Problem 27 - Basic Statistics for Business and Economics with Student

Question

Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79,64,84,82,92 and 77 . Instead of averaging the six scores, the instructor indicated he would randomly select two grades and compute the final percent correct based on the two percents. a. How many different samples of two test grades are possible? b. List all possible samples of size two and compute the mean of each. c. Compute the mean of the sample means and compare it to the population mean. d. If you were a student, would you like this arrangement? Would the result be different from dropping the lowest score? Write a brief report.

Step-by-Step Solution

Answer

15 samples possible; mean of sample means is 79.5, similar to population mean 79.67. Random selection is risky.

Step 1: Compute Number of Samples

To find out how many different samples of two test grades are possible, we use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n = 6 \) is the total number of tests, and \( k = 2 \) is the sample size. Therefore, \( C(6, 2) = \frac{6 \times 5}{2 \times 1} = 15 \). There are 15 possible samples of size two.

Step 2: List All Possible Samples and Their Means

List all the possible pairs of test scores and calculate the mean for each pair. The possible samples and their means are:1. (79, 64) - Mean: \( \frac{79 + 64}{2} = 71.5 \)2. (79, 84) - Mean: \( \frac{79 + 84}{2} = 81.5 \)3. (79, 82) - Mean: \( \frac{79 + 82}{2} = 80.5 \)4. (79, 92) - Mean: \( \frac{79 + 92}{2} = 85.5 \)5. (79, 77) - Mean: \( \frac{79 + 77}{2} = 78 \)6. (64, 84) - Mean: \( \frac{64 + 84}{2} = 74 \)7. (64, 82) - Mean: \( \frac{64 + 82}{2} = 73 \)8. (64, 92) - Mean: \( \frac{64 + 92}{2} = 78 \)9. (64, 77) - Mean: \( \frac{64 + 77}{2} = 70.5 \)10. (84, 82) - Mean: \( \frac{84 + 82}{2} = 83 \)11. (84, 92) - Mean: \( \frac{84 + 92}{2} = 88 \)12. (84, 77) - Mean: \( \frac{84 + 77}{2} = 80.5 \)13. (82, 92) - Mean: \( \frac{82 + 92}{2} = 87 \)14. (82, 77) - Mean: \( \frac{82 + 77}{2} = 79.5 \)15. (92, 77) - Mean: \( \frac{92 + 77}{2} = 84.5 \).

Step 3: Compute the Mean of the Sample Means

To find the mean of the sample means, add all the calculated means and divide by the number of samples, which is 15. \[ \text{Mean of sample means} = \frac{71.5 + 81.5 + 80.5 + 85.5 + 78 + 74 + 73 + 78 + 70.5 + 83 + 88 + 80.5 + 87 + 79.5 + 84.5}{15} = 79.5 \]

Step 4: Compute the Population Mean

To calculate the population mean, sum all the test scores and divide by the number of scores, which is 6. \[ \text{Population mean} = \frac{79 + 64 + 84 + 82 + 92 + 77}{6} = 79.67 \]

Step 5: Compare Means and Provide Conclusion

Compare the mean of the sample means (79.5) to the population mean (79.67). The two means are very close. As a student, this random sampling method introduces variability and may not be beneficial because it might select low scores. Dropping the lowest score generally raises the average, which may be more favorable.

Key Concepts

Population Mean

The population mean, denoted as \( \mu \), is the average of all data points in a given population. In our case, the 'population' consists of all six test scores received throughout the semester. To find the population mean, you add all the test scores together and then divide by the total number of tests. This is mathematically expressed as: \[ \mu = \frac{\sum X}{N} \]where \( \sum X \) is the sum of all test scores and \( N \) is the number of test scores.

In the given exercise, the population mean is computed using the scores: 79, 64, 84, 82, 92, and 77. Adding these scores gives 478, and then dividing by 6 (the number of scores) produces the population mean: 79.67.

Understanding the population mean is crucial because it serves as a reference point against which other statistical measures, such as sample means, can be compared.

Combinatorial Samples

Combinatorial sampling involves selecting items from a larger set where the order of selection does not matter. This is often achieved through combinations. In our case, we want to know how many ways we can select two test scores from a pool of six.

The number of possible combinations can be calculated using the formula for combinations:\[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \(!\) denotes a factorial, which is the product of an integer and all the integers below it.

For six tests and choosing two at a time, we set \( n = 6 \) and \( k = 2 \). This results in:\[ C(6, 2) = \frac{6 \times 5}{2 \times 1} = 15 \]Thus, there are 15 different pairs of test scores. This concept is essential in statistical analysis to understand how many distinct groupings or samples can be derived from a population.

Statistical Analysis

Statistical analysis is the process of collecting and analyzing data to identify patterns or trends, test hypotheses, and make informed decisions. In this exercise, the statistical analysis involves examining how the sample mean compares to the population mean.

The steps include:

• Listing all the possible sample pairs (combinatorial samples)
• Calculating the mean for each pair, known as the sample mean
• Calculating the mean of these sample means
• Comparing the mean of sample means to the population mean

The mean of the sample means in our exercise is calculated as 79.5, which is derived by averaging all the individual sample means, given the 15 combinations.

The significance of this analysis lies in its ability to demonstrate the central limit theorem. This theorem suggests that the sample mean of an adequately large number of samples will approximate the population mean. The comparison here shows a close approximation, reinforcing statistical principles.

Education Assessment

Education assessment is an important part of student evaluation and often involves various methods of calculating final grades. In this exercise, the instructor's approach is unique as it involves randomly selecting two out of six test scores. This is a different take on educational assessment from simply taking the average of all test scores or eliminating the lowest one.

Using random sampling to determine grades exemplifies a broader assessment technique but also introduces variability and uncertainty in the outcome. It departs from standard assessments by providing different potential outcomes based on random choices.

In conclusion, when analyzing this method of education assessment, it might not be the fairest approach for students due to its unpredictable nature. Typically, methods like dropping the lowest score are seen as more favorable for students as they tend to improve the overall average unless all scores are closely aligned. Navigating these assessment methods helps students and instructors understand the impact of decision-making on educational outcomes.