The College Board Reported Mean Scores For The

The College Board Reported The Following Mean Scores For The Three Par

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Assume that the population standard deviation on each part of the test is 100. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test (to 4 decimals)? What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)? What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)? (next question) The mean tax-return preparation fee H&R Block charged retail customers last year was $183 (The Wall Street Journal, March 7, 2012). Use this price as the population mean and assume the population standard deviation of preparation fees is $50. Round your answers to four decimal places. a. What is the probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean? b. What is the probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean? c. What is the probability that the mean price for a sample of 100 H&R Block retail customers is within $8 of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within $8 of the population mean?

Paper For Above instruction

The given problem involves calculating probabilities related to sample means within a specified margin of the population mean, utilizing properties of the normal distribution and the Central Limit Theorem. Both scenarios—SAT scores and tax preparation fees—require determining the likelihood that a sample mean falls within a certain range of the population mean, given the sample size, population standard deviation, and desired margin of error. The calculations hinge on understanding standard errors and converting the problem into a standard normal distribution to find the probabilities accurately.

Introduction

The Central Limit Theorem (CLT) plays a vital role in inferential statistics by permitting the use of normal distribution approximations for sample means, even when the underlying data are not normally distributed, provided the sample size is sufficiently large. This theorem states that the distribution of the sample mean approaches normality with a mean equal to the population mean and a standard deviation called the standard error, which depends on the population standard deviation and the sample size. This foundation allows us to evaluate probabilities about the sample mean's proximity to the population mean—crucial for quality control, research, and decision-making in various fields.

Part 1: SAT Scores Analysis

First, consider the SAT score data. The population standard deviation is given as 100, with sample sizes of 90, and interest in the probability that the sample mean is within 10 points of the population mean (502 for Critical Reading, 515 for Mathematics, and 494 for writing). The general formula involves computing the standard error (SE):

SE = σ / √n, where σ is the population standard deviation, and n is the sample size.

For each part, the probability that the sample mean falls within a range around the population mean (μ) can be computed as:

P( |X̄ - μ|

Converting to a standard normal variable (Z):

Z = (X̄ - μ) / (σ / √n)

Thus, the probability becomes:

P( -10

Calculating for each case:

Critical Reading Part

  • μ = 502
  • σ = 100
  • n = 90

Standard error:

SE = 100 / √90 ≈ 100 / 9.4868 ≈ 10.5332

Value of Z:

Z = 10 / 10.5332 ≈ 0.9491

Probability (within ±10 points):

P = 2  Φ(0.9491) - 1 ≈ 2  0.8299 - 1 ≈ 0.6598

Therefore, the probability that a sample of 90 test-takers yields a mean within 10 points of the population mean (502) is approximately 0.6598.

Mathematics Part

  • μ = 515
  • σ = 100
  • n = 90
SE = 100 / √90 ≈ 10.5332 (as above)
Z = 10 / 10.5332 ≈ 0.9491
P ≈ 2 * Φ(0.9491) - 1 ≈ 0.6598

The probability remains the same because the parameters are identical, except for the mean value, which affects the central value but the probability calculation for within 10 points remains similar, centered at the mean.

Writing Part

  • μ = 494
  • σ = 100
  • n = 100
SE = 100 / √100 = 100 / 10 = 10
Z = 10 / 10 = 1.0
P = 2  Φ(1.0) - 1 ≈ 2  0.8413 - 1 ≈ 0.6826

The probability that a sample of 100 test-takers produces a mean within 10 points of the population mean (494) is approximately 0.6826.

Part 2: Tax-return Preparation Fees Analysis

Next, examine the scenario involving tax-return preparation fees. The population mean is $183, with standard deviation of $50. The sample sizes are 30, 50, and 100, and the margin of error is $8. The objective is to find the probability that the sample mean falls within ±$8 of the population mean.

Similar to the prior analysis, the standard error is:

SE = 50 / √n

The probability calculation uses:

P( |X̄ - μ|

For n = 30

SE = 50 / √30 ≈ 50 / 5.4772 ≈ 9.1287
Z = 8 / 9.1287 ≈ 0.8766
P = 2  Φ(0.8766) - 1 ≈ 2  0.8097 - 1 ≈ 0.6194

For n = 50

SE = 50 / √50 ≈ 7.0711
Z = 8 / 7.0711 ≈ 1.1314
P ≈ 2  Φ(1.1314) - 1 ≈ 2  0.8729 - 1 ≈ 0.7458

For n = 100

SE = 50 / √100 = 50 / 10 = 5
Z = 8 / 5 = 1.6
P ≈ 2  Φ(1.6) - 1 ≈ 2  0.9452 - 1 ≈ 0.8904

Part 3: Recommendation of Sample Size for ≥ 0.95 Probability

To determine which sample size ensures at least a 95% probability that the sample mean is within $8 of the population mean, we set:

2 * Φ(z) - 1 ≥ 0.95

which implies:

Φ(z) ≥ 0.975

From standard normal distribution tables, the z-value corresponding to 0.975 is approximately 1.96.

Thus, we require:

8 / SE = 1.96 ⇒ SE = 8 / 1.96 ≈ 4.08

SE = 50 / √n ≥ 4.08 ⇒ √n ≤ 50 / 4.08 ≈ 12.25 ⇒ n ≥ (12.25)^2 ≈ 150.06


Therefore, a sample size of at least 151 is needed to have at least a 0.95 probability of the sample mean being within $8 of the population mean.


Conclusion


Overall, the statistical calculations reveal how sample size, population standard deviation, and acceptable margin of error influence the likelihood of accurately estimating the population mean. Larger sample sizes decrease the standard error, thus increasing the probability that the sample mean falls within a desired range. Adopting sufficiently large samples is vital for achieving high-confidence estimates, particularly in quality assurance, market research, and social sciences. These calculations underscore the importance of understanding the interplay between sample size and variability for effective statistical inference.


References
  • DeGroot, M. H., & Schervish, J. (2012). Probability and Statistics (4th ed.). Pearson Education.
  • Freund, J. E., & Williams, F. (2003). Modern Business Statistics. Prentice Hall.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W. H. Freeman.
  • Ross, S. M. (2014). Introduction to Probability and Statistics (5th ed.). Academic Press.
  • Wackerly, D., Mendenhall, W., & Scheaffer, R. (2008). Mathematical Statistics with Applications. Brooks/Cole.
  • Triola, M. F. (2018). Elementary Statistics (13th ed.). Pearson.
  • Hogg, R. V., & Tanis, E. A. (2009). Probability and Statistical Inference. Pearson Education.
  • Wall Street Journal. (2012). H&R Block average fee data. March 7, 2012.
  • The World Almanac. (2009). SAT scores data.
  • Statistics Canada. (2010). Sample Size Calculations and Confidence Intervals. Statistics Canada Publications.

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