The Securities and Exchange Commission requires companies to file annual reports
ID: 3205029 • Letter: T
Question
The Securities and Exchange Commission requires companies to file annual reports concerning their financial status. It is impossible to audit every account receivable. Suppose an auditor audits a random sample of 50 accounts receivable invoices and finds a sample average of $1,000 and a sample standard deviation of $400.
(a) Find a 95% confidence interval for the mean size of an accounts receivable invoice.
(b) What is the underlying assumption of your estimation in (a)?
(c) How large a sample is required to be 99% sure that the estimate of the mean invoice size is accurate within $30? Is there any assumption made in your estimation?
(d) Suppose the population contains 1000 accounts receivable in total, determine a 95% confidence interval for the total size of accounts receivable invoice from the given information of the sample.
Explanation / Answer
Given that,
sample size (Xbar) = $1000
sample standard deviation (s) = $400
sample size (n) = 50
(a) Find a 95% confidence interval for the mean size of an accounts receivable invoice.
C = confidence level = 95% = 0.95
Here we use z-interval.
95% confidence interval for population mean is,
Xbar - E < mu < Xbar + E
where E is margin of error.
E = (Zc*s) / sqrt(n)
where tc is the critical value for t-distribution.
One sample z-interval we can find by using TI-83 calculator.
steps :
STAT --> TESTS --> 7:ZInterval --> ENTER --> Highlight on Stats --> ENTER --> Input all the values --> Calculate --> ENTER
95% confidence interval for population mean (mu) is ($889.13, $1110.9).
(b) What is the underlying assumption of your estimation in (a)?
When the population standard deviation is unknown, one uses s to estimate the population standard deviation. The resulting interval is the one sample t-interval. Conditions to use the t-interval are:
(c) How large a sample is required to be 99% sure that the estimate of the mean invoice size is accurate within $30? Is there any assumption made in your estimation?
Given that,
confidence level (C) = 99% = 0.99
Margin of error (E) = $30
The sample size formula is,
n = ((Zc*s) / E)2
where tc ic the critical value for tdistribution.
tc we can find by using EXCEL.
syntax :
=NORMSINV(probability)
where probability = 1 - a/2
a = 1 - C
Zc = 2.58
n = [ (2.58*400) / 30 ]2 = 1179.54
(d) Suppose the population contains 1000 accounts receivable in total, determine a 95% confidence interval for the total size of accounts receivable invoice from the given information of the sample.
n = 1000
95% confidence interval for mu is ($975.21, $1024.8).
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