91Ó°ÊÓ

Consider college officials in admissions, registration, counseling, financial aid, campus ministry, food services, and so on. How much money do these people make each year? Suppose you read in your local newspaper that 45 officials in student services earned an average of \(\bar{x}=\$ 50,340\) each year (Reference: Cbronicle of Higher Education). (a) Assume that \(\sigma=\$ 16,920\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (b) Assume that \(\sigma=\$ 10,780\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (c) Assume that \(\sigma=\$ 4830\) for salaries of college officials in student services. Find a \(90 \%\) confidence interval for the population mean salaries of such personnel. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the standard deviation decreases, does the margin of error decrease? (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the standard deviation decreases, does the length of a \(90 \%\) confidence interval decrease?

Step by step solution

01

Understanding the Problem

We are tasked with calculating the 90% confidence intervals for the population mean salaries of college officials based on different standard deviations. We'll also discuss how the margin of error changes with different standard deviations and compare the interval lengths.

02

Formula for Confidence Interval

The formula for a confidence interval when the population standard deviation is known is given by:\[\text{Confidence Interval} = \bar{x} \pm Z \left(\frac{\sigma}{\sqrt{n}}\right)\]where \(Z\) is the Z-score corresponding to the desired confidence level, which is 1.645 for a 90% confidence interval, \(\bar{x}\) is the sample mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

03

Calculate the Margin of Error for Part (a)

For part (a), where \(\sigma = 16920\), we plug the values into the formula:\[M.E. = 1.645 \left(\frac{16920}{\sqrt{45}}\right) \approx 4157.59\]Thus, the margin of error is approximately $4157.59.

04

Calculate Confidence Interval for Part (a)

Using the margin of error calculated previously:\[\text{Confidence Interval} = 50340 \pm 4157.59\]This gives the interval: \(50340 - 4157.59 = 46182.41\) and \(50340 + 4157.59 = 54497.59\). So, the interval is (46182.41, 54497.59).

05

Calculate the Margin of Error for Part (b)

For part (b), where \(\sigma = 10780\), we calculate:\[M.E. = 1.645 \left(\frac{10780}{\sqrt{45}}\right) \approx 2647.66\]Thus, the margin of error is approximately $2647.66.

06

Calculate Confidence Interval for Part (b)

Using the margin of error found:\[\text{Confidence Interval} = 50340 \pm 2647.66\]Lower bound: \(50340 - 2647.66 = 47692.34\) and upper bound: \(50340 + 2647.66 = 52987.66\). Interval: (47692.34, 52987.66).

07

Calculate the Margin of Error for Part (c)

For part (c), with \(\sigma = 4830\), we calculate:\[M.E. = 1.645 \left(\frac{4830}{\sqrt{45}}\right) \approx 1186.21\]The margin of error is approximately $1186.21.

08

Calculate Confidence Interval for Part (c)

For this margin of error:\[\text{Confidence Interval} = 50340 \pm 1186.21\]Lower bound: \(50340 - 1186.21 = 49153.79\) and upper bound: \(50340 + 1186.21 = 51526.21\). Interval: (49153.79, 51526.21).

09

Analyze Margins of Error and Interval Lengths

From parts (a) to (c), we have the margins of error as $4157.59, $2647.66, and $1186.21, respectively. As the standard deviation decreases, the margin of error decreases. Likewise, the confidence intervals also become shorter: (46182.41, 54497.59), (47692.34, 52987.66), (49153.79, 51526.21).

10

Conclusion

In conclusion, both the margin of error and the length of the confidence interval decrease as the standard deviation decreases.

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Key Concepts

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

Margin of Error

When calculating confidence intervals for the salaries of college officials, it is important to understand the concept of the margin of error. Simply put, the margin of error represents the amount by which the sample mean could differ from the actual population mean. It is a measure of how much uncertainty there is in an estimate.
For example, if we have a salary average of $50,340, and a margin of error of $4,157.59, the true average salary could realistically be $4,157.59 more or less than $50,340.

• The margin of error is influenced by the sample size and the standard deviation of the population.
• A smaller standard deviation or a larger sample size will generally lead to a smaller margin of error, indicating more precision in our estimates.
• In our example, we calculated three different margins of error for the same average salary but using different standard deviations: $4,157.59, $2,647.66, and $1,186.21 respectively.

Understanding the margin of error helps us gauge the accuracy of our estimates.

Standard Deviation

Standard deviation is a key component when determining the spread or variability of salaries among college officials. It shows how much individual salaries differ from the average salary in the population.
The standard deviation is crucial for several reasons:

• A larger standard deviation means salaries are more spread out over a wider range, affecting the margin of error and widening the confidence interval.
• A smaller standard deviation suggests that the salaries are closely packed around the mean, resulting in a smaller margin of error and a tighter confidence interval.

In our exercise, the standard deviation values used were $16,920, $10,780, and $4,830. Each of these values impacted the confidence intervals differently, demonstrating how variability of data directly affects confidence measurements.

Population Mean

The population mean is a central value that estimates the average salary of all college officials in the relevant categories (such as admissions, registration, etc.). Determining this mean is crucial as it provides a benchmark for comparing individual or group salaries.
For an accurate estimate of the population mean:

• A representative sample and reliable data are required.
• Factor in the calculated margin of error to understand the range within which the true mean likely falls.

In our scenario, the population mean was estimated to be $50,340, calculated from a sample of 45 officials. This sample mean, plus or minus the margin of error, forms the confidence interval that suggests a range for the population mean.

College Officials' Salaries

College officials' salaries, particularly for those working in student services, can vary significantly. Understanding these salaries involves considering the many factors that determine pay levels, such as experience, location, and institutional budget.
Analyzing these salaries using statistical methods like confidence intervals allows us to understand more about the pay distribution across the sector.
Key insights include:

• Variability in salaries is a normal occurrence and can be captured using the standard deviation.
• The confidence intervals provide a range within which we are fairly confident the actual population mean of salaries lies, given our study samples.
• Using confidence levels, such as 90%, helps in decision-making and policy formulation regarding salary structures and adjustments.

This approach not only gives college officials an overview of current salaries but also aids in forecasting and planning for potential salary changes.