Paper For Above instruction
Introduction
The analysis of sample means in statistics provides crucial insights into the behavior of populations, particularly when direct measurement of the entire population is impractical. In the context of H&R Block's tax return preparation fees, understanding the probability that a sample mean falls within a specified range around the population mean aids in assessing the reliability of sample estimates. This paper explores the probability that the sample mean of preparation fees falls within $8 of the population mean, given different sample sizes, using properties of the sampling distribution of the sample mean under the assumption of known population standard deviation.
Background and Assumptions
The problem states that the population mean fee in 2012 was $183, with a population standard deviation of $50. It is assumed that the fees follow a normal distribution or that the sample sizes are large enough for the Central Limit Theorem to apply, ensuring the sampling distribution of the sample mean is approximately normal. The goal is to compute the probability that the sample mean is within $8 of the population mean for different sample sizes, specifically for n=30, n=50, and n=100.
Methodology
The probability calculation involves understanding the properties of the sampling distribution of the sample mean, which has a mean equal to the population mean ($183) and a standard deviation called the standard error (SE), calculated as:
\[ SE = \frac{\sigma}{\sqrt{n}} \]
where \(\sigma = 50\), and \(n\) is the sample size.
The probability that the sample mean is within $8 of the population mean is equivalent to:
\[ P(|\bar{X} - \mu| < 8) \]
which, for a normal distribution, can be written as:
\[ P(-8 < \bar{X} - \mu < 8) \]
Standardizing to a Z-score:
\[ P\left( -\frac{8}{SE} < Z < \frac{8}{SE} \right) \]
where
\[ Z = \frac{\bar{X} - \mu}{SE} \]
and \(Z\) follows a standard normal distribution.
Using the symmetry of the normal distribution:
\[ P(|Z| < z) = 2 \times P(Z < z) - 1 \]
we can find the probability once the Z-score is computed.
Calculations for each sample size
- For \(n=30\):
\[ SE_{30} = \frac{50}{\sqrt{30}} \approx 9.1287 \]
\[ z_{30} = \frac{8}{SE_{30}} = \frac{8}{9.1287} \approx 0.8768 \]
Probability:
\[ P(|\bar{X} - \mu| < 8) = 2 \times P(Z < 0.8768) - 1 \]
Using standard normal distribution tables or a calculator:
\[ P(Z < 0.8768) \approx 0.8090 \]
Thus:
\[ P \approx 2 \times 0.8090 - 1 = 0.6180 \]
- For \(n=50\):
\[ SE_{50} = \frac{50}{\sqrt{50}} \approx 7.0711 \]
\[ z_{50} = \frac{8}{7.0711} \approx 1.132 \]
Probability:
\[ P(Z < 1.132) \approx 0.8729 \]
Thus:
\[ P \approx 2 \times 0.8729 - 1 = 0.7458 \]
- For \(n=100\):
\[ SE_{100} = \frac{50}{\sqrt{100}} = 5 \]
\[ z_{100} = \frac{8}{5} = 1.6 \]
Probability:
\[ P(Z < 1.6) \approx 0.9452 \]
Thus:
\[ P \approx 2 \times 0.9452 - 1 = 0.8904 \]
Results Interpretation
The calculations reveal that increasing sample size enhances the probability that the sample mean will be within $8 of the population mean. Specifically, the probability grows from approximately 61.8% with 30 samples to about 89.0% with 100 samples.
Recommendation
for Sample Size
To achieve at least a 95% probability (0.95) that the sample mean is within $8 of the population mean, the sample size must be sufficiently large. Solving for \(n\):
\[ P(|\bar{X} - \mu| < 8) \geq 0.95 \]
which requires:
\[ z = \frac{8}{SE} \]
and
\[ P(|Z| < z) \geq 0.95 \]
From standard normal distribution, for 95% coverage:
\[ P(|Z| < z) = 0.95 \Rightarrow z \approx 1.96 \]
Therefore:
\[ SE \leq \frac{8}{1.96} \approx 4.0816 \]
Given \(SE = \frac{50}{\sqrt{n}}\), solve for \(n\):
\[ \frac{50}{\sqrt{n}} \leq 4.0816 \]
\[ \sqrt{n} \geq \frac{50}{4.0816} \approx 12.25 \]
\[ n \geq (12.25)^2 \approx 150.06 \]
Thus, a sample size of at least 151 is necessary.
Conclusion
This analysis underscores the importance of increasing sample size to reliably estimate population parameters within a specified margin of error. For a high confidence level (95%), a sample size exceeding 150 is recommended to ensure the sample mean falls within $8 of the actual mean with at least 95% probability. The calculations also illustrate how statistical principles guide decision-making in survey design and quality control in service industries, such as tax preparation services offered by H&R Block.
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