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In addition to teachers and administrative staff, schools also have many other employees, including bus drivers, custodians, and cafeteria workers. In Auburn, WA, the average hourly wage is \(\$ 16.92\) for bus drivers, \(\$ 17.65\) for custodians, and \(\$ 12.86\) for cafeteria workers. \({ }^{23}\) Suppose that a second school district employs \(n=36\) bus drivers who earn an average of \(\$ 13.45\) per hour with a standard deviation of \(s=\$ 2.84 .\) Find a \(95 \%\) confidence interval for the average hourly wage of bus drivers in school districts similar to this one. Does your confidence interval enclose the Auburn, WA average of \(\$ 16.92 ?\) What can you conclude about the hourly wages for bus drivers in this second school district?

Step by step solution

01

Calculate Standard Error

Using the given standard deviation (\(s = \$2.84\)) and the sample size (\(n=36\)), we can calculate the standard error as: Standard Error (SE) \(= \frac{s}{\sqrt{n}}\) SE \(= \frac{2.84}{\sqrt{36}}\)

02

Find Critical Value

For a 95% confidence interval, we will use a t-distribution. With a sample size of 36, we have 35 (36-1) degrees of freedom. Using a t-table or a calculator, we can find the corresponding critical value (t-score) for a two-tailed 95% confidence interval. For 35 df, it is approximately 2.03.

03

Calculate Margin of Error

Using the standard error and critical value, we can calculate the margin of error: Margin of Error (ME) \(= \text{Critical Value} \times \text{Standard Error}\) ME \(= 2.03 \times \frac{2.84}{\sqrt{36}}\)

04

Create Confidence Interval

We can create the confidence interval for the average hourly wage by adding and subtracting the margin of error from the given average wage (\(\mu = \$13.45\)). Confidence Interval \(= \mu \pm \text{Margin of Error}\) CI \(= 13.45 \pm (2.03 \times \frac{2.84}{\sqrt{36}})\)

05

Compare to Auburn, WA Average and Make Conclusion

Once we have our confidence interval, compare it to the Auburn, WA average of \(\$16.92\) to see if it falls within this interval. If it does, then there's no significant difference in hourly wages for bus drivers between the two school districts. If it does not, then there might be a significant difference in pay in the second school district compared to Auburn, WA.

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Key Concepts

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

Understanding Standard Error

The standard error is a key component in statistics, especially when working with confidence intervals. It represents the average distance that the observed values fall from the sample mean. In other words, it quantifies the precision of the sample mean estimate.

To calculate the standard error, you divide the sample standard deviation by the square root of the sample size:

• Formula: Standard Error (SE) = \( \frac{s}{\sqrt{n}} \)
• For the problem, with a standard deviation \(s = 2.84\) and sample size \(n = 36\), the SE is calculated as \( \frac{2.84}{\sqrt{36}} \).

This gives a smaller SE when the sample size is larger, meaning your sample mean is a more accurate estimate of the population mean.

A smaller standard error indicates less variability among the sample means and increases the reliability of the confidence interval.

Exploring the t-Distribution

The t-distribution is essential for constructing confidence intervals, especially when the sample size is small (typically below 30) or when the population standard deviation is unknown. It is similar to the normal distribution but accounts for greater variability in small samples.

Key aspects of the t-distribution include:

• It is symmetrical and bell-shaped like the normal distribution, but has heavier tails, which decrease as the sample size increases.
• The shape of the t-distribution changes with degrees of freedom, becoming more like the normal distribution as degrees of freedom increase.
• In a 95% confidence interval for large datasets (n > 30), it approaches a standard normal distribution.
• To determine the appropriate t-score for calculating a confidence interval, degrees of freedom (sample size minus one) are used.
• In the exercise, the degrees of freedom are 35, and the t-value is approximately 2.03 for a 95% confidence interval.

Decoding Margin of Error

The margin of error indicates how much you expect the true population parameter to differ from the sample statistic at a given confidence level. It provides a range of values that are likely to encompass the population parameter.

Here's how the margin of error is computed:

• Formula: Margin of Error (ME) = \( \text{Critical Value} \times \text{Standard Error} \)
• With a critical value from the t-distribution and the standard error calculated earlier, the margin of error can be determined.
• For our exercise, ME = \(2.03 \times \frac{2.84}{\sqrt{36}}\).

This calculation provides you a buffer around your sample mean, creating a range of plausible values for the population mean of bus drivers’ wages.

A smaller margin of error indicates more confidence in the interval's ability to capture the true parameter.

Understanding Critical Value

The critical value is a multiplier used to calculate the margin of error in a confidence interval. It indicates how many standard errors you need to extend in each direction from the sample mean to generate the confidence interval.

Here's what you should know:

• The critical value is based on the desired confidence level, which in this exercise is 95%.
• For a two-tailed t-distribution with 95% confidence and 35 degrees of freedom, the critical value is around 2.03.
• It quantifies the coverage probability needed, meaning how "confident" you can be that the true population parameter is within the interval.

The larger the critical value, the wider the confidence interval. This implies that for higher confidence levels, a larger critical value is essential to ensure the interval captures the population parameter.

Critical values are crucial in interpreting and constructing the confidence intervals accurately.