Question

In: Statistics and Probability

In examining the credit accounts of a department store, an auditor would like to estimate the...

In examining the credit accounts of a department store, an auditor would like to estimate the true mean account error (book value – audited value). To do so, the auditor selected a random sample of 50 accounts and found the mean account error to be $60 with a standard deviation of $30.

a. Construct a 95% confidence interval for the population mean account error. What conclusion can be made from this confidence interval?

b. How large a sample is actually needed to ensure with 95% confidence in estimating the population mean account error to within ±$5?

c. Among the 50 selected accounts, the auditor found that there were 6 of them consisted of an account error of more than $100. Based on this information, construct a 99% confidence interval for the population proportion of accounts with an error of more than $100. What conclusion can be made from this confidence interval?

Solutions

Expert Solution

Answer:

a)

Given,

To determine the 95% confidence interval

Sample size n = 50

xbar = 60

standard deviation = s = 30

degrees of freedom = n - 1

= 50 - 1

df = 49

alpha = 1 - 0.95 = 0.05

alpha/2 = 0.05/2 = 0.025

crotical value t (0.025;59) = 2.00958

Now for the 95% confidence interval for the population mean account error.

i.e.,

(xbar - t*s/sqrt(n) , xbar+t*s/sqrt(n) )

substitute values

= ( 60 - 2.00958 *30 /sqrt(50) , 60 - 2.00958 * 30 / sqrt(50))

= (51.474 , 68.526)

lower limit = 51.474

upper limit = 68.526

b)

To determine the sample size

z critical value for 95% confidence interval = 1.96

E = margin of error = 5

standard deviation = sigma = 30

n = (z*sigma/E)^2

now substitute values

= (1.96*30/5)^2

=138.2976

=138 (approximately)

sample size = 138

c)

To determine the 99% confidence interval

sample proportion = p^ = x/n

= 6/50

p^ = 0.12

z critical value for 99% = 2.576

99% confidence interval = (p^-z*sqrt(p^(1-p^)/n , p^+z*sqrt(p^(1-p^)/n )

= (0.12 - 2.576*sqrt(0.12*(1-.12)/50 , 0.12 + 2.576*sqrt(0.12*(1-.12)/50))

= (0.00162 , 0.23838)

So interval limit is 0.00162 < p < 0.23838

lower limit = 0.00162

upper limit = 0.23838

So the interval lies between 0.00162 and 0.23838

orchestra answered 2 years ago
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