Problem 52 - Basic Statistics for Business & Economics

Question

A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of 65.00 per hour. The sample standard deviation was 6.25 per hour. a. What is the population mean? What is the best estimate of the population mean? b. Develop a 99 % confidence interval for the population mean wage (including benefits) for these employees. c. How large a sample is needed to assess the population mean with an allowable error of 1.00 at 95%confidence?

Step-by-Step Solution

Answer

a. Best estimate is 65.00 per hour. b. 99% confidence interval: [61.78, 68.22]. c. Sample size needed: 151.

Step 1: Analyze Problem for Population Mean

We are asked for the population mean in part (a). The problem doesn't provide the population mean but instead gives the sample mean as \(\bar{x} = 65.00\) per hour. Therefore, the best estimate of the population mean is the sample mean itself, which is \(65.00\) per hour.

Step 2: Identify Confidence Interval Formula

For part (b), we need to determine a 99% confidence interval for the population mean. The formula for the confidence interval is:\[ \bar{x} \pm Z_{\alpha/2} \cdot \frac{s}{\sqrt{n}} \]Where:- \( \bar{x} = 65.00 \) (sample mean)- \( Z_{\alpha/2} \) is the z-value for the 99% confidence level.- \( s = 6.25 \) (sample standard deviation)- \( n = 25 \) (sample size)For a 99% confidence level, the critical z-value is approximately 2.576.

Step 3: Calculate Standard Error

We calculate the standard error (SE) using the formula:\[ SE = \frac{s}{\sqrt{n}} = \frac{6.25}{\sqrt{25}} = \frac{6.25}{5} = 1.25 \]

Step 4: Compute Confidence Interval

Now plug the values into the confidence interval formula:\[ 65.00 \pm 2.576 \times 1.25 \]Calculate the margin of error:\[ 2.576 \times 1.25 = 3.22 \]Thus, the confidence interval is:\[ 65.00 \pm 3.22 \]Which gives us an interval from 61.78 to 68.22.

Step 5: Determine Sample Size for Given Error

For part (c), to find the required sample size for a confidence level of 95% and allowable error of 1.00, we use:\[ n = \left( \frac{Z_{\alpha/2} \cdot s}{E} \right)^2 \]At 95% confidence, \( Z_{\alpha/2} \) is approximately 1.96. Thus:\[ n = \left( \frac{1.96 \times 6.25}{1.00} \right)^2 \]Calculate:\[ n = \left( 12.25 \right)^2 = 150.0625 \]Rounding up, the minimum required sample size is 151.

Key Concepts

Population Mean

When surveying a group to find out an average characteristic, like wage, of the entire population, the term *population mean* is crucial. The population mean, often denoted as \( \mu \), represents the average value for all individuals in the population. However, in many cases, it's not feasible to determine this by checking every participant in the population.
Instead, we rely on a *sample mean*, noted by \( \bar{x} \), which is calculated from a subset of the entire population. In our example, the sample mean was \( 65.00 \) per hour, based on a survey of 25 judges in Florida. Given the absence of the actual population mean, the sample mean serves as our best estimate.
This estimation works on the assumption that the sample truly reflects the population. For a more accurate estimate, you'd often increase the sample size, which helps ensure that the sample mean closely approximates the population mean.

Sample Size Determination

Determining the right sample size is important to achieve accurate results, especially when working toward a desired level of precision. The sample size influences the accuracy of our estimate for the population mean.
For corrective decision-making, we often choose a level of confidence and a margin of error we're willing to accept. In our exercise, an allowable error of 1.00 at a 95% confidence level was required. The formula for determining the sample size \( n \) is:

• \( n = \left( \frac{Z_{\alpha/2} \cdot s}{E} \right)^2 \)

In the formula, \( Z_{\alpha/2} \) is the z-value matching the desired confidence level, \( s \) is the sample standard deviation, and \( E \) is the allowable error. For a 95% confidence level, \( Z_{\alpha/2} \) is about 1.96. The calculated sample size ensures that the margin of error stays within the specified limit.
Using this formula with the provided data, the computed sample size needed was 151.

Standard Error

The Standard Error (SE) measures the average amount that the sample mean is expected to differ from the actual population mean. It's a concept that shows the variability or uncertainty involved when estimating a population parameter from a sample statistic. In simpler terms, it's like a margin around the sample mean where we expect the true population mean to lie more often than not.
This is calculated with the formula:

• \( SE = \frac{s}{\sqrt{n}} \)

Here, \( s \) represents the sample standard deviation, and \( n \) is the sample size. In this example, the standard deviation was \( 6.25 \) and the sample size was \( 25 \), leading to a standard error of \( 1.25 \).
The SE helps form the basis for confidence intervals, which provide a range in which the true population mean is believed to lie. Smaller standard errors indicate more reliable estimates of the population mean.