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The Mean Preparation Fee Hr Block Charged Retail Customers In 2012 Wa

The mean preparation fee H&R Block charged retail customers in 2012 was $183. Assume the population standard deviation is $50.

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 analysis of statistical data related to consumer fees provides a vital insight into the predictability and reliability of service costs. This paper explores the probability that sample means of preparation fees charged by H&R Block in 2012 are within a specific margin of the population mean, given different sample sizes. It examines the principles of the sampling distribution of the sample mean, calculates probabilities using the standard normal distribution, and evaluates the appropriate sample size needed for high-confidence estimates.

In 2012, H&R Block charged an average preparation fee of $183 to retail customers, with a population standard deviation of $50. To assess the likelihood that the sample mean falls within $8 of the population mean for different sample sizes (30, 50, and 100), we utilize the properties of the normal distribution. When sampling from a population with a known standard deviation, the sampling distribution of the sample mean is normally distributed with a standard error (SE) given by:

SE = σ / √n

where σ is the population standard deviation, and n is the sample size. The probability that the sample mean is within $8 of the population mean translates into calculating P(μ − 8 ≤ X̄ ≤ μ + 8).

For each case:

- n = 30:

SE = 50 / √30 ≈ 50 / 5.477 ≈ 9.128

The margin of error in terms of the z-score is:

z = (E) / (SE) = 8 / 9.128 ≈ 0.876

The probability corresponds to the area within ±0.876 standard errors, which is 2 * P(Z ≤ 0.876) - 1.

From the standard normal distribution table, P(Z ≤ 0.876) ≈ 0.81.

Thus, the probability is approximately 2 * 0.81 - 1 = 0.62 or 62%.

- n = 50:

SE = 50 / √50 ≈ 50 / 7.071 ≈ 7.07

z = 8 / 7.07 ≈ 1.13

P(Z ≤ 1.13) ≈ 0.87, so probability ≈ 2 * 0.87 - 1 ≈ 0.74 or 74%.

- n = 100:

SE = 50 / √100 = 50 / 10 = 5

z = 8 / 5 = 1.6

P(Z ≤ 1.6) ≈ 0.945, so probability ≈ 2 * 0.945 - 1 ≈ 0.89 or 89%.

Considering the probabilities calculated, as the sample size increases, the likelihood that the sample mean is within $8 of the population mean also increases. To determine which sample size provides at least 95% confidence, we need to find the minimum sample size n that satisfies:

z = (8) / (σ / √n) ≥ z_(0.975) ≈ 1.96

Rearranged: √n ≥ (z * σ) / 8

√n ≥ (1.96 * 50) / 8 ≈ 98 / 8 ≈ 12.25

n ≥ (12.25)^2 ≈ 150.06

Therefore, a sample size of at least 151 would yield a probability of at least 0.95 of the mean being within $8 of the population mean.

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