Problem 42 - Basic Statistics for Business and Economics with Student

Question

Human Resource Consulting (HRC) is surveying a sample of 60 firms in order to study health care costs for a client. One of the items being tracked is the annual deductible that employees must pay. The state Bureau of Labor reports the mean of this distribution is \(\$ 502\) with a standard deviation of \(\$ 100 .\) a. Compute the standard error of the sample mean for HRC. b. What is the chance HRC finds a sample mean between \(\$ 477\) and \(\$ 527 ?\) c. Calculate the likelihood that the sample mean is between \(\$ 492\) and \(\$ 512\). d. What is the probability the sample mean is greater that \(\$ 550 ?\)

Step-by-Step Solution

Answer

a. SE is approximately $12.91. b. Probability is 0.8950. c. Probability is 0.5608. d. Probability is less than 0.0001.

Step 1: Compute the Standard Error

The formula for the standard error of the sample mean (SE) is \( \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. For this problem, \( \sigma = 100 \) and \( n = 60 \). Plug the values into the formula:\[ SE = \frac{100}{\sqrt{60}} \approx \frac{100}{7.746} \approx 12.91 \]

Step 2: Compute the Probability for Mean Between $477 and $527

To find the probability of the sample mean being between \\(477 and \\)527, convert these to z-scores using the standard error found in Step 1. The z-score formula is \( z = \frac{X - \mu}{SE} \). Calculating the z-scores:- For \\(477: \( z = \frac{477 - 502}{12.91} \approx \frac{-25}{12.91} = -1.937 \)- For \\)527: \( z = \frac{527 - 502}{12.91} \approx \frac{25}{12.91} = 1.937 \)Using the standard normal distribution table, find the probabilities corresponding to these z-scores. The probability between \(-1.937\) and \(1.937\) is approximately \(0.9475 - 0.0525 = 0.8950\).

Step 3: Compute the Probability for Mean Between $492 and $512

Similarly, convert \\(492 and \\)512 to z-scores:- For \\(492: \( z = \frac{492 - 502}{12.91} \approx \frac{-10}{12.91} = -0.775 \)- For \\)512: \( z = \frac{512 - 502}{12.91} \approx \frac{10}{12.91} = 0.775 \)Find the probabilities corresponding to these z-scores. The probability between \(-0.775\) and \(0.775\) is approximately \(0.7804 - 0.2196 = 0.5608\).

Step 4: Compute the Probability for Mean Greater than $550

Calculate the z-score for \$550:\( z = \frac{550 - 502}{12.91} \approx \frac{48}{12.91} = 3.720 \)From the standard normal distribution table, find the probability of a z-score greater than \(3.720\). This is a very high z-score, and the probability is essentially close to 0 (less than 0.0001).

Key Concepts

Standard Error

The standard error (SE) is a crucial concept in statistics, particularly when dealing with sample data. It helps us understand the variability of a sample mean compared to the actual population mean.

To calculate the SE of the sample mean, we use the formula:

• \( SE = \frac{\sigma}{\sqrt{n}} \)

Here:

• \( \sigma \) is the standard deviation of the population, which reflects how much the data points deviate from the mean.
• \( n \) is the size of the sample.

By using this formula, we can determine how much the sample mean is expected to deviate from the actual population mean. If we take the specific case from the HRC exercise:- Population standard deviation, \( \sigma = 100 \) dollars.- Sample size, \( n = 60 \).
The calculation works out to \( SE \approx 12.91 \) dollars. This value indicates the average distance of the sample mean from the population mean. Understanding SE is vital because it provides the basis for making inferences about the population from which the sample is drawn. It tells us how much we can expect our sample mean to "hover" around the actual population mean. With a smaller standard error, the sample mean is more likely to be close to the population mean.

Z-Score

The Z-score is a statistical measure that tells you how many standard errors a data point is from the mean. It is incredibly useful in comparing data from different normal distributions.

To find a Z-score, you use the formula:

• \( z = \frac{X - \mu}{SE} \),

where \( X \) is the value of interest, \( \mu \) is the population mean, and \( SE \) is the standard error. This formula helps you translate raw scores into a common language: the standard normal distribution.For example, in the context of the HRC exercise, to find the probability of the sample mean being between \\(477 and \\)527, the Z-scores are calculated as follows:

• For \\(477: \( z = \frac{477 - 502}{12.91} = -1.937 \)
• For \\)527: \( z = \frac{527 - 502}{12.91} = 1.937 \)

Using these Z-scores, we consult a standard normal distribution table or calculator to find the probabilities associated with them, providing an understanding of how likely the sample mean falls within a certain range.Z-scores are valuable because they allow you to standardize different data sets, making them easier to compare. If you can convert your data into Z-scores, it's easy to calculate the probabilities of various outcomes using the properties of the standard normal distribution.

Normal Distribution

Normal distribution is a fundamental concept in statistics, often referred to as the "Gaussian distribution." It depicts how the values of a variable are distributed.

The distribution is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. Its probability density function is bell-shaped, showing that most occurrences take place around the average value.Key characteristics of the normal distribution include:

• It is defined by the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
• The total area under the curve of a normal distribution is equal to 1.
• About 68% of the data falls within one standard deviation from the mean.
• Approximately 95% falls within two standard deviations.
• Almost 99.7% lies within three standard deviations.

In the HRC exercise, the normal distribution is the basis for determining probabilities of different sample means. Once you convert data into Z-scores, you can make use of the properties of the normal distribution to assess probabilities over different ranges.The normal distribution is vital in statistics because it allows for the application of the Central Limit Theorem, which states that, regardless of the original distribution of data, the distribution of the sample mean will tend to be approximately normal, especially with a large enough sample size. This principle enables statisticians to make inferences about population parameters effectively.