The Mean Tax Return Preparation Fee At HR Block Charged Reta

The Mean Tax Return Preparation Fee Hr Block Charged Retail Customers

The mean tax-return preparation fee H&R Block charged retail customers in 2012 was $183. 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

Understanding the concepts of sampling distribution, standard error, and probability is essential in interpreting data about tax preparation fees, as exemplified in this case involving H&R Block's charges. The question aims to determine the likelihood that a sample mean falls within a specified range of the population mean, given different sample sizes, leveraging the properties of the normal distribution.

Given the population mean (\(\mu\)) of $183 and the population standard deviation (\(\sigma\)) of $50, we recognize that the sampling distribution of the sample mean (\(\bar{x}\)) follows a normal distribution, especially for sufficiently large samples, due to the Central Limit Theorem. The standard error (SE) of the mean is calculated as \(\mathrm{SE} = \sigma / \sqrt{n}\), where \(n\) is the sample size.

First, the probability that the sample mean is within $8 of the population mean corresponds to calculating \(P(|\bar{x} - \mu|

Part a: Sample size of 30

Calculate the standard error: \(\mathrm{SE} = 50 / \sqrt{30} \approx 50 / 5.4772 \approx 9.1287\).

The z-score for the margin of $8: \(z = 8 / 9.1287 \approx 0.8768\).

The probability that \(\bar{x}\) is within $8 of \(\mu\) is \(P(-0.8768

Using standard normal tables or software, \(\Phi(0.8768) \approx 0.8092\).

Thus, probability \(= 2 \times 0.8092 - 1 = 0.6184\).

Part b: Sample size of 50

Standard error: \(\mathrm{SE} = 50 / \sqrt{50} \approx 50 / 7.0711 \approx 7.0711\).

z-score: \(8 / 7.0711 \approx 1.1314\).

Calculate \(2 \times \Phi(1.1314) - 1\). From standard normal tables, \(\Phi(1.1314) \approx 0.8980\).

Probability \(= 2 \times 0.8980 - 1 = 0.7960\).

Part c: Sample size of 100

Standard error: \(\mathrm{SE} = 50 / \sqrt{100} = 50 / 10 = 5\).

z-score: \(8 / 5 = 1.6\).

From standard normal tables, \(\Phi(1.6) \approx 0.9452\).

Probability \(= 2 \times 0.9452 - 1 = 0.8904\).

Part d: Recommendations for at least 0.95 probability

To achieve a probability of at least 0.95 that the sample mean is within $8 of the population mean, we need the z-score corresponding to this probability. Since \(P(|Z|

Set \(z = 1.96 = \frac{8}{\mathrm{SE}}\), solving for \(\mathrm{SE}\):

\(\mathrm{SE} = \frac{8}{1.96} \approx 4.0816\).

Recall that \(\mathrm{SE} = 50 / \sqrt{n}\), so

\(\sqrt{n} = 50 / 4.0816 \approx 12.253\), and \(n = (12.253)^2 \approx 150.0\).

Thus, a sample size of approximately 150 or more is needed to have at least a 95% probability that the sample mean is within $8 of the population mean.

Hence, among the given sample sizes, only a sample of 150 or larger (not provided directly in parts a, b, c) would suffice. Since 100 is below 150, it does not meet this criterion, but larger samples such as 200 would be recommended for even higher certainty.

Conclusion

Analyzing the probabilities demonstrates how sample size influences the confidence that the sample mean is close to the population mean. Larger samples reduce the standard error, increasing the probability of the sample mean falling within a specified margin of the true mean. For a 0.95 probability threshold within $8, a sample size of approximately 150 or more is required when the population standard deviation remains constant at $50.

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