# 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

Ans:

X: Billing error

: Mean Billing error

n: Number of Billing error sampled

=sample size

= 8

: 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

$t_{\alpha /2,n-1} = t_{8-1, 0.01}$ =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

sample standard deviation s

Calculations:

sample mean

= 26/8

=3.25

sample standard deviation s

= 38.6514

99% confidance interval for Mean Billing error

(-44.565, 51.065)