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

## EXPERT ANSWER

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

=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

Observation numbers | X | (X- )^{2} |

1 | 32 | 826.5625 |

2 | -45 | 2328.063 |

3 | 66 | 3937.563 |

4 | 2 | 1.5625 |

5 | -8 | 126.5625 |

6 | -51 | 2943.063 |

7 | 12 | 76.5625 |

8 | 18 | 217.5625 |

Total | 26 | 10457.5 |

Calculations:

sample mean

= 26/8

=3.25

sample standard deviation s

= 38.6514

**99% confidance interval for Mean Billing error**

(-44.565, 51.065)