95% critical value, which is approximately 1.96.
07

Calculate the Sample Size for 95% Confidence Level

Substitute the values into the sample size formula: \[n = \left( \frac{1.96*13.4}{2} \right)^2\] Calculate the value: \[n = \left( \frac{26.264}{2} \right)^2 = 172.0081\] Rounding up, n \approx 173\text{ subjects}.
08

Compare the Sample Sizes Required

With a 99% confidence level, we need approximately 298 subjects. With a 95% confidence level, we need approximately 173 subjects. Reducing the confidence level decreases the number of subjects required.

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Key Concepts

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

Understanding Confidence Level
When estimating a population parameter, the confidence level indicates how confident we are that the true population parameter lies within our calculated interval. It is expressed as a percentage. For example, if we talk about a 99% confidence level, it means that if we were to take 100 different samples and compute the confidence interval for each sample, we expect that 99 of those intervals will contain the true population mean. Higher confidence levels give us more certainty but require larger sample sizes.
To connect with our example, a 99% confidence level means that we need a higher number of subjects to ensure the provided interval captures the actual mean serum HDL cholesterol.
The Importance of Margin of Error
The margin of error (E) represents the range within which we believe the true population parameter lies. It's the buffer we give to our result. In Dr. Oswiecmiski's problem, the margin of error given is 2 points. This means that we want our estimate to be within 2 points of the true mean serum HDL cholesterol for 20 - to 29-year-old females.
Smaller margins of error will require larger sample sizes because we need more data to be more precise. In our example, reducing the margin of error from 2 points to a lower value would require increasingly more subjects to maintain the same confidence level.
The Role of Standard Deviation
Standard deviation (s) is a measure of how dispersed the values in a dataset are around the mean. In our problem, the standard deviation is given as 13.4. Higher standard deviation means greater variability in the data.
When calculating the required sample size, the larger the standard deviation, the larger the sample size needed to estimate the mean accurately. This effect is directly seen in our formula for sample size estimation: The sample size increases as the standard deviation increases, reflecting the need for more data to accurately pin down the mean in a more variable population.
Understanding Z-Score
A Z-score represents the number of standard deviations a data point is from the mean of a dataset. In the context of confidence intervals, different Z-scores correspond to different confidence levels.
For example, in our exercise, the Z-score for a 99% confidence level is approximately 2.576, while for a 95% confidence level, it is approximately 1.96. These critical Z-values help us determine how far our sample mean can be from the true mean. In the sample size formula: A higher Z-score means a higher requirement for data, hence a larger sample size. This means with increased confidence levels, like moving from 95% (Z=1.96) to 99% (Z=2.576), more subjects will be needed to maintain the same precision, as seen in the given calculations where the required sample size decreases significantly from about 298 to 173 as we reduce the confidence level.

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Most popular questions from this chapter

The Sullivan Statistics Survey I asks, "Would you be willing to pay higher taxes if the tax revenue went directly toward deficit reduction?" Treat the survey respondents as a random sample of adult Americans. Go to www.pearsonhighered.com/sullivanstats to obtain the data file SullivanSurveyI using the file format of your choice for the version of the text you are using. The column "Deficit" has survey responses. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would be willing to pay higher taxes if the revenue went directly toward deficit reduction.

True or False: To construct a confidence interval about the mean, the population from which the sample is drawn must be approximately normal.

Construct the appropriate confidence interval. A simple random sample of size \(n=40\) is drawn from a population. The sample mean is found to be \(\bar{x}=120.5,\) and the sample standard deviation is found to be \(s=12.9\). Construct a \(99 \%\) confidence interval for the population mean.

A simple random sample of size \(n\) is drawn from a population that is normally distributed. The sample mean, \(\bar{x},\) is found to be \(50,\) and the sample standard deviation, \(s,\) is found to be \(8 .\) (a) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is 20 (b) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(15 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(95 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is 20. Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, \(E\) ? (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why?

A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without (a) Obtain a point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (b) Verify that the requirements for constructing a confidence interval about \(p\) are satisfied. (c) Construct and interpret a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without. (d) Is it possible that a supermajority (more than \(60 \%\) ) of adult Americans believe that television is a luxury they could do without? Is it likely? (e) Use the results of part (c) to construct a \(95 \%\) confidence interval for the population proportion of adult Americans who believe that televisions are a necessity.

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