The mean preparation fee H&R Block charged retail customers last year was $183. Use this price as the population mean and assume the population standard deviation of preparation fees is $50. Q1) Now we randomly select 30 H&R Block retail customers. What are the values of the mean and the standard deviation of the sampling distribution of the sample mean?

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The mean preparation fee H&R Block charged retail customers last year was $183. Use this price as the population mean and assume the population standard deviation of preparation fees is $50.

Q1) Now we randomly select 30 H&R Block retail customers. What are the values of the mean and the standard deviation of the sampling distribution of the sample mean?

Q2) Let's assume we don’t know the population mean; but we still know the population standard deviation to be σ=$50. We randomly sampled 40 H&R Block retail customers and the mean price is $183. What is the probability that the population mean is within $5 of the sample mean?

Q3) We randomly sampled 100 H&R Block retail customers and the mean price is $183, assuming the population standard deviation is still $50. Construct a 90%, 95%, and 99% confidence interval of the population mean, respectively.

Q3 a) Provide a practical interpretation of the above 90% confidence interval. What conclusions can you draw based on the 90%, 95%, and 99% confidence intervals you constructed above?

Q4)

When the population standard deviation is unknown, the best we can do is to replace it with the sample standard deviation, s. Just like the sample mean is a random variable, so is the sample standard deviation s. The replacement of σ with s adds more variability. Some adjustment to the Central Limit Theorem is thus necessary. It turns out that when the population standard deviation is unknown and the sample size n is sufficiently large, the sample statistic t=(X ̅-μ)/(s/√n) approximately follows a t distribution with a degree of freedom n-1. As a result, when we look for a confidence interval with σ unknown, we will replace normal distributions with t distribution. Use this result to find a 92% confidence interval for the population mean price that the retail customers pay for, given the sample mean is $183, the sample standard deviation is $50, and the sample size is 36. Comparing this result to that in question g), you should notice that the margin of error is slightly larger when the population standard deviation is unknown.

 

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