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. Use this price as the population mean and assume the population standard deviation of preparation fees is $50. Calculate the following probabilities:
a) The probability that the mean price for a sample of 30 H&R Block retail customers is within $8 of the population mean.
b) The probability that the mean price for a sample of 50 H&R Block retail customers is within $8 of the population mean.
c) 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
This paper aims to analyze the probabilities associated with sampling distributions of preparation fees charged by H&R Block in 2012, focusing specifically on the likelihood that sample means fall within a specified range of the population mean. Given a population mean of $183 and a standard deviation of $50, the task involves calculating probabilities for different sample sizes. These calculations are essential for understanding statistical variability, confidence intervals, and the adequacy of sample sizes for estimating population parameters with high certainty.
The central statistical concept underpinning this analysis is the sampling distribution of the sample mean, which, according to the Central Limit Theorem, approaches a normal distribution when the sample size is sufficiently large, regardless of the population distribution. For a given sample size n, the standard error (SE) of the mean is calculated as the population standard deviation divided by the square root of n (SE = σ/√n). With known population parameters, the probability that the sample mean falls within a certain distance of the population mean can be determined using z-scores, which describe how many standard errors a particular value is from the mean.
In part a), with a sample size of 30, the standard error is SE = 50/√30 ≈ 9.12. The within $8 range translates to a z-score of z = 8/9.12 ≈ 0.878. Consulting the standard normal distribution table, the
probability that the z-score falls between -0.878 and +0.878 corresponds to approximately 71.65%. This indicates that there is about a 71.65% chance that the sample mean for 30 customers is within $8 of the population mean.
Part b) involves a larger sample size of 50, with SE = 50/√50 ≈ 7.07. The z-score for an $8 deviation is 8/7.07 ≈ 1.13. The probability between -1.13 and +1.13 standard deviations from the mean is approximately 74.87%. This shows a higher likelihood, roughly 74.87%, that the mean for 50 customers is within $8 of the population mean.
Part c) considers an even larger sample size of 100, resulting in SE = 50/√100 = 5. The z-score becomes 8/5 = 1.6. The cumulative probability between -1.6 and +1.6 standard deviations is approximately 89.44%. This suggests an even greater probability, about 89.44%, that the sample mean of 100 customers will fall within $8 of the population mean.
Finally, in part d), determining the appropriate sample size to achieve at least a 95% probability involves calculating the required n such that the z-score corresponding to the $8 range reaches approximately 1.96, the critical value for 95% confidence. Solving for n yields n = (σ / (distance/ z))^2 = (50 / (8/1.96))^2 ≈ (50 / 4.08)^2 ≈ (12.25)^2 ≈ 150.06. Therefore, a sample size of approximately 151 would ensure at least a 95% probability that the sample mean falls within $8 of the population mean.
In conclusion, as the sample size increases, the standard error decreases, and the probability of the sample mean being within a specified interval of the population mean increases. To achieve a high confidence level of 95%, a sample of at least 151 customers is recommended. These findings highlight the importance of choosing adequate sample sizes in statistical analysis to ensure reliable estimation and decision-making.
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