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Compare and Contrast Confidence Intervals for Samples

P9.43

The SEC requires companies to file annual reports concerning their financial status. It is impossible to audit every account receivable. Suppose we audit a random sample of 49 accounts receivable invoices and find a sample average of $128 and a sample standard deviation of $53.

a) Find a 99% confidence interval for the mean size of an accounts receivable invoice. Does your answer require that the sizes of the accounts receivable invoices follow a normal distribution?

b) How large a sample do we need if we want to be 99% sure that we can estimate the mean invoice size within $5?

c) Take new sample size of 100, with sample mean equal to $210 and sample standard deviation of $24. Compute 99.8% confidence interval. Then, how large a sample size do we need to be sure of that 99.8% mean to within 5$?
Compare and contrast the results of a,b) with results of c)

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I have attached my original problem in word format and my answers in excel format.
Could the OTA please check my work and offer suggestions where mistakes are obvious as this is the premise for me to build on in my coursework.
Also, I am looking for detailed help in understanding what this data means/implies. For example, how would I read the data to myself in my head gleaning meaning from it? Or, the big, so what does this really say or imply? In six weeks for my final exam I will have to interpret problems as well as do the formulas and I need to begin practicing on work that is relevant for me. I decided to start with some role modeling from you. I hope I have offered payment compensatory for this added necessary feedback.

P9.43

The SEC requires companies to file annual reports concerning their financial status. It is impossible to audit every account receivable. Suppose we audit a random sample of 49 accounts receivable invoices and find a sample average of $128 and a sample standard deviation of $53.

a) Find a 99% confidence interval for the mean size of an accounts receivable invoice. Does your answer require that the sizes of the accounts receivable invoices follow a normal distribution?

Using the formula:
SE = SD/SQRT(N) = 53/ SQRT(49) = 7.57
Two-tailed t (0.01, 48) = 2.7
Then the Confidence intervals: lower and upper bounds of the range of values that contain the population parameter with a given probability.
The confidence interval = Mean +- t*SE = 128 +- 2.7*7.57 = (107.69, 148.31)
So your calculation is correct in this ...

Solution Summary

The expert compares and contrasts the confidence intervals for samples. The expert estimates the mean invoice size of a sample.

$2.19