Population Consists Of The Following Five Values: 2244 And

6 A Population Consists of the Following Five Values 2244 And 8

Evaluate the statistical properties of a small population and its samples, including calculating sample means, population mean, and dispersion measures. Extend the analysis to real-world sales data, compare sample distributions, and analyze probabilities using normal distribution concepts.

Paper For Above instruction

The study of small populations and their sampling distributions provides valuable insights into the behavior of data, variability, and inferential statistics. This paper discusses the analysis of a population with specified values, examines sampling methods and their outcomes, and explores probability calculations in the context of normal distributions, applying these concepts to real-world sales data and IRS tax preparation times.

Analysis of a Small Population

Consider a population consisting of five values: 2, 2, 4, 4, and 8. To understand the characteristics of this population, we first list all possible samples of size two. Since the population size is small, listing all combinations is feasible. The total number of such samples can be calculated using combinations: nCr = 5C2 = 10. These samples include:

  • (2, 2)
  • (2, 4)
  • (2, 4)
  • (2, 8)
  • (2, 4) — from the other 4
  • (2, 8) — from the other 8
  • (4, 4)
  • (4, 8)
  • (4, 8) — second occurrence
  • (4, 8) — third occurrence

Calculating the mean of each sample yields:

  • (2, 2): (2+2)/2 = 2
  • (2, 4): 3
  • (2, 4): 3
  • (2, 8): 5
  • (2, 4): 3
  • (2, 8): 5
  • (4, 4): 4
  • (4, 8): 6
  • (4, 8): 6
  • (4, 8): 6

The mean of the distribution of sample means can be computed by averaging these sample means: (2+3+3+5+3+5+4+6+6+6)/10 = 4.3. The population mean is calculated as (2+2+4+4+8)/5 = 4.0. Comparing the two, the mean of the sampling distribution (4.3) is close to the population mean (4.0), illustrating the principle that the sampling distribution mean approximates the population mean, especially with larger samples.

Regarding dispersion, the variance within the population can be computed as follows: first find the deviations from the mean, square them, and average:

Deviations squared:

  • (2 - 4)^2 = 4
  • (2 - 4)^2 = 4
  • (4 - 4)^2 = 0
  • (4 - 4)^2 = 0
  • (8 - 4)^2 = 16

Population variance = (4+4+0+0+16)/5 = 24/5 = 4.8. The dispersion of sample means can be assessed via the standard error, which diminishes as sample size increases, reflecting less variability in larger samples.

Application to Sales Data of Car Representatives

Transitioning from theoretical populations to practical applications, consider five sales representatives at Mid-Motors Ford and their last week's car sales: Peter Staller (8), Connie Stallter (6), Juan Lopez (4), Ted Barnes (10), and Peggy Chu (6). To determine potential sampling behaviors, first, determine how many samples of size 2 are possible:

Number of possible samples: 5C2 = 10. List all sample pairs and compute their means:

  • (Peter, Connie): (8+6)/2=7
  • (Peter, Juan): (8+4)/2=6
  • (Peter, Ted): (8+10)/2=9
  • (Peter, Peggy): (8+6)/2=7
  • (Connie, Juan): (6+4)/2=5
  • (Connie, Ted): (6+10)/2=8
  • (Connie, Peggy): (6+6)/2=6
  • (Juan, Ted): (4+10)/2=7
  • (Juan, Peggy): (4+6)/2=5
  • (Ted, Peggy): (10+6)/2=8

The sample means range from 5 to 9, with an average of (7+6+9+7+5+8+6+7+5+8)/10= 6.8, closely approximating the population mean of (8+6+4+10+6)/5=6.8. This confirms the efficiency of sampling in representing population characteristics.

Graphing these sample means would typically show a distribution centered around the population mean, with dispersion depending on the variability of individual values and sample size. The distribution of sample means tends to be more normal-shaped as per the Central Limit Theorem, especially with larger samples, even if the population distribution is not normal.

Analysis of a Larger Sales Population

Scrapper Elevator Company provides data for 20 sales representatives last month. To analyze this population, plotting a histogram enables visual assessment of the distribution shape. Calculating the mean involves summing all sales figures and dividing by 20, providing insights into the average sales per representative. Selecting random samples of five representatives allows for the computation of sample means, which upon comparison to the population mean, should, on average, be similar if sampling is random and unbiased.

By drawing a histogram of the sample means, one would observe a distribution that approximates a normal curve, especially with multiple samples, in accordance with the Central Limit Theorem. The shape and spread of this distribution are typically narrower than the population distribution, illustrating decreased variability of the mean as sample size increases.

Such analyses highlight the importance of sampling techniques and the implications for inferential statistics, enabling estimation of population parameters with known margins of error, which is fundamental in business decision-making and statistical inference.

Probability Calculations Using Normal Distribution

In the context of normally distributed data, probabilities related to sample means can be calculated using z-scores. For example, consider a normal population with a mean of 75 and a standard deviation of 5. When sampling 40 individuals, the standard error of the mean (SEM) is calculated as:

SEM = σ / √n = 5 / √40 ≈ 0.79.

To find the probability that the sample mean is less than 74, calculate the z-score:

z = (74 - 75)/0.79 ≈ -1.27. Consulting the standard normal table, this corresponds to a probability of approximately 0.10, indicating a 10% chance the sample mean is below 74.

Similarly, probabilities for other intervals are computed by deriving their respective z-scores and referencing the standard normal distribution. For instance, the probability between 74 and 76 involves calculating z-scores for both bounds and finding the area between them.

In a further example, the IRS study reports a mean of 330 minutes and a standard deviation of 80 minutes for tax preparation time. For a sample of 40 taxpayers, the standard error is:

SEM=80/√40≈12.65 minutes.

If assessing the likelihood that the sample mean exceeds 320 minutes, the z-score is:

z = (320 - 330)/12.65 ≈ -0.79, which gives a probability of approximately 0.21. Calculations for other ranges follow similarly, providing vital insights into process efficiencies and resource allocation.

Conclusion

This comprehensive analysis demonstrates fundamental statistical concepts, including the calculation of sample means, understanding the relationship between population and sample distributions, and applying normal distribution properties to real-world data. These principles underpin effective decision-making in business and public policy by enabling accurate inference about populations based on samples.

References

  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (8th ed.). Brooks/Cole, Cengage Learning.
  • Rice, J. A. (2007). Mathematical Statistics and Data Analysis (3rd ed.). Duxbury Press.
  • Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics (8th ed.). Pearson Education.
  • Wackerly, D., Mendenhall, W., & Scheaffer, R. (2014). Mathematical Statistics with Applications (7th ed.). Cengage Learning.
  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Learning from Data (4th ed.). Pearson.
  • Everitt, B. (2002). The Cambridge Dictionary of Statistics. Cambridge University Press.
  • Freedman, D., Pisani, R., & Purves, R. (2007). Statistics. Norton & Company.
  • Bluman, A. G. (2018). Elementary Statistics: A Step by Step Approach (9th ed.). McGraw-Hill Education.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W. H. Freeman.
  • Lohr, S. (2010). Sampling: Design and Analysis. Cengage Learning.

Related Assignments