Question

An auditor in an accounting firm checks a sample of n = 250 randomly selected transactions from among the thousands processed in an office. Twenty percent contain errors in crediting or debiting the appropriate account, i.e. p ̂ = 0.20

a.Does this situation meet the conditions required for the use of normal distribution for a z-interval for the proportion? Explain.

b.Construct a 95% confidence interval for the true population proportion of all transactions processed in this office that have these errors. Interpret your findings.

c.Managers in this office claim that the proportion of errors in about 10%. Based on your findings in part b, does the claim seem reasonable? Explain.

Answer 1

a)

np = 250 * 0.20 = 50 >= 10

n(1 - p) = 250 * 0.80 = 200 >= 10

The conditions required to use normal distribution are met.

b)

95% confidence interval for p is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.20 - 1.96 * sqrt ( 0.20 * 0.80 / 250) < p < 0.20 + 1.96 * sqrt ( 0.20 * 0.80 / 250)

0.150 < p < 0.250

95% CI is ( 0.150 , 0.250 )

c)

Since claimed proportion 0.10 is not in the confidence interval, we do not have sufficient evidence to

support the claim.