An auditor of a certain company, found that the billing errors that occurred on a random sample of invoices were: R32, -R45, R66, R2, -R8, -R51, R12 and R18. Positive errors meant that the customers were billed too much and negative ones meant that the customers were billed too little. Construct a 99% confidence interval for the mean billing error. Assume that the population of billing errors is normally distributed. The t(n – 1); 1-a/2 = 3.499. The 99% interval is

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An auditor of a certain company, found that the billing errors that occurred on a random sample of invoices were: R32, -R45, R66, R2, -R8, -R51, R12 and R18. Positive errors meant that the customers were billed too much and negative ones meant that the customers were billed too little. Construct a 99% confidence interval for the mean billing error. Assume that the population of billing errors is normally distributed. The t(n – 1); 1-a/2 = 3.499. The 99% interval is

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Ans:

X: Billing error

 : Mean Billing error

n: Number of Billing error sampled

=sample size

= 8

X : Mean error of sampled Billing

: Sample mean

s=standard deviation of error of sampled Billing i.e samlple standard deviation

Confidance Interval

(1- )100% confidance interval for population mean 

S
S
(X - -1,1-a/2 *
X+m–1,1-a/2 *
이트

ta/2,n-1 = t3-1,0.01
= =3.499 where t is the t distribution variable having n-1 = 8-1 = 7 degrees of freedom

Let us compute the sample mean and sample standard deviation

sample mean

X X -
ΣΧ
η

sample standard deviation s

S S =
Σ(Xi - X)2
.
η -1
Observation numbersX(X- X)2
132826.5625
2-452328.063
3663937.563
421.5625
5-8126.5625
6-512943.063
71276.5625
818217.5625
Total2610457.5

Calculations:

sample mean

X X -
ΣΧ
η

= 26/8

=3.25

sample standard deviation s

S S =
Σ(Xi - X)2
.
η -1
10457.5
V 8-1

= 38.6514

99% confidance interval for Mean Billing error

(-44.565, 51.065)